Chapter 11 Resource Masters
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Contents
Teacher’s Guide to Using the Chapter 11 Resource Masters ........................................... iv
Chapter ResourcesStudent-Built Glossary ....................................... 1Anticipation Guide (English) .............................. 3Anticipation Guide (Spanish) ............................. 4
Lesson 11-1Descriptive StatisticsStudy Guide and Intervention ............................ 5Practice .............................................................. 7Word Problem Practice ..................................... 8Enrichment ........................................................ 9TI-Nspire Activity ............................................. 10
Lesson 11-2Probability DistributionsStudy Guide and Intervention .......................... 11Practice ............................................................ 13Word Problem Practice ................................... 14Enrichment ...................................................... 15
Lesson 11-3The Normal DistributionStudy Guide and Intervention .......................... 16Practice ............................................................ 18Word Problem Practice ................................... 19Enrichment ...................................................... 20Graphing Calculator Activity ............................ 21
Lesson 11-4The Central Limit TheoremStudy Guide and Intervention .......................... 22Practice ............................................................ 24Word Problem Practice ................................... 25Enrichment ...................................................... 26
Lesson 11-5Confidence IntervalsStudy Guide and Intervention .......................... 27Practice ............................................................ 29Word Problem Practice ................................... 30Enrichment ...................................................... 31
Lesson 11-6Hypothesis TestingStudy Guide and Intervention .......................... 32Practice ............................................................ 34Word Problem Practice ................................... 35Enrichment ...................................................... 36
Lesson 11-7Correlation and Linear RegressionStudy Guide and Intervention .......................... 37Practice ............................................................ 39Word Problem Practice ................................... 40Enrichment ...................................................... 41
AssessmentChapter 11 Quizzes 1 and 2 ........................... 43Chapter 11 Quizzes 3 and 4 ........................... 44Chapter 11 Mid-Chapter Test .......................... 45Chapter 11 Vocabulary Test ........................... 46Chapter 11 Test, Form 1 ................................. 47Chapter 11 Test, Form 2A ............................... 49Chapter 11 Test, Form 2B ............................... 51Chapter 11 Test, Form 2C .............................. 53Chapter 11 Test, Form 2D .............................. 55Chapter 11 Test, Form 3 ................................. 57Chapter 11 Extended-Response Test ............. 59Standardized Test Practice ............................. 60
Answers ........................................... A1–A28
Chapter 11 iii Glencoe Precalculus
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Teacher’s Guide to Using theChapter 11 Resource Masters
The Chapter 11 Resource Masters includes the core materials needed for Chapter 11. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.
Chapter ResourcesStudent-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 11-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.
Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.
Lesson Resources Study Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Guided Practice exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.
Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.
Word Problem Practice This master includes additional practice in solving word problems that apply to the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.
Enrichment These activities may extend the concepts of the lesson, offer an historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.
Graphing Calculator, TI–Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.
Chapter 11 iv Glencoe Precalculus
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Assessment OptionsThe assessment masters in the Chapter 11 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.
Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.
Mid-Chapter Test This one-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.
Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.
Leveled Chapter Tests
• Form 1 contains multiple-choice questions and is intended for use with below grade level students.
• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.
• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.
• Form 3 is a free-response test for use with above grade level students.
All of the above mentioned tests include a free-response Bonus question.
Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers are included for evaluation.
Standardized Test Practice These three pages are cumulative in nature. It includes two parts: multiple-choice questions with bubble-in answer format andshort-answer free-response questions.
Answers• The answers for the Anticipation Guide
and Lesson Resources are provided as reduced pages.
• Full-size answer keys are provided for the assessment masters.
Chapter 11 v Glencoe Precalculus
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(continued on the next page)
This is an alphabetical list of key vocabulary terms you will learn in Chapter 11. As you study this chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Precalculus Study Notebook to review vocabulary at the end of the chapter.
Vocabulary TermFound
on PageDefinition/Description/Example
binomial distribution
Central Limit Theorem
confidence interval
correlation
correlation coefficient
critical values
empirical rule
hypothesis test
inferential statistics
level of significance
Student-Built Glossary11
Chapter 11 1 Glencoe Precalculus
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Vocabulary TermFound
on PageDefinition/Description/Example
normal distribution
percentiles
probability distribution
P-value
random variable
regression line
residual
standard error of the mean
t-distribution
z-value
Student-Built Glossary11
Chapter 11 2 Glencoe Precalculus
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11
Before you begin Chapter 11
• Read each statement.
• Decide whether you Agree (A) or Disagree (D) with the statement.
• Write A or D in the first column OR if you are not sure whether you agree ordisagree, write NS (Not Sure).
After you complete Chapter 11
• Reread each statement and complete the last column by entering an A or a D.
• Did any of your opinions about the statements change from the first column?
• For those statements that you mark with a D, use a piece of paper to write anexample of why you disagree.
STEP 1 A, D, or NS
StatementSTEP 2 A or D
1. In a negatively skewed distribution, the mean is less than the median.
2. Percentiles divide a distribution into 100 equal groups.
3. A continuous random variable takes on a countable number of possible values.
4. In a probability distribution, the sum of P(X) must be 1.
5. In a normal distribution, the mean, median, and mode are equal and are located at the center of the distribution.
6. A z-value represents the number of standard deviations that a given data value is from the mean.
7. If the sample size is adequately large, the distribution of the sample means will a have a mean and standard deviation equal to the population mean and standard deviation.
8. As the probability of success increases to 0.5, the shape of the binomial distribution begins to resemble the normal distribution.
9. A confidence level gives the probability that the interval estimate will include a given parameter.
10. The null hypothesis states that there is a difference between the sample value and the population parameter.
11. Extrapolation uses an equation to make predictions over the range of the data.
12. A correlation coefficient of -0.95 shows a strong negative correlation between two variables.
Anticipation GuideInferential Statistics
Step 2
Step 1
Chapter 11 3 Glencoe Precalculus
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Ejercicios preparatoriosEstadística inferencial
11
Capítulo 11 4 Precálculo de Glencoe
Después de completar el Capítulo 11
• Relee cada enunciado y escribe A o D en la última columna.
• Compara la última columna con la primera. ¿Cambiaste de opinión sobre alguno de los enunciados?
• En los casos en que hayas estado en desacuerdo con el enunciado, escribe en una hoja aparte un ejemplo de por qué no estás de acuerdo.
Paso 2
Antes de comenzar el Capítulo 11
• Lee cada enunciado.
• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.
• Escribe A o D en la primera columna O si no estás seguro(a), escribe NS (no estoy seguro(a)).
Paso 1
PASO 1 A, D o NS
EnunciadoPASO 2 A o D
1. La media es menor que la mediana en una distribución asimétrica negativa.
2. Los percentiles dividen una distribución en 100 grupos iguales.
3. Se puede contar el número de valores posibles que puede tener una variable aleatoria continua.
4. La suma de P(X) debe ser igual a 1 en una distribución probabilística.
5. La media, la mediana y la moda de una distribución normal son iguales y están ubicadas en el centro de la distribución.
6. El valor de z representa el número de desviaciones estándar que separan la media de un valor dado de los datos.
7. Si la muestra es de un tamaño suficientemente grande, la distribución de las medias de la muestra tendrá una media y una desviación estándar igual a la media y la desviación estándar de la población.
8. A medida que la probabilidad de éxito se acerca a 0.5, la forma de la distribución binomial se empieza a parecer a la forma de la distribución normal.
9. Los niveles de confianza indican la probabilidad de que el intervalo del estimado incluya el parámetro dado.
10. La hipótesis nula establece que hay una diferencia entre el valor de la muestra y el parámetro de la población.
11. La extrapolación utiliza una ecuación para predecir valores más allá del rango de los datos.
12. Un coeficiente de correlación de -0.95 indica una fuerte correlación negativa entre dos variables.
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Chapter 11 5 Glencoe Precalculus
Study Guide and InterventionDescriptive Statistics
11-1
Describing Distributions When a data set has a symmetrical distribution, the data are evenly distributed on both sides of the mean. In a negatively skewed distribution, the left side of the distribution extends farther than the right and the mean is less than the median. In a positively skewed distribution, the right side extends farther than the left and the mean is greater than the median. If a distribution is reasonably symmetric, the mean and standard deviation can be used to describe it. Otherwise, the five-number summary is better.
The table shows the number of passengers who boarded planes at 36 airports in the United States in one year.
Number of Passengers
30,526 30,372 26,623 22,722 16,287 15,246 14,807 14,117 14,054
13,547 12,916 12,616 11,906 11,622 11,489 10,828 10,653 10,008
9703 9594 9463 9348 9125 8572 7300 6772 6549
6126 5907 5712 5287 4848 4832 4820 4750 4684
a. Construct a histogram and use it to describe the shape ofthe distribution.
Enter the data in L1, turn on Plot1, and select histogram.Press GRAPH . Press ZOOM and choose ZoomStat.Press TRACE to see how many data value are in each bar. The graph is positively skewed.
b. Summarize the center and spread of the data using eitherthe mean and standard deviation or the five-number summary. Justify your choice.Use the five-number summary since the data are skewed.
The number of passengers range from 4684 to 30,526, andthe median number of passengers is about 9856. Half of theairports had between 6338 and 13,800 passengers.
Exercise 1. A restaurant manager advertises a 20-ounce sirloin steak on his menu.
The weights of a sample of the steaks is shown.
Steak Weights (ounces)
20 20 20 18 20 18 18 19 19 19
21 20 17 20 19 21 18 19 20 20
a. Construct a histogram and use it to describe the shapeof the distribution.
b. Summarize the center and spread of the data usingeither the mean and standard deviation or the five-numbersummary. Justify your choice.
Example
[4000, 35,000] scl: 3000 by [0, 15] scl: 1
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Chapter 11 6 Glencoe Precalculus
11-1 Study Guide and Intervention (continued)
Descriptive Statistics
Measures of Position The quartiles given by the five-number summary specify the positions of data values within a distribution. For this reason, box plots are most useful for side-by-side comparisons of two or more distributions.
The ages of Oscar-winning actresses from two 20-year periods are shown. Construct side-by-side box plots of the data sets. Then use this display to compare the distributions.
1969–1988
61 35 34 34 26 37 42 41 35 31
41 33 30 74 33 49 38 61 21 41
1989–2008
26 80 42 29 33 36 45 49 39 34
26 25 33 35 35 28 30 29 61 32
Input the data into L1 and L2. Turn on Plot1 and Plot2 andchoose a box plot with outliers shown as the type of graph.
From the graph, we can see that the median age for the first20 years is just slightly higher than the median age for the second20 years. The range of ages for the middle 50% of the data setsis greater for the actresses from the second 20-year period.
Both data sets have outliers greater than the rest of the data. Ignoring the outliers, the distribution for the second 20-year period is more symmetric than the first and the range of data for the second is less than that of the first.
Exercise
The number of credit cards owned by students in a statisticsclass is shown. Construct side-by-side box plots of the data sets.Then use this display to compare the distributions.
Example
[15, 85] scl: 5 by [0, 0.5] scl: 0.125
Females
1 8 0 1 5
4 4 0 5 0
0 0 3 0 1
Males
0 4 0 0 2
0 5 7 0 0
3 5 0 0 1
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Chapter 11 7 Glencoe Precalculus
11-1 PracticeDescriptive Statistics
1. WEATHER The average wind speeds recorded at various weather stations in the United States are listed below.
StationSpeed
(mph)Station
Speed
(mph)Station
Speed
(mph)
Albuquerque 8.9 Anchorage* 7.1 Atlanta* 9.1
Baltimore* 9.1 Boston* 12.5 Chicago 10.4
Dallas– Ft. Worth 10.8 Honolulu* 11.3 Indianapolis 9.6
Kansas City 10.7 Las Vegas 9.3 Little Rock 7.8
Los Angeles* 6.2 Memphis 8.8 Miami* 9.2
Minneapolis– St. Paul 10.5 New Orleans 8.1 New York City* 9.4
Philadelphia* 9.5 Phoenix 6.2 Seattle* 9.0
Source: National Climatic Data Center
a. Construct a histogram and use it to describe the shapeof the distribution.
b. Summarize the center and spread of the data using eitherthe mean and standard deviation or the five-numbersummary. Justify your choice.
2. OCEANS The ten weather stations with an asterisk haverelatively close proximity to either the Atlantic Ocean orPacific Ocean. Construct side-by-side box plots of the datasets. Then use this display to compare the distributions.
3. SCORES The table gives the frequency distribution of thescores on a test.
a. Construct a percentilegraph of the data.
b. Estimate the percentile ranka score of 62 would have in thisdistribution. Interpret itsmeaning.
10
0
20
30
40
Cum
ulat
ive
Perc
enta
ge
50
60
70
80
90
100
Score50 60 70 80 90 100
Class Boundaries f
55.5–60.5 3
60.5–65.5 8
65.5–70.5 12
70.5–75.5 5
75.5–80.5 9
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Chapter 11 8 Glencoe Precalculus
11-1 Word Problem PracticeDescriptive Statistics
1. SHOES A shoe store employee designs a display by placing shoe boxes in ten stacks. The number of boxes in each stack are 5, 7, 9, 11, 13, 10, 9, 8, 7, and 5.
a. Construct a box plot and use it to describe the shape of the distribution.
b. Summarize the center and spread of
the data using either the mean and standard deviation or the five-number summary. Justify your choice.
2. MEDICINE A histogram for the number of patients treated at 50 U.S. cancer centers in one year is shown.
Source: U.S. News Online
a. Which is greater, the mean or the median of the data set? Explain.
b. Sketch the general shape of a box plot that would represent the histogram.
3. EXAMS The table gives the frequencies of the final exam scores of 50 students in two precalculus classes.
Class BoundariesFrequency
f
42.5–52.5 4
52.5–62.5 10
62.5–72.5 15
72.5–82.5 13
82.5–92.5 7
92.5–102.5 1
a. Construct a percentile graph ofthe data.
10
0
20
30
40
Cum
ulat
ive
Perc
enta
ge50
60
70
80
90
100
Score5040 60 70 80 90 100
b. Estimate the percentile rank a test score of 77 would have in this distribution. Interpret its meaning.
4. MOVIES The ages of movie patrons ina theater are 25, 47, 16, 45, 54, 17, 14, 16, 16, 39, 48, 48, 18, 12, 13, 62, 51, 46, and 18. Summarize the distribution of the data.
Patients in U.S. Cancer Centers
20
30
10
0Num
ber o
r Can
cer C
ente
rs 40
Number of Patients
3500500 1000 1500 2000 2500 3000
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Chapter 11 9 Glencoe Precalculus
11-1 Enrichment
Means
The average of a data set is known as the arithmetic mean. There are other means.
A trimmed mean is the arithmetic mean of a data set after the top 10% and bottom 10% of the data values are trimmed off. Therefore, if there are 100 data values, the least 10 values and greatest 10 values are removed.
A geometric mean is the nth root of the product of n data values. It is often used in economics to find an average rate of growth.
A harmonic mean H is the number of data values divided by the sum of the multiplicative inverses of all the data values. It is often used in averaging speeds. It cannot be used if 0 is a data value.
H = n −
∑
1 − x , where n is the number of data values and x is a data value.
Exercises 1. Find the trimmed mean of the data set.
32, 24, 56, 102, 54, 12, 27, 49, 35, 23, 44, 51, 66, 36, 52, 16, 63, 75, 21, 41
2. Find the arithmetic mean and median of the data set in Exercise 1. What is the benefit of using a trimmed mean instead of an arithmetic mean?
3. WATER The EPA recommends that quality freshwater be monitored for e.coli concentration. The geometric mean of the e.coli concentrations in 5 or more samples taken over 30 days should be less than 126 per 100 milliliters. Do the data {225, 181, 110, 118, 107} meet this criterion? Explain.
4. TRAVEL If you travel at one speed for half the distance of a trip and you travel at a different speed for the other half of the distance, then your average speed for the trip is the harmonic mean of the speeds. If Greg drove at 60 mph for half a trip and 70 mph when the speed limit increased for the second half, what was his average speed for the trip?
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Chapter 11 10 Glencoe Precalculus
11-1 TI-NspireTM Activity
Graphing DataYou can use a TI-Nspire to graph data sets and describe the shape of the distribution.
The lengths of fish caught on a one-dayfishing trip are shown in the table. Make a histogramand box plot of the data.
Step 1: Add a Lists & Spreadsheet page. List the datain column A and title the column lengths.
Step 2: With the cursor in column A, press b andchoose Data > Quick Graph. Press b andchoose Plot Type > Histogram. Press bagain and choose Plot Properties >Histogram Properties > Bin Settings to changethe width of the bars. Press b and chooseWindow/Zoom > Zoom-Data to automatically sizethe axes to see all the data.
You can change the vertical category to percentby pressing b and choosing Plot Properties >Histogram Properties > Histogram Scale >Percent.
Step 3: To change the graph to a box plot, press b andchoose Plot Type > Box Plot. Move the cursor overthe graph to view the five-number summary.
Exercises
Open a new Lists & Spreadsheet page. Title column A as data. Use it to complete the following.
1. Make a histogram of the data set: 44.5, 13.7, 29.4, 22.0, 32.5, 45.8, 38.6, 24.3,18.1, and 50.3. Use the graph to describe the shape of the distribution.
2. Change the histogram in Exercise 1 to a box plot. List the five-number summary.
3. Open a new Lists & Spreadsheet page. Title column A as data1 and column Bas data2. Enter the data shown below. Use Quick Graph in each column to makebox plots. Describe and compare the distributions.
Data1: 102, 116, 132, 111, 124, 103, 101, 108, 129, 103, 109, 115, 111 Data2: 115, 129, 101, 128, 125, 115, 101, 124, 124, 118, 121, 119, 120
Lengths of Fish (in.)
14 16 18 18 14
21 26 23 35 15
32 30 9 22 12
Example
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Chapter 11 11 Glencoe Precalculus
Probability Distribution A discrete random variable X can take on acountable number of values. A probability distribution links each possiblevalue of X with its probability of occurring.
Mean of probability distribution of X: μ = Σ [X � P(X)]
Standard deviation of probability distribution of X: σ = √
�������
Σ [(X - μ)2 � P(X)
]
The number of books bought by each of 100 randomly selected bookstore customers during one week is shown.
a. Construct a probability distribution for X.
There are 100 customers, so P(0) = 0.45, P(1) = 0.30, P(2) = or 0.15, and P(3) = 0.10.
b. Find and interpret the mean in the context of the problem situation.Find the variance and standard deviation.
Organize your calculations in a table.
The mean of the distribution is 0.9. On average, each customer bought one book.The variance = 0.99, so the standard deviation is √
��
0.99 or about 0.995.
Exercise
1. The table shows the number of medical tests that15 randomly selected patients entering a particularhospital received one day.
a. Construct a probability distribution for X.
b. Find and interpret the mean in the context of the problem situation. Find thevariance and standard deviation.
11-2 Study Guide and InterventionProbability Distributions
Example
Books, X Frequency
0 45
1 30
2 15
3 10Books, X 0 1 2 3
Frequency 0.45 0.30 0.15 0.10
Books, X P(X) X � P(X) (X - μ)2 (X - μ)2 � P(X)
0 0.45 0 � 0.45 = 0 (0 - 0.9)2 = 0.81 0.81 � 0.45 = 0.3645
1 0.30 1 � 0.30 = 0.30 (1 - 0.9)2 = 0.01 0.01 � 0.30 = 0.003
2 0.15 2 � 0.15 = 0.30 (2 - 0.9)2 = 1.21 1.21 � 0.15 = 0.1815
3 0.10 3 � 0.10 = 0.30 (3 - 0.9)2 = 4.41 4.41 � 0.10 = 0.441
μ = 0.9 σ2 = 0.99
Tests, X Frequency
0 6
1 5
2 3
3 1
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Chapter 11 12 Glencoe Precalculus
Binomial Distribution In a binomial experiment, the outcomes are success or failure. There are a fixed number of independent trials n and the random variable X represents the number of successes. The probability of success p and the probability of failure q or 1 - p remain constant. The probability of X successes in n independent trials is
P(X) = nCx px qn - x = n! −
(n - x)!x! px qn - x .
A survey found that 20% of Americans have visited a doctor in the past six months. Five people will be selected at random and asked if they visited a doctor in the past six months. Construct and graph a binomial distribution for the random variable X, representing the number of people who say yes. Then find the probability that at least four of these people say yes.
For this binomial experiment, n = 5, p = 0.2, and q = 1 - 0.2 or 0.8. Compute each possible value of X using the Binomial Probability Formula.P(0) = 5C0 � 0.20 � 0.85 ≈ 0.328 P(3) = 5C3 � 0.23 � 0.82 ≈ 0.051P(1) = 5C1 � 0.21 � 0.84 ≈ 0.410 P(4) = 5C4 � 0.24 � 0.81 ≈ 0.006P(2) = 5C2 � 0.22 � 0.83 ≈ 0.205 P(5) = 5C5 � 0.25 � 0.80 ≈ 0.000
To find the probability that at least four people said yes, find the sum of P(4) and P(5).P(X ≥ 4) = P(4) + P(5) = 0.006 + 0.000 or 0.6%
Exercise
A survey found that 60% of American victims of health-care fraud were senior citizens. Six victims of health-care fraud will be chosen at random and their ages will be recorded. Construct and graph a binomial distribution for the random variable X, representing the number of senior citizens chosen. Then find the probability that at least three of the victims will be senior citizens.
Study Guide and Intervention (continued)
Probability Distributions
11-2
Example
X P(X)
0 0.328
1 0.410
2 0.205
3 0.051
4 0.006
5 0.000
00 1 2 3 4 5
0.1
0.2
0.3
0.4
0.5
People
Prob
abili
ty
P(x)
x
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Chapter 11 13 Glencoe Precalculus
Classify each random variable X as discrete or continuous. Explain your reasoning.
1. X represents the time it takes a randomly selected classroom to reach 68°Ffrom 60°F. .
2. X represents the number of photographs taken by a photographer at a randomlyselected wedding.
3. The table shows the number of cell phones owned by 100 randomly selected households. Construct and graph a probability distribution for X. Then find and interpret the mean in the context of the problem situation. Find the variance and standard deviation.
4. RACE A resort is planning a bicycle race. The cost of sponsoring the race is$8000. The resort expects to make $15,000 on the event. There is a 30% chanceof a hurricane arriving the day of the race. If this happens, the race will becancelled and will not be rescheduled. What is the resort’s expected profit?
5. COMMUTE In a recent poll, 45% of a town’s citizens said they use the bus toget to work. Five of these citizens will be randomly chosen and asked if theyuse the bus to get to work.
a. Construct a binomial distribution for the random variable X, representing thepeople who say yes.
b. Find the mean, variance, and standard deviation of this distribution. Interpretthe mean in the context of the problem situation.
11-2 PracticeProbability Distributions
Phones, X Frequency
0 2
1 30
2 48
3 13
4 7
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Chapter 11 14 Glencoe Precalculus
1. RETAIL A store manager made the probability distribution shown below. It shows the probability of selling Xswimsuits on a randomly selected dayin June.
Find the mean, variance, and standard deviation of the distribution.
2. INSURANCE An insurance company insures a painting worth $20,000 against theft for $300 per year. The company has assessed the probability of the painting being stolen in a given year as 0.002. What is the insurance company’s expected annual profit?
3. RESTAURANT A survey found that 25% of all parties at a restaurant were groups of five or larger. Eighteen parties are randomly selected.
a. Find the probability that exactlyfive parties are made up of five or more people.
b. Find the probability that 5, 6, or7 parties are made up of five ormore people.
4. PETS According to one poll, about 63% of American households include at least one pet. Six new homes are built and sold.
a. Construct a binomial distribution for the random variable X, representing the number of these homes that will have at least one pet.
b. Find the mean, variance, and standard deviation of this distribution.
c. Find the probability that at least half of the new homes have pets.
5. TESTING Mr. Hanlon distributed a 5-question multiple choice quiz to his students. There were 5 choices for each question. Ashley guesses the answer on each question.
a. What is Ashley’s probability of guessing exactly 3 questions correctly?
b. What would be the probability inpart a if there were 4 choices foreach question?
c. What would be the probability inpart a if the quiz contained onlytrue/false questions?
11-2 Word Problem PracticeProbability Distributions
Swimsuits, X 19 20 21 22 23
P(X) 0.20 0.20 0.30 0.20 0.10
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Chapter 11 15 Glencoe Precalculus
The Poisson distribution is a probability distribution for an event occurringx times over a given interval. The interval is usually a time interval, such as the number of people entering a store during one hour. The distribution is a discrete distribution, so x must be a whole number.
The probability of an event occurring x times over a given interval is λx · e-λ
−
x! ,
where λ is the average number of times the event occurs during the interval.
A bank manager determined that an average of 9.2 customers use a certain ATM every hour. You can create a Poisson distribution by using 9.2 for λ. For example, the probability that 5 customers use the ATM in a given hour
is given by 9.25 · e-9.2 −
5! . Use your calculator to verify P(5) ≈ 0.055 or
about 5.5%.
On a graphing calculator, you can also find the probability by using the poissonpdf( command found by pressing [DISTR]. In the parentheses, enter λ, x.
1. The bank manager determined that an average of 1.8 customers inquire about opening a new account at a certain branch every hour. Complete the table.
2. Notice that unlike a binomial distribution, x has no upper limit. The table in Exercise 2 stopped at x = 5, but it is possible for 6 or more people to inquire about opening a new account. How can you find P(x > 2)?
3. On average, the bank receives 4.2 online loan applications per day. Find each probability.
a. The bank receives 3 online loan applications one day.
b. The bank receives 4 online loan applications one day.
c. The bank receives 5 online loan applications one day.
d. The bank receives 6 online loan applications one day.
e. The bank receives 7 online loan applications one day.
4. Make a conjecture about the general shape of any Poisson distribution.
11-2 EnrichmentThe Poisson Distribution
X 0 1 2 3 4 5
P(X)
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Chapter 11 16 Glencoe Precalculus
The Normal Distribution A normal distribution is a continuous probability distribution. The graph of a normal distribution is symmetric and bell-shaped. It approaches but never touches the x-axis. It includes 100% of the data, so the area under the curve is 1. The z-value represents the number of standard deviations that a given data value is from the mean.
z = X - μ
−
σ , where X is the data value, μ is the mean, and σ is the standard deviation.
The standard normal distribution has a mean of 0 and a standard deviation of 1.
On his last 20 airline trips, an employee had an average layover of 82 minutes with a standard deviation of7.5 minutes. Find the number of layovers that were less than75 minutes.
First, find the z-value.
z = X - μ
−
σ Formula for z -values
= 75 - 82 −
7.5 or about -0.93 X = 75, μ = 82, and σ = 7.5
Use a graphing calculator to find the area under the curve that is to the left of 75. Press 2nd [DISTR] and choose normalcdf(. Enter the lower value (you can use -4 instead of negative infinity) and the upper value as -0.93. The resulting area is 0.176. This means that about 17.6% of the data values are less than-0.93 standard deviations from the mean.
Because there are 20 flights, about 20 � 0.176 or about 4 flights had layover times that were less than 75 minutes.
Exercises
1. At a restaurant, the average time between when an order is placed and when the entree is served is 12.5 minutes with a standard deviation of1.2 minutes. Out of 100 randomly selected customers, how many will be served their entrees within 14 minutes of ordering?
2. Mrs. Quan, a full professor at a community college, earns a salary of $48,600. The average salary for a full professor at the college is $52,000 with a standard deviation of $3600. How many of the 45 full professors earn less than Mrs. Quan?
3. During one October, the average water temperature of a pond was 53.2° with a standard deviation of 2.3°. How many days was the temperature greater than 50°?
Study Guide and InterventionThe Normal Distribution
11-3
Example
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Chapter 11 17 Glencoe Precalculus
Probability and the Normal Distribution The area under the normal curve corresponds to the probability of data values falling within a given interval.
The average monthly temperatures for a city for one year were normally distributed with μ = 65° and σ = 5. FindP(50 < X < 70). Use a graphing calculator to sketch the corresponding area under the curve.
Find the z-values for X = 50 and X = 70.
z = X - μ
−
σ z =
X - μ −
σ
= 50 - 65 −
5 = -3 = 70 - 65 −
5 = 1
Find the area between z = -3 and z = 1. On your calculator, press 2nd [DISTR] and choose ShadeNorm under the DRAW menu. Enter the lower and upper z-values and press enter.
P(50 < X < 70) ≈ 84%
Therefore, approximately 84% of the temperatures were between 50° and 70°.
Exercises 1. The average age of the swimmers on a master swim team is normally
distributed with μ = 56 and σ = 4. Find each probability. Use a graphing calculator to sketch the corresponding area under the curve.
a. P(53 < X < 59)
b. P(X < 53)
2. Students who score in the bottom 5% of a physical education test will be enrolled in a supplemental physical education program. The scores of all of the students who took the test are normally distributed with μ = 122.6 and σ = 18. What is the greatest score that a student who enrolled in the supplemental program could have received?
11-3 Study Guide and Intervention (continued)
The Normal Distribution
Example
[–4, 4] scl: 1 by [0, 0.5] scl: 0.125
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Chapter 11 18 Glencoe Precalculus
1. TREES The heights of 200 trees in a nursery are normally distributed with a mean of 120 inches and a standard deviation of 16 inches.
a. Approximately how many trees are more than 136 inches tall?
b. What percent of the trees are between 88 inches and 104 inches tall?
Find each of the following.
2. z if X = 65, μ = 50, and σ = 10 3. X if z = -0.4, μ = 40, and σ = 5
Find the interval of z-values associated with each area.
4. the middle 60% of the data
5. the outside 30% of the data
6. DOGS The weights of the 42 full-grown German shepherds at a kennel are normally distributed. The mean weight is 86 pounds and the standard deviation is 3 pounds.
a. Determine the number of German shepherds that weigh more than 82 pounds.
b. How many German shepherds weigh less than 88 pounds?
7. HOTELS The prices of rooms at hotels around an airport are normally distributed with μ = $120 and σ = $20. Find each probability.
a. The cost of a room is greater than $150.
b. The cost of a room is between $110 and $130.
c. The cost of a room is between $90 and $100.
d. If only the most expensive 10% of the rooms are available, what is the least amount you will pay for a room?
8. EXAMS A student scored 65 on a biology exam with μ = 50 and σ = 10. She scored 30 on a literature exam with μ = 25 and σ = 5. Compare her scores on each test. Assume that both sets of scores were normally distributed.
11-3 PracticeThe Normal Distribution
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Chapter 11 19 Glencoe Precalculus
1. REAL ESTATE The average price of a one-bedroom condominium listed by a realtor is $145,500 with a standard deviation of $1500. The prices are normally distributed. Determine the probability that a randomly selected condominium costs between $143,580 and $147,420.
2. COMMUTING The average times spent commuting to work in a certain city are normally distributed with a mean of25.5 minutes and a standard deviationof 6.1 minutes. What is the probability that a randomly selected commute to work takes longer than a half hour?
3. SIGHTSEEING The times people spend viewing certain ancient ruins are normally distributed with a mean of 96 minutes with a standard deviation of 17 minutes.
a. Find the probability that a sightseer will spend at least two hours at the ruins.
b. Find the probability that a sightseer will spend at most 80 minutes at the ruins.
c. If a tour bus drops off a group of sightseers at 9 A.M., what time should the bus pick up the sightseers? Explain.
4. TRAINING To qualify for a security position, candidates must take a physical fitness test. The scores on the test are normally distributed with a mean of 400 and a standard deviation of 100.
a. Candidates scoring in the top 3% are later recruited as trainers in the program. What is the minimum score a candidate needs in order to be recruited later as a trainer?
b. Candidates scoring in the bottom 1.5% must retake the physical training program. What is the minimum score a candidate would need to avoid retaking the training program?
5. JUICE The amount of juice pouredinto bottles in a factory is normally distributed with a mean of 16 ouncesand a standard deviation of 0.3 ounce. A shipment contains 280 bottles.
a. How many bottles are expected to contain more than 16.5 ounces of juice?
b. How many bottles are expected to contain less than 15.75 ounces of juice?
11-3 Word Problem PracticeThe Normal Distribution
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Chapter 11 20 Glencoe Precalculus
Before graphing calculators, students had to find the area under a normal distribution curve by using a table of values. You can still find these tables in many statistics books.It is good to know how to use them, in case you find yourself in a situation without a graphing calculator.
Here is part of a standard normal probability table. It gives the area that is between the mean, 0, and the z-score. To find the area between a z-score of 0 and a z-score of 1.35, find 1.3 in the vertical column and 0.05 in the top row. It is 0.4115.
To find the area that is to the left of a certain positive z-score, remember to add the areathat is to the left of the mean, 0.5.
Because a normal distribution is symmetric about the mean, you can use the samechart to find the area between a negative z-score and the mean. For example, thearea between a z-score of -1.35 and 0 is the same as between 0 and 1.35: 0.4115.
Use the table to answer each question.
1. What is the area between a z-score of 0 and a z-score of 1.02?
2. What is the area between a z-score of 1.18 and a z-score of 1.43? Explain howyou found it.
3. Find the area to the right of a z-score of 1.17. Explain how you found it.
4. What is the area to the left of a z-score of 1.41?
5. What is the area between a z-score of -1.25 and a z-score of 0?
6. What is the area to the left of a z-score of -1.33?
7. What is the area between a z-score of -1.05 and a z-score of 1.38?
11-3 EnrichmentFinding Areas by Using a Table
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621
1.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .3830
1.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .4015
1.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177
1.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319
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Chapter 11 21 Glencoe Precalculus
The graph shown at the right is known as the standard normal
curve. The standard normal curve is the graph of f(x) = 1 −
√
��
2π e -
x2 −
2 .
You can use a graphing calculator to investigate properties of this function and its graph. Enter the function for the normal curve in the Y = list of a graphing calculator.
1. The standard normal curve models a probability distribution. As a result, probabilities for intervals of x-values are equal to areas of regions bounded by the curve, the x-axis, and the vertical lines through the endpoints of the intervals. The calculator can approximate the areas of such regions. To find the area of the region bounded by the curve, the x-axis, and the vertical lines x = -1 and x = 1, go to the CALC menu and select 7: ∫f(x) dx. Move the cursor to
the point where x = -1. Press ENTER . Then move the cursor to the
point where x = 1 and press ENTER . The calculator will shade the region and display its approximate area. What number does the calculator display for the area of the shaded region?
2. Enter 2nd [DRAW] 1. This causes the calculator to clear the shading and redisplay the graph. Find the area of the region bounded by the curve, the x-axis, and the vertical lines x = -2 and x = 2.
3. Find the area of the region bounded by the curve, the x-axis, and the vertical lines x = -3 and x = 3.
4. Without using a calculator, estimate the area of the region bounded by the curve, the x-axis, and the vertical lines x = -4 and x = 4 to four decimal places. Verify your conjecture.
11-3 Graphing Calculator ActivityThe Standard Normal Curve
[-4.7, 4.7] scl: 1 by [-0.2, 0.5] scl: 0.1
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Chapter 11 22 Glencoe Precalculus
11-4 Study Guide and InterventionThe Central Limit Theorem
The Central Limit Theorem The Central Limit Theorem states that as the sampling size n increases:
• the shape of the distribution of the sample means of a population with mean μand standard deviation σ will approach a normal distribution and
• the corresponding distribution will have a mean μ and standard deviation σ
−
x = σ −
√
�
n .
The z-value for a sample mean in a population is given by z = − x - μ
− σ − x , where − x is the
sample mean, μ is the mean of the population, and σ
−
x is the standard error.
A study was done in which parents reported the number of hours per day that their children, between the ages of 2.0 and 5.9, watched television or videos. The mean age of the children was 3.3 years with a standard deviation of 0.9 year. Assume that the variable is normally distributed. If a random sample of 30 children in this study is selected, find the probability that the mean age is less than 3 years.
The distribution of sample means will be approximately normal with μ = 3.3 and
σ
−
x = 0.9 −
√
��
30 or about 0.164. Find the z-value.
z = − x - μ
− σ − x z-value for a sample mean
z = 3 - 3.3 −
0.164 or about -1.83 − x = 3, μ = 3.3, σ − x = 0.164
The area to the left of a z-value of -1.83 is 0.03359.
The probability that the mean age of the sampleis less than 3 years or P( − x < 3) is about 3.36%.
Exercises 1. The mean pitch of the expert slopes at the ski resorts in a certain region
is 25˚ with a standard deviation of 3˚. Assume that the variable is normally distributed. If 20 expert slopes are chosen at random, what is the probability that the mean of the slopes will be greater than 26.3˚?
2. A veterinarian reports that the average age of the cats that she treats is 96 months with a standard deviation of 16 months. If a random sample of 36 of her cat patients is selected, find the probability that the mean age is between 90 and 100 months.
Example
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Chapter 11 23 Glencoe Precalculus
11-4 Study Guide and Intervention (continued)
The Central Limit Theorem
The Normal ApproximationThe normal distribution can be used to approximate a binomial distribution if:
• the original variable is normally distributed or n ≥ 30 and
• np ≥ 5 and nq ≥ 5,where n is the number of trials, p is the probability of success, and q is theprobability of failure.When using the normal distribution to approximate a binomial distribution, thecontinuity correction factor must be used. This involves adding 0.5 unit to or subtracting 0.5 unit from a given discrete boundary.
Ten percent of the members of a golf league are youngerthan 30. If 200 members are randomly selected, find the probability thatmore than 10 will be younger than 30.
Step 1 Find the mean μ and standard deviation σ of the binomial distribution.
μ = np σ = √
��
npq
= 200 · 0.10 or 20 = √
�������
200 · 0.10 · 0.90 or about 4.24
Since np = 20 and nq = 180, the normal distribution can be used.
Step 2 Write the problem in probability notation: P (X > 10) .
Step 3 Rewrite the problem with the continuity factor included. Since the question is asking for the probability that more than 10 members will be younger than 30, add 0.5 to 10.
P (X > 10) = P (X > 10 + 0.5) or P (X > 10.5)
Step 4 Find the corresponding z-value for X = 10.5.
z = X - μ
−
σ
= 10.5 - 20 −
4.24 ≈ -2.24
Step 5 Find the corresponding area above z = -2.24. It isabout 0.987. The probability is about 98.7%.
Exercises 1. A study found that 40% of all of a town’s citizens approve of a new train station.
If 50 citizens are randomly selected, find the probability that fewer than 20 citizens approve of a new train station.
2. A baseball player gets a hit 32% of the time. Find the probability that theplayer will get at least 26 hits in his next 100 times at bat.
Example
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Chapter 11 24 Glencoe Precalculus
11-4 PracticeThe Central Limit Theorem
1. NURSING The mean salary for nurses in a city is $52,129 with a standard deviationof $1800. What is the probability that the mean salary for a randomly selectedgroup of 50 nurses in the city is greater than $52,500?
2. UTILITIES The average electric bill in a residential area in June is $72. Assumethis variable is normally distributed with a standard deviation of $6. Find theprobability that the mean electric bill for a randomly selected group of 15 residentsis less than $75.
3. SHOES The prices of shoes in a store are normally distributed with a mean of$93 and a standard deviation of $18. If nine pairs of shoes are randomly selected,find the probability that the mean cost is between $100 and $110.
4. COLLEGE Of the total population at a small college, 20% are from the Mid-Atlantic states. If 200 students are randomly selected, find the probability that at least50 are from the Mid-Atlantic states.
5. WEATHER Kyle has researched the average annual precipitation in his city forseveral years and calculated the mean as 19.32 inches. Assume the averageprecipitation is normally distributed and the standard deviation is 2.44 inches.
a. If one year from the time period that Kyle researched is randomly selected,what is the probability that the precipitation is more than 18 inches?
b. If five years from the time period that Kyle researched are randomly selected,what is the probability that the mean precipitation is more than 18 inches?
6. GROCERY Eighty-five percent of the shoppers at a grocery store have afrequent-buyer card. Thirty-five shoppers are randomly selected for a tastetest. What is the probability that at least 25 and at most 30 of the tastetesters have a frequent-buyer card?
Find the minimum sample size needed for each probability so that the normal distribution can be used to approximate the binomial distribution.
7. p = 0.6 8. p = 0.15
Chapter 11 24 Glencoe Precalculus
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Chapter 11 25 Glencoe PrecalculusChapter 11 25 Glencoe Precalculus
Word Problem PracticeThe Central Limit Theorem
11-4
1. HEATING Workers at a public utilities company surveyed 200 of their customers and found that the mean temperature at which the customers’ thermostats were set in January was 70˚ with a standard deviation of 1.8˚. If 30 customers are randomly selected for a follow-upsurvey, find the probability that the mean temperature at which they set their thermostats in January is lessthan 69.5˚.
2. MILK A food study in a city found that the price of a gallon of whole milk in its stores was $3.72. This variable is normally distributed with a standard deviation of $0.08.
a. If a gallon of whole milk is randomly selected from one store, find the probability that it costs less than $3.70.
b. If a gallon of milk is randomly selected from each of 40 different stores, find the probability that the mean cost of the sample is less than $3.70.
3. BICYCLES Thirty-seven percent of the residents in a neighborhood own bicycles. If a group of 40 residents are randomly selected, what is the probability that at least half own bicycles?
4. TESTING The average time it takes a group of students to complete a reading test is 46.2 minutes with a standard deviation of 8 minutes. The times are normally distributed.
a. A group of 10 students is randomly selected. Find the probability that the mean time to complete the test is more than 45 minutes.
b. How does your answer to part achange if the group consists of15 students?
5. CAFETERIA College freshmen were asked if they ate breakfast on Sunday morning in the cafeteria. The graph shows the percent of males and females who said yes.
FemaleMale
40
60
20
0
Perc
ent
80
a. Out of 30 randomly selected female freshmen, what is the approximate probability that at least 25 of them ate breakfast in the cafeteria on Sunday morning?
b. Out of 30 randomly selected male freshmen, what is the approximate probability that at least 25 of them ate breakfast in the cafeteria on Sunday morning?
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Chapter 11 26 Glencoe Precalculus
11-4
Experience the Central Limit TheoremThe Central Limit Theorem is best appreciated when experienced. This activitywill allow you this privilege. You may wish to work with a partner. Simulate spinning a spinner with equal sections labeled 1–10two times by selecting the randInt( command from thePRB menu after pressing MATH . The screen shown indicatesthe spinner landed on 7 on the first spin and 1 on the second spin.The mean of these spins is 4.
1. Complete 25 simulations and list the means of the spins.
2. Find the mean and standard deviation of the means in Exercise 1.
Now simulate spinning the same spinner four times.
3. Complete 25 simulations and list the means of the spins.
4. Find the mean and standard deviation of the means in Exercise 3.
Now simulate spinning the same spinner six times.
5. Complete 25 simulations and list the means of the spins.
6. Find the mean and standard deviation of the means in Exercise 5.
7. What do you notice about the means in Exercises 2, 4, and 6?
8. What do you notice about the standard deviations in Exercises 2, 4, and 6?
9. Find σ −
√
�
n for n = 2, 4, and 6 given σ = 2.87. What do you notice?
10. If n = 1, that is the spinner is spun once, what is the distribution of the samples?How does it change as n increases?
Enrichment
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Chapter 11 27 Glencoe Precalculus
Normal Distribution The confidence level c is the probability that an interval estimate contains a given parameter, such as a population mean. The maximum error of estimate E is the maximum difference between the point estimate and the actual value of the parameter. A confidence interval CI is a specific interval estimate of a parameter. For a population mean:
CI = − x ± E or − x ± z · σ −
√
�
n .
z is the critical value that corresponds to a certain confidence level. The most common are shown below.
Confidence Level 90% 95% 99%
z-Value 1.645 1.960 2.576
When σ is unknown, the sample standard deviation s can be substituted for σ, provided the sample size n is greater than or equal to 30.
In a sample of 50 people who buy magazines, a researcher finds the mean amount spent per month to be $12. Assume a standard deviation of $4.50. Find the 95% confidence interval for the mean amount spent for magazines each month.CI = − x ± z · σ
−
√
�
n Confidence Interval for the Mean
= 12 ± 1.96 · 4.50 −
√
��
50 − x = 12, z = 1.96, σ = 4.50, and n = 50
≈ 12 ± 1.25 Simplify.
Add and subtract the margin of error.
Left Boundary: 12 - 1.25 = 10.75 Right Boundary: 12 + 1.25 = 13.25
The 95% confidence interval is 10.75 < μ < 13.25. Therefore, with 95% confidence, the mean amount spent on magazines per month is between $10.75 and $13.25.
Exercises
The number of days with temperatures above freezing for a sample of 35 cities had a mean of 190.7 days and a sample standard deviation of 54.2 days.
1. Find the 95% confidence interval for the mean number of days with temperatures above freezing.
2. Find the 90% confidence interval for the mean number of days with temperatures above freezing.
Study Guide and InterventionConfidence Intervals
Example
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Chapter 11 28 Glencoe Precalculus
t-Distribution To find a confidence interval when σ is unknown and the sample size n is less than 30, use the t-distribution, provided the variable is approximately normally distributed. The formula for using the t-distribution to construct a confidence interval is
CI = − x ± t · s −
√
�
n ,
where − x is the sample mean, t is a critical value with n - 1 degrees of freedom, s is the sample standard deviation, and n is the sample size.
The mean price of 9 randomly selected televisions at an electronics store is $783 with a standard deviation of $116. Find the 90% confidence interval of the mean price of all of the televisions at the store. Assume that the variable is normally distributed.
Because the population standard deviation is unknown and n < 30, use the t-distribution. Because n = 9, there are 9 - 1 or 8 degrees of freedom.
To find the t-value, determine the area in the lower tail of the distribution. Since 100% - 90% or 10% is in the tails, then 5% is in each tail. Press 2nd DISTR and choose invT(. Type thearea to the left of each critical value, as a decimal, followed by the degrees of freedom.
CI = − x ± t · s −
√
�
n Confidence interval using t-distribution
≈ 783 ± 1.860 · 116 −
√
�
9 − x = 783, t ≈ 1.860, s = 116, and n = 9
≈ 783 ± 71.92 Simplify.
Therefore, the 90% confidence interval is $711.08 < μ < $854.92.
Exercises
1. At a manufacturing plant, the threads on ten randomly selected screws have a mean depth of 0.32 inch and a standard deviation of 0.03 inch. Find the 95% confidence interval of the mean depth of all the screws, assuming that the variable is normally distributed.
2. An environmental study involves counting the number of light bulbs in randomly selected rooms of a building. A building manager counts the bulbs in 16 rooms and finds the mean number of bulbs to be 21 with a standard deviation of 4.8. Find the 99% confidence interval of the mean number of bulbs in all the rooms of the building, assuming the variable is normally distributed.
Study Guide and Intervention (continued)
Confidence Intervals
Example
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Chapter 11 29 Glencoe Precalculus
1. CHILD CARE A random sample of 50 parents of young children found that they spend a mean of $7648 each year for child care. The standard deviation of the sample was $630. Find the 90% confidence interval of the mean annual cost of child care.
2. FITNESS Twenty-eight people who enrolled in a fitness program lost a mean of 14.3 pounds with a standard deviation of 2 pounds. Find the 95% confidence interval of the mean weight loss in pounds of all of the members enrolled in the program.
3. COMMUTE The average number of minutes it takes 8 people to commute to and from work during rush hour is shown. Assume that the times are normally distributed.
Commuting Time (minutes)
55 70 60 55 60 56 55 60
a. Decide the type of distribution that can be used to estimate the commuting time mean. Explain your reasoning.
b. Calculate the mean and standard deviation to the nearest hundredth.
c. Construct a 90% confidence interval for the average commuting time in minutes for a commuter from this city.
4. TRAINING In a survey of 442 employees at a call center, the mean time that employees felt was needed for adequate training for their jobs was 7 days. The sample standard deviation was 1.5 days. Construct a 98% confidence interval for the amount of training time that employees felt was adequate to begin their jobs.
Determine whether the normal distribution or t-distribution should be used for each question. Then find each confidence interval given the following information.
5. 95%; − x = 115, s = 6, n = 6
6. 96%; − x = 18.5, s = 1.2, n = 40
7. 99%; − x = 236, σ = 8, n = 45
8. FOOD The owners of a sandwich shop want to find the 95% confidence interval of the true mean cost of a hamburger in their city. How large should their sample be if they want to be accurate to within 15 cents? In an earlier survey, the standard deviation of the price was 26 cents.
9. SPORTS A teacher wants to estimate the average number of hours per week that her students are at practice or at games for a sports team. The standard deviation from a previous year was 6.2 hours. How large of a sample must she select if she wants to be 99% confident of finding the average amount of time students spent participating in sports within 1.5 hours?
PracticeConfidence Intervals
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Chapter 11 30 Glencoe Precalculus
1. READING A sample of normally distributed reading scores of forty eighth-grade students has a mean of 82 and a standard deviation of 15. Find the 95% confidence interval for the mean of all of the reading scores.
2. CHOLESTEROL The serum cholesterol level was collected for a group of525 college women. The mean of the sample was 191.7
mg −
100 mL with a
standard deviation of 41.0.
a. Construct a 90% confidence level for the mean serum cholesterol level.
b. Construct a 95% confidence level for the mean serum cholesterol level.
c. Suppose you hear a claim that the mean serum cholesterol level for women in college is 200. What would be your reaction based on your answers to parts a and b? Why?
3. CAR POLLUTION Seven cars were tested for nitrogen-oxide emissions. The results are shown in the table.
Emissions (grams per mile)
0.05 0.12 0.16 0.15 0.14 0.19 0.14
a. Find the mean of the data.
b. Find the standard deviation of the data.
c. Find the 99% confidence interval of the mean nitrogen-oxide emissions.
4. FUEL CONSUMPTION The mean and standard deviation for city and highway fuel consumption in miles per gallon for 33 randomly selected pre-owned cars on a dealer’s lot is shown. Assume the variables are normally distributed.
− x s
City 21.35 4.13
Highway 29.65 3.65
a. Find the 98% confidence interval for the mean fuel consumption in the city.
b. Find the 98% confidence interval for the mean fuel consumption on the highway.
c. Compare the confidence intervals in parts a and b.
5. INTELLIGENCE QUOTIENT Suppose managers of a corporation want to estimate the IQ score for their employees. How many employees must be randomly selected for IQ tests if the managers want to be 95% confident that the mean is within 2 IQ points of the population mean? They know from previous studies that the standard deviation is 15 points.
Word Problem PracticeConfidence Intervals
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Chapter 11 31 Glencoe Precalculus
The Population Correction FactorThe formula for the maximum error of estimate, E = z · σ
−
√
�
n , assumes a very large
population or sampling with replacement. If the population size is N and the sample number is n, the formula should be modified with a correction factor if n > 0.05N.
The population correction factor is √
���
N - n −
N - 1 . Therefore, if n > 0.05N, the maximum
error of estimate is E = z · σ −
√
�
n ·
√
���
N - n −
N - 1 .
Determine if the population correction factor would be used when determininga confidence interval. Justify your answer.
1. 29 of the 140 employees in a company are surveyed
2. 75 of the town’s residents are surveyed and the town population is 2998
3. 10 of the 80 pages printed by a new printer are examined
4. There are 250 students in a school. A sample of 35 randomly selected students finds that their mean daily study time is 52 minutes with a standard deviationof 3.3 minutes. Find the 95% confidence interval for the mean daily study timeof all of the students. Round E to the nearest hundredth.
5. What would be the confidence interval in Exercise 4 if there were 750 students in the school?
6. Forty of the 160 potatoes in a bin are randomly selected for weight. The meanweight of the sample is 150 grams with a standard deviation of 6.5 grams. Findthe 90% confidence interval for the mean weight of all of the potatoes. Round Eto the nearest hundredth. Assume the variable is normally distributed.
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Chapter 11 32 Glencoe Precalculus
Hypotheses A hypothesis test allows you to evaluate a claim about a population.
The null hypothesis or H0 states that there is not a significant difference between the sample value and the population parameter and it always contains a statement of equality.The alternative hypothesis or Ha states that there is a significant difference betweenthe sample value and the population parameter and it always contains a statement of inequality.
For each statement, write the null and alternative hypotheses. State which hypothesis represents the claim.
a. A doctor claims that the pulse rate of a patient changes from 82 beats per minute after taking a new medication.
This claim becomes μ ≠ 82 and is the alternative hypothesis since it includes an inequality symbol. The complement is μ = 82.
H0: μ = 82 Ha: μ ≠ 82 (claim)
b. A track member claims that he can run at least 10 miles that day.
This claim becomes μ ≥ 10 and is the null hypothesis since it includes an equality symbol. The complement is μ < 10.H0: μ ≥ 10 (claim) Ha: μ < 10
c. A contractor claims that installing a particular kind of insulation willlower the average July cooling bill of $68 in her area.
This claim becomes μ < $68 and is the alternative hypothesis since it includesan inequality symbol. The complement is μ ≥ $68.H0: μ ≥ $68 Ha: μ < $68 (claim)
Exercises
Write the null and alternative hypotheses for each statement. State which hypothesis represents the claim.
1. On a scale of 1 to 10, patients describe their anxiety levels as 8 during dentalprocedures. A dental assistant thinks this level is lower when soft music isplayed during the procedures.
2. A real estate agent claims that the average home price in an area is $250,000.
3. A hiker claims that the average trail length in a park is at most 10 miles.
4. A restaurant owner claims that the average age of diners in a certain area is greater than 40.
Study Guide and InterventionHypothesis Testing
Example
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Chapter 11 33 Glencoe Precalculus
Significance and Tests To validate a claim, the null hypothesis is always tested at a chosen level of significance, α, which defines the area of the critical region. If the test statistic z- or t-value is in the critical region, then H0 should be rejected. Otherwise, H0 should not be rejected.
An employee at a sporting goods store claims that the average price of a pair of baseball cleats is less than $80. Another employee randomly selects a sample of 36 pairs and finds − x = 75 and s = 19.2. Determine if theresults are statistically significant at α = 0.10.Step 1 State the null and alternative hypotheses and identify the claim. H0: μ ≥ $80 Ha: μ < 80 (claim)
Step 2 Determine the critical value and region.Because n ≥ 30, use the z-value. The test is left-tailed since μ < 80. All of the critical region, with an area of 0.10, is to the left of the critical value. Use 2nd DISTR invNorm to see that the criticalvalue is -1.28. The critical region is the area to the left of z = -1.28.
Step 3 Calculate the test statistic. Use σ − x = 19.2 −
√
��
36 = 3.2.
z = − x - μ
− σ − x Formula for z-statistic
= 75 - 80 −
3.2 or -1.5625 − x = 75, μ = 80, and σ
−
x = 3.2
Step 4 Accept or reject the null hypothesis. H0 is rejected since the test statistic of -1.56 falls within the critical region. There is enough evidence to support the claim that the average cost of baseball cleats is less than $80.
Exercises
1. Managers of a large department store claim that the average salary for theirpart-time employees is $24,000. A sample of 10 part-time employees has a meansalary of $23,450 and a standard deviation of $1400. Determine whether thereis enough evidence to support the claim at α = 0.05.
2. In Exercise 1, suppose that the mean of $23,450 was taken from a sample of50 part-time employees. Would there be enough evidence to support the claimat α = 0.05?
11-6 Study Guide and Intervention (continued)
Hypothesis Testing
Example
-1.28
Critical region
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Chapter 11 34 Glencoe Precalculus
Write the null and alternative hypotheses for each statement. State which hypothesis represents the claim.
1. A florist claims that a certain type of flower has a vase life of at least 7 days.
2. A brand of cereal claims that a serving contains less than 2 grams of sugar.
3. Robert claims that he swims 100 laps each week.
For each claim k, use the specified information to calculate the test statistic and determine whether there is enough evidence to reject the null hypothesis. Then make a statement regarding the original claim.
4. k: μ ≥ 60, α = 0.10, − x = 58.88, s = 5.08, n = 8
5. k: μ = 8, α = 0.05, − x = 8.2, s = 0.6, n = 32
6. MAIL A mail carrier claims that the average number of pieces of mail received daily by households in a certain neighborhood is 7. Sample data for 15 households is collected. The average number of mail pieces was 6.5 with a standard deviation of 0.8.
a. Is there enough evidence to reject the mail carrier’s claim at α = 0.05?
b. Is there enough evidence to reject the mail carrier’s claim at α = 0.01?
7. WEDDINGS A wedding planner wants to test the claim that the average wedding has 125 guests. In a random sample of 35 weddings, he found the mean to be 110 guests with a standard deviation of 30 guests. Find the P-value and determine whether there is enough evidence to support the claim at α = 0.01.
PracticeHypothesis Testing
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Chapter 11 35 Glencoe Precalculus
1. BOTTLED WATER The average volume in ounces of a random sample of 36 bottles of water at a packaging plant was found to be 12.19 ounces with a standard deviation of 0.11 ounce. The floor supervisor made the claim that the mean volume was greater than 12 ounces. Test her claim at α = 0.01.
a. Write the null and alternative hypotheses and state which hypothesis represents the claim.
b. Is there enough evidence to reject the null hypothesis? Why?
c. Make a statement regarding the original claim.
2. TEMPERATURE It is a long-established claim that the mean body temperature for humans is 98.6°F. In a random sample of 10 people, the mean body temperature was 98.3°F with a standard deviation of 0.23°F. Test the claim atα = 0.02.
a. What are the critical values?
b. What is the test statistic?
c. Is this evidence enough to supportthe claim that the mean body temperature for humans is 98.6°?
3. PULSE RATES An aerobics instructor had a pulse rate of 110 beats per minute after a warm-up routine with her class. Students recorded their pulse rates at the same time. Their rates are recorded in the table below. The instructor claims that their average pulse rate is lower than hers. Test her claim at α = 0.05.
80 70 90 75 110
105 120 110 85 115
95 95 105 90 70
105 95 100 105 90
a. Write the null and alternative hypotheses and state which hypothesis represents the claim.
b. What are the critical values and test statistic?
c. Does the data support the instructor’s claims?
4. MOUNTAIN CLIMBING A mountain climbing instructor claims that his students take longer than 5 minutes to pack their backpacks. In a random sample of 80 students, the average time it took to pack a backpack was 5.3 minutes with a standard deviationof 1.7 minutes. Find the P-value and determine whether there is enough evidence to support the claim atα = 0.05.
11-6 Word Problem PracticeHypothesis Testing
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Chapter 11 36 Glencoe Precalculus
Hypothesis TestingYou have tested hypotheses by finding a test statistic and seeing if it falls within the critical region. Alternatively, you can test a hypothesis by finding a confidence interval and seeing if the claim falls within that interval.
A condominium superintendent buys 5-pound bags of salt for melting ice.He weighs the salt in 50 randomly selected bags and finds the mean to be4.6 pounds with a standard deviation of 0.7 pound. Use a confidence levelto test the superintendent’s claim that the mean weight is not 5 pounds at α = 0.05.
Write the hypotheses: H0: μ = 5 Ha: μ ≠ 5
Use the normal distribution since n > 30. Since α = 0.05, the confidence level is 1 - α = 0.05, or 0.95. The z-value associated with a 95% confidence level is 1.96.
Find the 95% confidence interval for the mean weight of the salt in the bags.CI = − x ± z · σ
−
√
�
n Confidence interval for the mean
= 4.6 ± 0.194 − x = 4.6, and z · σ
−
√
�
n = 1.96 · 0.7
−
√
��
50 or about 0.194
So, the 95% confidence interval is 4.406 < μ < 4.794.
The confidence interval for the population does not contain the claim of 5, so we reject the null hypothesis. Evidence supports the claim that the mean is not 5 pounds.
Exercises
1. SOCCER A soccer coach claims that the mean weight of the players on the opposing teams is 200 pounds. A random sample of 10 players from opposing teams has a mean of 198.2 pounds and a standard deviation of 3.3 pounds. At α = 0.05, can the claim be supported? Support your answer with a confidence interval.
2. CALORIES A teacher walked by a vending machine and claimed that the average number of Calories in the snacks was 250. Students who overheard her wanted to prove her incorrect. They bought a random sample of 8 snacks and found the mean number of Calories of those snacks to be 210 with a standard deviation of 24. At α = 0.10, does the data support the teacher or the students? Support your answer with a confidence interval.
3. HOMEWORK A teacher claims her students spend an average of 45 minutes on her homework each night. To test the claim, she asks 35 students how much time, on average, they spend on her homework each night and the results show a mean of 40 minutes with a standard deviation of 9 minutes. At α = 0.05, does the data support her claim? Support your answer with a confidence interval.
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Chapter 11 37 Glencoe Precalculus
11-7
Correlation To determine if the correlation coefficient r is significant,
perform a hypothesis test with H0 : ρ = 0, and t as the test statistic, t = r √
���
n - 2 −
1 - r2 ,
where the degrees of freedom is n - 2.
The table shows the number of absences of 7 students from a psychology class and their final grades.
Absences 9 10 15 2 8 2 6
Grades 70 58 43 90 78 87 79
a. Make a scatter plot of the data and identify the relationship. Then calculate and interpret the correlation coefficient.
Enter the data into L1 and L2. Turn on Plot1 and choose a scatter plot. The data appears to have a negative linear correlation.
Press STAT and select LinReg(ax + b) under the Calc menu. The correlation coefficient r is about -0.9609. Because r is close to -1, this suggests a strong negative correlation.
b. Test the significance of the correlation coefficient from part a at the 5% level.
Step 1 State the hypotheses. H0: ρ = 0 Ha: ρ ≠ 0
Step 2 Determine the critical values using n - 2 or 5 degrees of freedom. The critical values are ±2.6.
Step 3 Calculate the test statistic.
t = -0.9609 √
������
5 −
1 - (-0.9609)2 ≈ -7.7597
Step 4 Since t < -2.6, reject the null hypothesis. The evidence supports a significant correlation between the number of absences and a student’s grade.
Exercises
The table shows data about geysers at Yellowstone National Park.
Duration (min) 3.25 3 5 7.5 17.5 3.5 6 35 10 6.5 10 1 20 4
Height (ft) 184 25 150 20 75 75 10 100 160 60 5 25 75 5
Source: National Park Service
1. Make a scatter plot of the data and identify the relationship. Then calculate and interpret the correlation coefficient.
2. Test the significance of the correlation coefficient from Exercise 1 at the 5% level.
Study Guide and InterventionCorrelation and Linear Regression
Example
[1, 17] scl: 1 by [35, 100] scl: 5
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Chapter 11 38 Glencoe Precalculus
Linear Regression If the correlation between two variables is significant, you can determine the least-squares regression line, which is a line of best fit. Then you can use the equation of the line to make predictions over the range of data.
The table shows the data from the previous page, which showed a significant correlation. Find the equation of the regression line for the data. Interpret the slope and intercept in context. Use the equation to predict the expected grade for a student with 4 absences and state whether this prediction is reasonable. Explain.
Absences 9 10 15 2 8 2 6
Grades 70 58 43 90 78 87 79
Press STAT and select LinReg(ax + b) under the Calc menu. The equation is approximately y = -3.48x + 97.993.
The slope indicates that for every additional absence a student has, his or her grade will drop by about 3.5 points. The y-intercept indicates that a student with no absences will have a grade of about 98.
Evaluate the regression equation for x = 4 and calculate y.y = -3.48x + 97.993 Regression equation
= -3.48(4) + 97.993 x = 4
= 84.073 Simplify.
We expect that a student with 4 days of absences would have a final grade of about 84. This is reasonable because 4 is in the range of the given x-values and 84 is in the range of the given y-values.
Exercises
A sports writer has already determined that there is a negative correlation between the batting average and the number of home runs for third basemen whose statistics are shown in the table.
Average 0.307 0.328 0.305 0.294 0.306 0.311 0.271 0.267 0.267 0.320
Home Runs 96 118 317 65 78 103 512 268 548 58
1. Find the equation of the regression line for the data.
2. Use the equation to predict the number of home runs for a third baseman with a batting average of 0.350. State whether this prediction is reasonable. Explain.
Study Guide and Intervention (continued)
Correlation and Linear Regression
11-7
Example
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Chapter 11 39 Glencoe Precalculus
A supervisor of a cleaning business has the data in the table that shows the age of her workers and the number of sick days they take each year. She wonders if there is a significant linear relationship between the age of an employee and the number of sick days he or she takes each year.
Age 38 25 60 18 45 54 19 22 43 34
Days 9 12 2 16 4 5 15 17 3 1
1. Make a scatter plot of the data and identify the relationship. Then calculate and interpret the correlation coefficient.
2. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels. Explain your reasoning.
3. If the correlation is significant at the 10% level, find the least-squares regression equation and interpret the slope and y-intercept in context.
4. Graph and analyze the residual plot.
5. Identify any influential outliers. Describe the effect the outlier has on the strength of the correlation and on the slope and intercept of the original regression line.
6. If any data were removed, reassess the significance of the correlation at the 10% level and, if still appropriate, recalculate the regression equation.
7. If appropriate, predict the number of sick days for employees who are 30, 50, and 70 years old. Interpret your results and state whether the predictions are reasonable. Explain your reasoning.
PracticeCorrelation and Linear Regression
11-7
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Chapter 11 40 Glencoe Precalculus
1. COLLEGE The data in the table represent the American College Test (ACT) composite scores and grade point averages (GPA) of 20 randomly selected students after their first semester in college. A college counselor wants to determine if there is a correlation between ACT scores and first semester GPAs.
ACT 27 18 17 15
GPA 3.9 2.9 3.3 3.0
ACT 22 20 17 21
GPA 3.6 2.7 2.9 3.4
ACT 25 17 25 18
GPA 3.5 3.1 4.0 3.0
ACT 23 19 20 29
GPA 3.6 2.6 3.0 3.4
ACT 23 28 22 20
GPA 1.8 4.0 3.0 4.0
a. Calculate and interpret the correlation coefficient.
b. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels.
c. Graph and analyze the residual plot.
2. SALES A sales associate wants to know if there is a relationship between the average number of times his coworkers contact clients each month and the average monthly sales volume in thousands of dollars. He collected the data shown in the table.
Clients 21 23 48 50 46
Sales 30 30 95 110 80
Clients 12 55 14 50 16
Sales 15 130 25 90 30
a. Make a scatter plot of the data and identify the relationship. Then calculate and interpret the correlation coefficient.
b. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels. Explain your reasoning.
c. If the correlation is significant at the 10% level, find the least-squares regression equation and interpret the slope and y-intercept in context.
11-7 Word Problem PracticeCorrelation and Linear Regression
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Chapter 11 41 Glencoe Precalculus
Multiple RegressionIn multiple regression, there are two or more independent variables and one dependent variable. The multiple regression equation is y = a + b1x1 + b2x2 + … + bkxk, where x1, x2, …, xk are the independent variables.
The strength of the relationship between the independent variables and the dependent variable is measured by the multiple correlation coefficient R. This value can range from 0 to 1, where values closer to 1 indicate a stronger relationship than those closer to 0. The formula for a multiple correlation coefficient with two independent variables is
R =
√
�������������
(
ry x 1 ) 2 +
(
r yx2 )
2 - 2 r yx1 (
r yx2 )
(
r x1x2 )
−−
√
����
1 - (
r x1x2 )
2 ,
where ry x 1 is the correlation coefficient for y and x1; ry x 2
is the correlation coefficient for y and x2 ; and r x1x2
is the correlation coefficient for x1 and x2.
Hospital administrators wish to see if a nursing applicant’s GPA and age are related to the nurse’s score on the state nursing board exam. The table shows these statistics for five nurses.
1. Find the values of ry x 1 , ry x 2
, and r x1x2 .
2. Find and interpret R.
3. A researcher is conducting a study to see if a person’s age x1 and cholesterol level x2 are related to his or her systolic blood pressure y. Find and interpret R if ry x 1
= 0.681,
ry x 2 = 0.872, and r x1x2
= 0.746.
A track coach wants to see if the 5k times for the first two meets of the season are related to the 5k times at the state meet. The table shows these statistics, in minutes, for five runners.
4. Find the values of r yx1 , r yx2
, and r x1x2 .
5. Find and interpret R.
Enrichment11-7
GPA
x1
Age
x2
State Board
Exam y
3.5 28 675
2.7 28 570
3.2 22 560
2.2 23 490
2.4 24 550
Meet 1
x1
Meet 2
x2
State Meet
y
17.4 16.5 16.5
15.4 16.1 15.8
18 18.2 16.9
16.5 16.3 16.5
17.2 17.5 16.1
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Chapter 11 43 Glencoe Precalculus
11 Chapter 11 Quiz 1(Lessons 11-1 and 11-2)
11 Chapter 11 Quiz 2(Lessons 11-3 and 11-4)
1. MULTIPLE CHOICE Find X if z = 2.88, μ = 43, and σ = 5.2.
A -10.6 B -7.7 C 43.6 D 58.0
2. Find the interval of z-values associated with the middle 60%.
Wait-listed students at a certain college had a mean ACT reading score of 21.3 with a standard deviation of 6.
3. If a sample of 50 students is selected, find the probability that the mean score of the sample is less than 20.
4. If a sample of 20 students is selected, find the probability that the mean score is between 19 and 21.
5. For a normal approximation of a binomial distribution, how would you rewrite P(X < 16) to account for the continuity correction factor?
1.
2.
3.
4.
5.
X 1 2 3 4
P(X) 0.18 0.34 0.21 0.27
The exam scores of 19 students are shown.
78 92 76 66 88
52 76 86 84 74
74 72 78 82 82
62 96 76 68
1. Construct a histogram.
2. Describe the shape of the distribution.
3. Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice.
4. What is the mean of the probability distribution?
5. MULTIPLE CHOICE One of the competitors in a dart competition hits the center 90% of the time. Find the probability that he will hit the center on at least 4 of his next 5 throws.
A 0.328 B 0.590 C 0.919 D 0.950
1.
2.
3.
4.
5.
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Chapter 11 44 Glencoe Precalculus
11
11
A random sample of 60 college students showed a mean commute of 10.22 miles, with a standard deviation of 2.4 miles. 1. Find the maximum error of estimate given a
95% confidence level. 2. Write the 95% confidence interval for the mean
commuting time. 3. A doctor wants to estimate the mean weight of all
7-year old boys to within 1 pound with 95% confidence. How large should her sample size be if a previous study showed a standard deviation of 3 pounds?
Kurt claims that the mean car rental rate in his city is at least$60 per day. A sample of rates from 10 local companies gives a mean of $63.50 and a standard deviation of $2.75. 4. MULTIPLE CHOICE Determine the critical value to test
the claim at α = 0.05.A 1.36 C 2.26B 1.83 D 2.31
5. Is there sufficient evidence to reject Kurt’s claim? Explain.
Chapter 11 Quiz 3(Lessons 11-5 and 11-6)
Students who participated in a bully awareness project were given tolerance surveys before and after the project. The pre- and post-scores are shown.
Pre-score 18 14 11 23 19 21 21 11 22
Post-score 17 17 10 25 20 15 24 10 24
1. Calculate and interpret the correlation coefficient.
2. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels. Explain your reasoning.
3. If the correlation is significant at the 10% level, find the least-squares regression equation.
4. If appropriate, use the regression equation to predict a student’s post-score if their pre-score is 15.
5. MULTIPLE CHOICE Which data point is influential?
A (14, 17) C (19, 20)
B (11, 10) D (21, 15)
1.
2.
3.
4.
5.
Chapter 11 Quiz 4(Lesson 11-7)
1.
2.
3.
4.
5.
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Chapter 11 45 Glencoe Precalculus
11
Write the letter for the correct answer in the blank at the right of each question.
1. Which random variable X is a discrete variable?
A the number of degrees a liquid rises after being heated
B the number of female babies in the birth of twins
C the thickness of a bolt
D the height of a tomato plant 20 days after germination
2. BREAD The mean weight of a loaf of Italian bread at a bakery is 482 grams with a standard deviation of 18 grams. In a random sample of 40 loaves, what is the probability that the mean of the sample will be less than 478 grams?
F 0.08 G 0.16 H 0.41 J 0.92
3. MEDICINE Seventy-five percent of those surveyed reported that a certain medicine was effective in reducing arthritis pain. If 7 respondents are randomly selected, what is the probability that 3 thought the medicine was effective?
A 0.058 B 0.071 C 0.093 D 0.173
Part II
DEXTERITY The data show the time, in minutes, it takes people with carpal tunnel syndrome to complete a manual task. Half of the group are taking medication and half are taking a placebo.
Placebo
2.9 4.1 2.7 3.5 5.6 4.0 13.3 5.2
5.6 6.5 5.2 4.8 4.8 4.0 5.7 5.5
Medicine
1.4 3.9 2.9 3.3 3.4 3.2 4.0 4.4
3.1 2.6 5.3 2.6 2.3 2.8 3.4 2.9
4. Construct side-by-side box plots of the data sets.
5. Use the box plots to compare the center and spread of the distributions.
6. Find the interval of z-values associated with the outside 35% of the data.
7. TOOLS Weights of hammers are normally distributed withμ = 2.2 pounds and σ = 0.3 pound. If the heaviest 15% of hammers are on sale, what is the lightest weight a hammer can be and still be on sale?
Mid-Chapter Test(Lessons 11-1 through 11-4)
Part I
1.
2.
3.
4.
5.
6.
7.
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alternative hypothesis
binomial distribution
Central Limit Theorem
confidence interval
continuous random variable
correlation
correlation coefficient
critical values
discrete random variable
empirical rule
explanatory variable
extrapolation
hypothesis test
inferential statistics
interpolation
least squares regression line
negatively skewed distribution
normal distribution
null hypothesis
percentiles
positively skewed distribution
probability distribution
random variable
regression line
sampling distribution
sampling error
standard normal distribution
symmetrical distribution
t-distribution
z-value
Choose a term from the vocabulary list above to complete each sentence.
1. The graph of a(n) _______________ is a bell-shaped curve that is symmetric about the mean of the data.
2. A(n) _______________ represents a numerical value assigned to an outcome of a probability experiment.
3. A(n) _______________ assesses evidence provided by data about a claim concerning a population parameter.
4. A(n) _______________ can be found when the maximum error of the estimate is added to and subtracted from the sample mean.
5. A(n) _______________ is used when the population standard deviation is unknown and n < 30.
6. In _______________, a sample of data is analyzed and conclusions are made about the entire population.
7. The _______________ represents the number of standard deviations that a given data value is from the mean in a normal distribution.
8. A(n) _______________ gives the type and strength of a linear relationship.
Define each term in your own words. 9. Central Limit Theorem
10. probability distribution
Pdf Pass
Chapter 11 46 Glencoe Precalculus
11 Chapter 11 Vocabulary Test
1.
2.
3.
4.
5.
6.
7.
8.
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Write the letter for the correct answer in the blank at the right of each question.
CARS The table shows the number of cars sold by 20 salespeople in one week.
Number of Cars
10 7 6 9 7 3 5 6 8 4
8 2 7 5 7 9 11 5 7 10
1. Which best describes the shape of the distribution?
A fairly symmetric C positively skewed B bimodal D negatively skewed
2. What is the median of the data? F 6.8 G 6 H 7 J 7.5
3. What is the mean of the data?
A 6.5 B 6.8 C 7 D 7.2
For Questions 4 and 5, use the probability distribution.
4. What is the mean of the distribution? F 1.58 G 3 H 3.34 J 13.66
5. What is the standard deviation of the distribution? A 1.58 B 2.50 C 3 D 3.34
6. INVESTMENT If an investment of $10,000 is successful, the investor makes $50,000. Otherwise, he or she loses everything. Which is the expected value if the probability of success is 40%?
F $4000 G $6000 H $10,000 J $14,000
7. What is z if X = 237, μ = 220, and σ = 12.3? A -4.55 B -1.38 C 1.38 D 4.55
8. LICENSES In a certain region, the ages of licensed drivers is normally distributed with a mean of 44.5 years and a standard deviation of 9.1 years. Find the probability that a randomly selected driver is younger than 25.
F 1.6% G 2.7% H 4.9% J 88.4%
9. If there are 42,000 drivers in the region described in Question 8, what is the approximate number of drivers younger than 25?
A 672 B 1134 C 2058 D 2625
10. In a normal distribution with μ = 120 and σ = 4, a random sample of 35 values is chosen. Find the probability that the sample mean is between 119 and 121.
F 11.2% G 40.1% H 50% J 86.1%
11. In a binomial distribution, n = 50, p = 0.20, and q = 0.80. Find P(X < 12).
A 70.2% B 76% C 79.8% D 81.2%
X 1 2 3 4 5 6
P(X) 0.15 0.20 0.18 0.22 0.13 0.12
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Pdf Pass
Chapter 11 47 Glencoe Precalculus
11 Chapter 11 Test, Form 1
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EXAMS A random sample of 1000 exams resulted in an average score of 500 points. Assume that the standard deviation is 80 points.
12. Find the maximum error of estimate for a 99% confidence level.
F 4.96 G 6.52 H 30.99 J 40.7
13. Find a 99% confidence interval for the mean population score.
A 499–501 B 495–505 C 479–520 D 493–507
14. For a maximum error of ±4 points and a 90% confidence level, what is the minimum number of samples that should be selected?
F 1008 G 1083 H 1537 J 2655
RAIN Ryan believes that the mean monthly rainfall in his town is more than 4 inches. He randomly selects monthly rainfall amounts, as shown below, to test his claim at α = 0.05. Use this data for Questions 15–17.
Monthly Average Rainfall (inches)
8.8 6.6 5.9 3.0 2.5 0.9 0.1 0.0 0.6 3.4 5.9 9.0
15. Identify the null and alternative hypotheses and the claim.
A H0: μ > 4 (claim); Ha: μ ≤ 4 C H0: μ < 4; Ha: μ ≥ 4 (claim)
B H0: μ ≥ 4 (claim); Ha: μ < 4 D H0: μ ≤ 4; Ha: μ > 4 (claim)
16. What is the value of the test statistic?
F -0.381 G -0.034 H -0.116 J -0.11
17. What is the critical value?
A -2.2 B -1.8 C -1.6 D -1.4
HEALTH For Questions 18–20, use the data given in the table.
18. What is the correlation coefficient?
F -0.983 G -0.923 H 0.923 J 0.983
19. At which level is the correlation coefficient significant for H0: ρ = 0?
A 1% B 5% C 10% D all of these
20. Find the least-squares regression line for the data.
F ^y = -0.81x + 440.03 H ^y = 0.81x + 440.03 G ^y = -0.81x - 440.03 J ^y = 0.81x - 440.03
Bonus In a binomial distribution, n = 15 and q = 0.55. Find P(X > 1).
NAME DATE PERIOD
Calories Sodium (mg) Calories Sodium (mg) Calories Sodium (mg)
130 320 120 315 180 300
160 340 128 335 45 410
60 395 90 370 150 310
145 322 80 385 100 355
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Pdf Pass
Chapter 11 48 Glencoe Precalculus
11 Chapter 11 Test, Form 1 (continued)
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Write the letter for the correct answer in the blank at the right of each question.
MOVIES The playing times for 20 movies are recorded in the table. Use the table for Questions 1–3.
Playing Time of Movies (minutes)
102 128 123 132 104 95 109 121 108 124
92 140 117 102 124 115 113 89 111 108
1. Which best describes the shape of the distribution?
A symmetric B bimodal C skewed D none of these
2. What is the standard deviation of the data?
F 5.7 G 10.8 H 13.5 J 20
3. Which is not true?
A IQR = 20.5 B P50 = 112 C Q1 = 89 D Q3 = 123.5
For Questions 4 and 5, use the probability distribution.
4. What is the mean of the distribution?
F 1.92 G 2.45 H 3.5 J 4.5
5. What is the standard deviation of the distribution? A 1.28 B 1.65 C 2.45 D 7.65
6. RAFFLE You buy a raffle ticket for $2. If your ticket is selected from the500 sold, you will win $750. What is the expected value of your net gain?
F -$0.75 G -$0.50 H -$0.25 J $0.50
7. What is X if z = 2.88, μ = 43, and σ = 5.2?
A -42 B 28 C 44 D 58
8. STRESS Recruits for a security position must take a stress test. The scores are normally distributed with a mean of 400 and a standard deviation of 100. What percent of the recruits scored higher than a recruit who scored 644?
F 0.73% G 7.3% H 49.3% J 99.3%
9. Out of 500 randomly selected recruits in Question 8, about how many scored higher than a recruit who scored 644?
A 4 B 37 C 247 D 497
10. In a normal distribution with μ = 88 and σ = 1.8, a random sample of 40 values is selected. Find the probability that the sample mean is between 87 and 87.5.
F 1.8% G 3.9% H 10.1% J 28.9%
11. In a binomial distribution, n = 35 and p = 0.20. Find P(X < 5).
A 14.5% B 20.0% C 26.3% D 32.8%
X 1 2 3 4 5 6
P(X) 0.21 0.46 0.13 0.10 0.07 0.03
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Chapter 11 49 Glencoe Precalculus
11 Chapter 11 Test, Form 2A
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TIRES A random sample of 30 similar tires found an average life span of 36,200 miles. Assume that the standard deviation is 3800 miles.
12. Find the maximum error of estimate for a 99% confidence level.
F 248.27 G 326.29 H 1359.81 J 1787.18
13. Find a 99% confidence interval for the mean population lifespan. A 34,413-37,987 C 35,506-36,894 B 34,840-37,560 D 35,952-36,448
14. For a maximum error of ±1500 miles and a 90% confidence level, what is the minimum number of samples that should be chosen?
F 10 G 15 H 18 J 26
WEATHER A travel agent believes that the mean monthly temperature of a resort town is at least 70°. She randomly chose the mean monthly temperatures as shown to test her claim at α = 0.05.
Mean Monthly Temperatures
75.0 54.5 61.7 68.7 50.7 81.3 84.0 52.7 79.0 70.0 59.9 84.0
15. Identify the null and alternative hypotheses and the claim. A H0: μ > 70 (claim); Ha: μ ≤ 70 C H0: μ < 70; Ha: μ ≥ 70 (claim)
B H0: μ ≥ 70 (claim); Ha: μ < 70 D H0: μ ≤ 70; Ha: μ > 70 (claim)
16. What is the value of the test statistic? F -5.7 G -1.54 H -0.43 J -0.13
17. What is the critical value? A -2.2 B -1.8 C -1.6 D -1.4
FOOD For Questions 18–20, use the data given in the table.
Calories Sugar (g) Calories Sugar (g) Calories Sugar (g)
80 1 120 2.2 150 5
160 7 145 5 100 2
128 2.5 90 2 85 1.2
130 3 180 10 60 0
18. What is the correlation coefficient? F -0.927 G -0.860 H 0.860 J 0.927
19. At which level is the correlation coefficient significant for H0: ρ = 0? A 1% B 5% C 10% D all of these
20. Find the least-squares regression line for the data.
F y^ = -0.07x + 5.26 H y^ = 0.07x + 5.26
G y^ = -0.07x - 5.26 J y^ = 0.07x - 5.26
Bonus In a standard normal distribution, if P(z > a) = 0.0865, what is a?
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Pdf Pass
Chapter 11 50 Glencoe Precalculus
11 Chapter 11 Test, Form 2A (continued)
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Write the letter for the correct answer in the blank at the right of each problem.
MUSIC The playing times of 20 songs on a radio station are recorded in the table. Use the table for Questions 1–3.
Playing Time (seconds)
212 204 174 187 292 225 159 189 214 206
215 257 243 232 310 299 187 240 187 176
1. Which best describes the shape of the distribution?
A fairly symmetric C positively skewed B bimodal D negatively skewed
2. What is the standard deviation of the data?
F 6.5 G 11 H 36 J 42.7
3. Which is not true?
A Q1 = 159 B P50 = 213 C IQR = 54.5 D Q3 = 241.5
For Questions 4 and 5, use the probability distribution below.
4. What is the mean of the distribution?
F 1.48 G 2.48 H 3.19 J 3.5
5. What is the standard deviation of the distribution?
A 1.34 B 1.80 C 2.24 D 3.25
6. RAFFLE You sold your friend a raffle ticket for $1. If his ticket is selected from the 1000 sold, he will win $500. What is the expected value of his net gain?
F -$0.99 G -$0.50 H -$0.10 J $0.05
7. What is X if z = 1.38, μ = 220, and σ = 12.3?
A -203 B -211 C 229 D 237
8. WRITING Applicants for an editorial position must take a grammar test. The scores are normally distributed with a mean of 150 and a standard deviation of 8. What percent of the applicants scored higher than an applicant who scored 163?
F 2.3% G 5.2% H 94.8% J 97.7%
9. Out of 200 randomly selected applicants in Question 8, about how manyscored higher than an applicant who scored 163?
A 10 B 14 C 20 D 24
10. In a normal distribution with μ = 112 and σ = 5.3, a random sample of 50 values is chosen. Find the probability that the sample mean is between 110 and 111.
F 0.4% G 8.8% H 9.1% J 40.9%
11. In a binomial distribution, n = 30 and p = 0.60. Find P(X > 16).
A 68.1% B 71.2% C 77.2% D 82.4%
X 1 2 3 4 5 6
P(X) 0.25 0.18 0.16 0.11 0.14 0.16
Pdf Pass
Chapter 11 51 Glencoe Precalculus
11 Chapter 11 Test, Form 2B
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DRINKS A random sample of 200 drink machines found the average amount dispensed to be 8.1 ounces. Assume that the standard deviation is 0.75 ounce. 12. Find the maximum error of estimate for a 99% confidence level.
F 0.087 G 0.104 H 0.137 J 1.23
13. Find a 99% confidence interval for the population mean. A 7.35-8.85 B 7.96-8.24 C 8.00-8.20 D 8.10-8.11
14. For a maximum error of ±0.15 ounce and a 90% confidence level, what is the minimum number of samples that should be chosen?
F 68 G 97 H 166 J 178
CARS Sasha believes that the mean number of cars sold per week by her staff is greater than 7. She randomly selected the sales for one week, as shown, to test her claim at α = 0.05.
Number of Cars
10 7 6 9 7 3 5 6 8 4
8 2 7 5 7 9 11 5 7 10
15. Identify the null and alternative hypotheses and the claim. A H0: μ > 7 (claim); Ha: μ ≤ 7 C H0: μ < 7; Ha: μ ≥ 7 (claim)
B H0: μ ≥ 7 (claim); Ha: μ < 7 D H0: μ ≤ 7; Ha: μ > 7 (claim)
16. What is the value of the test statistic? F -3.77 G -0.52 H -0.38 J -0.08
17. What is the critical value? A -2.3 B -2.1 C -1.9 D -1.7
MOVIES For Questions 18–20, use the data given in the table.
Weekly Income vs. Number of Movie Rentals
Income ($) Movies Income ($) Movies Income ($) Movies
500 9 900 7 1225 3
1150 4 650 8 850 8
400 10 950 7 1650 2
825 9 1800 1 1000 5
18. What is the correlation coefficient? F -0.950 G -0.903 H 0.903 J 0.950
19. At which level is the correlation coefficient significant for H0: ρ = 0? A 1% B 5% C 10% D all of these
20. Find the least-squares regression line for the data. F y^ = -0.007x + 12.8 H y^ = 0.007x + 12.8 G y^ = -0.007x - 12.8 J y^ = 0.007x - 12.8
Bonus In a standard normal distribution, if P(z > a) = 0.1118, what is a?
NAME DATE PERIOD
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Pdf Pass
Chapter 11 52 Glencoe Precalculus
11 Chapter 11 Test, Form 2B (continued)
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Chapter 11 53 Glencoe Precalculus
11
MOVIES Fifty randomly selected people going to a science-fiction movie were asked their ages. The results are recorded in the table. Use the table for Questions 1–3.
Ages of Science-Fiction Moviegoers (years)
17 42 21 78 16 21 31 29 29 16
49 19 81 16 69 69 18 31 22 14
21 75 42 78 18 41 22 16 18 80
42 42 42 16 16 21 19 18 44 18
22 14 49 17 16 18 18 18 17 23
1. Construct a histogram of the data.
2. Describe the shape of the distribution of the data.
3. Find the mean and median of the distribution.
For Exercises 4 and 5, use the probability distribution.
X P(X) X P(X)
1 0.02 6 0.02
2 0.08 7 0.30
3 0.16 8 0.24
4 0.03 9 0.07
5 0.05 10 0.03
4. Find the mean of the distribution.
5. Find the standard deviation of the distribution.
6. RAFFLE You sold your friend a raffle ticket for $5. If his ticket is selected from the 1000 sold, he will win $1000. What is the expected value of his net gain?
7. Find X if z = 2.15, μ = 185, and σ = 7.5.
8. TRANSPORTATION The average time it takes a trolley to travel between two particular stops is 7.2 minutes with a standard deviation of 1.5 minutes. The times are normally distributed. What is the probability that on any given trip the trolley takes more than 8 minutes to travel between the two stops?
9. From a sample of 200 trolley trips described in Question 8, about how many trips will be less than 7 minutes long?
10. In a normal distribution with μ = 58 and σ = 3, a random sample of 30 values is selected. Find the probability that the sample mean is between 58.5 and 59.
Chapter 11 Test, Form 2C
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Chapter 11 54 Glencoe Precalculus
11
11. In a binomial distribution, n = 35 and p = 0.45. Find P(X < 15).
CLOCKS For Questions 12–14, a random sample of 225 homes found an average of 5.2 clocks per home. Assume from past studies the standard deviation is 0.8. .
12. Find the maximum error of estimate for a 99% confidence level.
13. Find a 99% confidence interval for the mean number of clocks in all the homes.
14. For a maximum error of ±0.25 clock and a 90% confidence level, what is the minimum number of samples that should be taken?
HOCKEY Jabir believes that the mean save percentages for NHL goalies is greater than 0.910. He randomly selected the save percentages for some goalies in a recent season, as shown below to test his claim at α = 0.05. Use these data for Questions 15–17.
0.940 0.937 0.930 0.929 0.926 0.925 0.925 0.923 0.923 0.923
0.921 0.920 0.919 0.918 0.916 0.914 0.911 0.911 0.910 0.910
0.909 0.907 0.907 0.904 0.903 0.903 0.902 0.900 0.898 0.895
15. Identify the null and alternative hypotheses and the claim for Jabir’s hypothesis test.
16. What is the critical value and the test statistic?
17. Is there sufficient evidence to reject the null hypothesis? Explain.
FABRIC For Questions 18–20, use the data in the table. It gives the amount of each fabric sold at a craft store and its price.
Sq. Yd Sold Price/Sq. Yd ($) Sq. Yd Sold Price/Sq. Yd ($)
900 8.99 965 32.99
1000 15.99 855 38.99
975 21.99 750 40.99
950 25.99 740 42.99
950 28.99 732 45.99
960 30.99 600 55.99
18. Calculate the correlation coefficient.
19. Is the correlation coefficient significant at α = 0.05 for ρ = 0? Explain.
20. Find the least-squares regression line for the data.
Bonus In a standard normal distribution, if P(z > a) = 0.2115, what is a?
Chapter 11 Test, Form 2C (continued)
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Chapter 11 55 Glencoe Precalculus
11
TELEVISION Fifty students recorded the number of hours that the television was on during one week in their homes. The results are in the table. Use the table for Questions 1–3.
Weekly Television Hours (to the nearest hour)
54 28 9 15 3 54 35 32 0 34
72 57 62 33 58 23 57 53 24 27
36 63 3 58 53 13 12 75 66 57
18 53 53 46 77 26 32 42 43 88
44 71 22 57 45 73 44 11 45 34
1. Construct a histogram of the data.
2. Describe the shape of the distribution of the data.
3. Find the mean and median of the distribution.
For Questions 4 and 5, use the probability distribution.
X P(X) X P(X)
1 0.08 6 0.10
2 0.10 7 0.06
3 0.30 8 0.02
4 0.20 9 0.01
5 0.12 10 0.01
4. Find the mean of the distribution.
5. Find the standard deviation of the distribution.
6. RAFFLE You sold your friend a raffle ticket for $3. If his ticket is selected from the 600 sold, he will win $750. What is the expected value of his net gain?
7. Find X if z = 1.65, μ = 125, and σ = 5.5.
8. BIRD FEEDER The average number of birds that visit a bird feeder per day is normally distributed with a mean of 26.8 and a standard deviation of 4.7. What is the probability that on any given day the number of birds is greater than 30?
9. From a sample of 80 randomly selected days described in Question 8, about how many days will the number of birds be less than 20?
10. In a normal distribution with μ = 425 and σ = 24, a random sample of 40 values is chosen. Find the probability that the sample mean is between 420 and 428.
Chapter 11 Test, Form 2D
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Chapter 11 56 Glencoe Precalculus
11
11. In a binomial distribution, n = 42 and p = 0.72. Find P(X > 28).
FOOD For Questions 12–14, a random sample of256 people found that they ate fast food an average of2.6 times per week. Assume from past studies the standard deviation is 0.4.
12. Find the maximum error of estimate for a 99% confidence level.
13. Find a 99% confidence interval for the mean number of times people eat fast food each week.
14. For a maximum error of ±0.1 times and a 90% confidence level, what is the minimum sample that should be taken?
WORK A supervisor believes that the mean pay increase for her employees is 4.5%. She randomly selected the increases for 13 employees, as shown below, to test her claim at α = 0.05. Use these data for Questions 15–17.
Percent Pay Raise
3.2% 4.4% 4.1% 3.8% 1.5% 2.4% 4.6%
3.3% 1.7% 9.2% 4.5% 4.2% 5.1%
15. Identify the null and alternative hypotheses and the claim for her hypothesis test.
16. What is the critical value and the test statistic?
17. Is there sufficient evidence to reject the null hypothesis? Explain.
BUDGET For Questions 18–20, use the data in the table.
Cost of Gas vs. Number of Meals Out of House
Cost of gas
($ per gal)
Number of
Meals Out (wk)
Cost of gas
($ per gal)
Number of
Meals Out (wk)
1.79 6 2.45 3
1.84 5 2.51 2
2.20 6 3.15 2
2.28 5 3.89 1
2.34 4 4.05 1
2.39 3 4.10 0
18. Calculate the correlation coefficient.
19. Is the correlation coefficient significant at α = 0.05for ρ = 0? Explain.
20. Find the least-squares regression line for the data.
Bonus In a standard normal distribution, if P(z > a) = 0.0794, what is a?
Chapter 11 Test, Form 2D (continued)
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Chapter 11 57 Glencoe Precalculus
11 Chapter 11 Test, Form 3
SPEEDS The speeds of 50 randomly selected cars crossing a state border are recorded in the tables. Use the tables for Questions 1–3.
Massachusetts Speeds
(miles per hour)
New Hampshire Speeds
(miles per hour)
66 69 70 67 74 88 71 75 68 73
69 72 67 69 68 78 68 58 69 77
72 65 72 59 63 73 72 65 76 77
58 63 73 66 77 76 82 72 73 81
64 71 72 67 65 73 64 71 65 84
1. Construct side-by-side box plots of the data.
2. Compare the distributions.
3. Describe the shape of the Massachusetts distribution.
For Questions 4 and 5, use the probability distribution.
X P(X) X P(X)
5 0.06 55 0.24
15 0.22 65 0.23
25 0.26 75 0.18
35 0.24 85 0.23
45 0.22 95 0.12
4. Find the mean of the distribution.
5. Find the variance and standard deviation of the distribution.
6. FIRE INSURANCE A company sells a one-year home fire insurance policy to a customer for $765. The probability that no claim for fire will be made is 0.996. If there is a fire, assume a pay-out to the customer for $165,000. What is the company’s expected profit?
7. Find X if z = -2.15, μ = 16.9, and σ = 1.85.
8. FURNITURE A decorator finds that retail costs of sofas are normally distributed with a mean of $950 and a standard deviation of $215. What is the probability that a randomly selected sofa costs between $700 and $800?
9. From a sample of 50 randomly selected sofas described in Question 8, how many will cost more than $1250?
10. In a normal distribution with μ = 102 and σ = 8.6, a random sample of 42 values is selected. Find the probability that the sample mean is between 101.5 and 103.
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Chapter 11 58 Glencoe Precalculus
11
11. In a binomial distribution, n = 22 and q = 0.178. Find P(X ≥ 19).
BATTERIES For Questions 12–14, a random sample of 85 batteries found a mean battery life of 450 minutes. Assume from past studies the standard deviation is 18.4 minutes.
12. Find the maximum error of estimate for a 99% confidence level.
13. Find the 99% confidence interval for the mean battery life of all the batteries.
14. For a maximum error of ±6 minutes and a 90% confidence level, what is the minimum number of samples to be taken?
LAW ENFORCEMENT Myra believes that the average number of traffic tickets issued per day in her town is no more than 10. She finds the number of traffic tickets issued on 20 random days, as shown below, to test her claim at a = 0.05. Use these data for Questions 15–17.
7 12 10 8 7 12 15 10 10 7
14 6 10 12 9 8 8 2 9 12
15. Identify the null and alternative hypotheses and the claim for Myra’s hypothesis test.
16. State the critical value, test statistic, and P-value.
17. Is there sufficient evidence to reject the null hypothesis? Explain.For Questions 18–20, use the data in the table.
Amount of Garbage Produced by Selected Households
Garbage
(lb)
Number in
HouseholdGarbage (lb)
Number in
Household
35.6 4 38.1 6
49.0 5 21.8 1
27.6 3 28.0 4
21.9 2 33.3 6
10.8 2 20.0 3
18. Calculate and interpret the correlation coefficient.
19. Determine whether the correlation coefficient is significant at the 1%, 5%, and 10% levels. Explain.
20. Use the least-squares regression line for the data to predict the size of a household that produces 60 pounds of garbage.
Bonus In a standard normal distribution, if P(-a < z < a) = 0.4775, what is a?
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Chapter 11 Test, Form 3 (continued)
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Chapter 11 59 Glencoe Precalculus
11 Extended-Response Test
Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solutions in more than one way or investigate beyond the requirements of the problem.
1. The table shows the weights of 37 dogs in a park’s large-dog play area.
Weights of Large Dogs (lb)
78 76 88 68 87 78 84 83 71 8285 91 86 92 80 93 98 73 93 93
96 80 94 64 97 91 84 100 89 81
86 74 87 97 90 91 101
a. Construct a histogram and use it to describe the shape of the distribution.
b. Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice.
2. The combined test scores for all of the advanced mathematics classes in a school are normally distributed. The mean score is 85 and the standard deviation is 6. There are 200 students in the classes.
a. Those who had scores above 94 were given a grade of A. How many students received an A? Explain your reasoning.
b. Students who score in the top 30% are eligible to participate in a regional competition. What is the minimum score required for eligibility? Explain.
c. If a random sample of 35 students are chosen, what is the probability that the mean of the sample is between 83 and 84?
3. Company executives report that their cereal boxes contain 14 ounces of cereal. A journalist weighs a random sample of 15 boxes and finds the mean weight to be 13.9 ounces with a standard deviation of 0.17 ounce. Use a 5% level of significance to test the executives’ claim.
4. The table shows the volume and cost of dried backpacking food.
Volume (oz) 6 3.75 21 7.13 1.4 3.5 4.8 4.6 2.1 5 3
Cost ($) 5.5 4.5 14.99 6.5 3 3.25 5.99 5.99 3.99 7.5 3.99
a. Calculate and interpret the correlation coefficient.
b. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels.
c. Find the least-squares regression equation. Interpret the slope and intercept in context.
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NAME DATE PERIOD
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Pdf Pass
Chapter 11 60 Glencoe Precalculus
11
1. Which best describes the distribution of the data in the table?
790 750 840 617 687 700
599 545 527 840 905 813
A fairly symmetric C positively skewed
B bimodal D negatively skewed 1.
2. The cost of renting a washer and dryer is $30 for one month, $55 for two months, and $20 per month for more than two months, up to one year. Which function represents the situation, where m represents the number of months?
F c(m) =
⎧
⎨
⎩
30 if 0 < m ≤ 1
55 if 1 < m ≤ 2
20m if 2 < m ≤ 12
H c(m) =
⎧
⎨
⎩
30 if 0 < m < 1
55 if 1 ≤ m < 2
20m if 2 ≤ m
G c(m) = 20m + 30 J c(m) = 30(m + 1) + 55(m + 2) + 20(m + 3) 2.
3. Find the distance between (
2, π
−
6 )
and (
5, 2π
−
3 )
.
A 0 B 3.39 C 7.68 D 9.43 3.
4. Use a half-angle identity to find the exact value of sin 67.5°.
F √
2 + √
2 −
2 G
√
2 - √
2 −
2 H -
√
2 + √
2 −
2 J -
√
2 - √
2 −
2 4.
5. Which is not a possible rational root of 2x3 - 9x2 - 11x + 8 = 0?
A -2 B -
1 −
2 C 1 −
4 D 8 5.
6. Find the partial fraction decomposition of 6x - 14 −
x2 -2x - 15 .
F 2 −
x + 3 + 4 −
x - 5 G 2 −
x + 3 + -4 −
x - 5 H 4 −
x + 3 + 2 −
x - 5 J 4 −
x + 3 + -2 −
x - 5 6.
7. What is -4 + 3i in polar form?
A 5(cos 53.13 - i sin 53.13) C 25(cos 53.13 + i sin 53.13)
B 25(cos 36.89 + i sin 36.89) D 5(-cos 36.89 + i sin 36.89) 7.
8. What is the coefficient of the third term in the expansion of (2x - y)5?
F -80 G 32 H 40 J 80 8.
9. Solve 19e4x = 95. A 0.19 B 0.25 C 0.40 D 1.16 9.
Part 1: Multiple Choice
Instructions: Fill in the appropriate circle for the best answer.
Standardized Test Practice(Chapters 1–11)
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
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NAME DATE PERIOD
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Chapter 11 61 Glencoe Precalculus
11
10. Which statement is an identity?
F tan θ −
4
= tan
θ −
4
H sin 2θ = 2 sin θ cos θ
G csc (
θ - π −
2 )
= sec θ J 3 tan2 θ - 1 = sec2 θ 10.
11. Find the angle between u = ⟨4, -3⟩ and v = ⟨1, 2⟩.
A 78.7° B 79.7° C 100.3° D 101.3° 11.
12. Which is the inverse of f(x) = x2 - 3? What are the restrictions on the domain?
F f -1(x) = ± √
�
x + 3, x ≠ 0 H f -1(x) = ± √
�
x - 3, x ≠ 0
G f -1(x) = ± √
���
x - 3 , x ≥ 3 J f -1(x) = ± √ ���
x + 3 , x ≥ -3 12.
13. Expand log 5x8y-2.
A log 5 + 8 log x - 2 log y C 5 log x + 8 log x - 2 log y
B log 1 −
5 + log 8 x - log 2 y D 1 −
5 log x + log 8x - log 2y 13.
14. Given f(x) = 2x2 and g(x) = 3x, find (g ◦ f)x.
F (g ◦ f)x = 6x2 G (g ◦ f)x = 9x2 H (g ◦ f)x = 18x2 J (g ◦ f)x = 8x4 14.
15. Find the exact value of cos (-210°).
A -
√
�
3 −
3 B -
√
�
3 −
2 C
√
�
3 −
2 D
√
�
3 −
3 15.
16. A car salesman is testing the gas mileage of cars in his lot. He knows from previous tests that the standard deviation is 4 miles per gallon. If he wants results that are accurate to within ±3 miles per gallon, with a 95% confidence level, what is the minimum number of cars he must test?
F 5 G 7 H 12 J 15 16.
17. What is the mean of the probability distribution?
X 1 2 3 4
P(X) 0.4 0.25 0.15 0.2
A 2.15 B 2.5 C 2.75 D 2.85 17.
18. What is the sum of the infinite geometric series 18 + 6 + 2 + … ?
F 27 G 36 H 45 J 54 18.
Standardized Test Practice (continued)
(Chapters 1–11)
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
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NAME DATE PERIOD
Pdf Pass
Chapter 11 62 Glencoe Precalculus
11
19. Find three arithmetic means between 16 and 52.
20. State the amplitude, period, frequency, phase shift,
and vertical shift of y = -sin (
x + π −
2 )
+ 2.
21. Find the dot product of u = ⟨4, -5⟩ and v = ⟨10, 8 ⟩. Then determine if u and v are orthogonal.
22. Identify, the vertex, focus, axis of symmetry, and directrix of the graph of y2 - 12x + 2y = -37.
23. Find all solutions to tan2 θ + tan θ = 0 on the interval [0, 2π].
24. Find the inverse matrix required
to solve ⎡
⎢
⎣
12 5 7 3 ⎤
⎦
·
⎡
⎢
⎣
x1 x2
y1 y2
⎤
⎦
= ⎡
⎢
⎣
2 3
-1
2
⎤
⎦
.
25. An engineer has developed a new microprocessing chip that he believes has a longer usable life than his company’s existing chip. The existing chip has a usable life of 20,000 hours with a standard deviation of 3200 hours. The engineer has found that in a randomly selected sample of 1000 new chips, the mean usable life is 20,500 hours. He wishes to test his claim at α = 0.01.
a. Identify the hypotheses and the claim.
b. Find the critical value.
c. Find the test statistic.
d. Is there sufficient evidence to show the new chip has a longer life? Justify your answer.
Part 2: Short Response
Instructions: Write your answers in the space provided.
19.
20.
21.
22.
23.
24.
25a.
25b.
25c.
25d.
Standardized Test Practice (continued)
(Chapters 1–11)
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Chapter 11 A1 Glencoe Precalculus
An
swer
s
Answers (Anticipation Guide and Lesson 11-1)
Pdf Pass
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Lesson 11-1
Ch
ap
ter
11
5
Gle
ncoe
Pre
calc
ulus
Stud
y Gu
ide
and
Inte
rven
tion
Desc
rip
tive
Sta
tist
ics
11-1
Des
crib
ing
Dis
trib
utio
ns W
hen
a da
ta s
et h
as a
sym
met
rica
l dis
trib
utio
n, t
he
data
are
eve
nly
dist
ribu
ted
on b
oth
side
s of
the
mea
n. I
n a
nega
tive
ly s
kew
ed
dist
ribu
tion
, the
left
sid
e of
the
dis
trib
utio
n ex
tend
s fa
rthe
r th
an t
he r
ight
and
the
mea
n is
less
tha
n th
e m
edia
n. I
n a
posi
tive
ly s
kew
ed d
istr
ibut
ion,
the
rig
ht s
ide
exte
nds
fart
her
than
the
left
and
the
mea
n is
gre
ater
tha
n th
e m
edia
n. I
f a d
istr
ibut
ion
is
reas
onab
ly s
ymm
etri
c, t
he m
ean
and
stan
dard
dev
iati
on c
an b
e us
ed t
o de
scri
be it
. O
ther
wis
e, t
he fi
ve-n
umbe
r su
mm
ary
is b
ette
r.
T
he t
able
sho
ws
the
num
ber
of p
asse
nger
s w
ho b
oard
ed
plan
es a
t 36
air
port
s in
the
Uni
ted
Stat
es in
one
yea
r.N
um
be
r o
f P
as
se
ng
ers
30
,52
63
0,3
72
26
,62
32
2,7
22
16
,28
71
5,2
46
14
,80
71
4,1
17
14
,05
4
13
,54
71
2,9
16
12
,61
61
1,9
06
11
,62
21
1,4
89
10
,82
81
0,6
53
10
,00
8
97
03
95
94
94
63
93
48
91
25
85
72
73
00
67
72
65
49
61
26
59
07
57
12
52
87
48
48
48
32
48
20
47
50
46
84
a. C
onst
ruct
a h
isto
gram
and
use
it t
o de
scri
be t
he s
hape
of
the
dist
ribu
tion
.
Ent
er t
he d
ata
in L
1, t
urn
on P
lot1
, and
sel
ect
hist
ogra
m.
Pres
s G
RA
PH
. Pre
ss
ZO
OM
and
cho
ose
Zoom
Stat
.Pr
ess
TR
AC
E t
o se
e ho
w m
any
data
val
ue a
re in
eac
h ba
r.
The
grap
h is
pos
itiv
ely
skew
ed.
b. S
umm
ariz
e th
e ce
nter
and
spr
ead
of t
he d
ata
usin
g ei
ther
the
mea
n an
d st
anda
rd d
evia
tion
or
the
five
-num
ber
sum
mar
y. J
usti
fy y
our
choi
ce.
Use
the
five
-num
ber
sum
mar
y si
nce
the
data
are
ske
wed
.
The
num
ber
of p
asse
nger
s ra
nge
from
468
4 to
30,
526,
and
the
med
ian
num
ber
of p
asse
nger
s is
abo
ut 9
856.
Hal
f of t
heai
rpor
ts h
ad b
etw
een
6338
and
13,
800
pass
enge
rs.
Exer
cise
1. A
res
taur
ant
man
ager
adv
erti
ses
a 20
-oun
ce s
irlo
in s
teak
on
his
men
u.Th
e w
eigh
ts o
f a s
ampl
e of
the
ste
aks
is s
how
n.
S
tea
k W
eig
hts
(o
un
ce
s)
20
20
20
18
20
18
18
19
19
19
21
20
17
20
19
21
18
19
20
20
a. C
onst
ruct
a h
isto
gram
and
use
it t
o de
scri
be t
he s
hape
of t
he d
istr
ibut
ion.
Th
e g
rap
h i
s n
eg
ati
ve
ly s
ke
we
d.
b. S
umm
ariz
e th
e ce
nter
and
spr
ead
of t
he d
ata
usin
gei
ther
the
mea
n an
d st
anda
rd d
evia
tion
or
the
five-
num
ber
sum
mar
y. J
usti
fy y
our
choi
ce.
Be
ca
us
e t
he
dis
trib
uti
on
is
sk
ew
ed
, u
se
th
e f
ive
-nu
mb
er
su
mm
ary
. T
he
we
igh
ts r
an
ge
fro
m 1
7 o
un
ce
s t
o
21
ou
nc
es
an
d t
he
me
dia
n w
eig
ht
is 1
9.5
ou
nc
es
. H
alf
th
e w
eig
hts
a
re b
etw
ee
n 1
8.5
ou
nc
es
an
d 2
0 o
un
ce
s.
Exam
ple
[ 17,
22]
scl:
1 b
y [ 0
, 10]
scl:
1
[ 400
0, 3
5,00
0] s
cl: 3
000
by [ 0
, 15]
scl:
1
005_
042_
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MC
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Chapter Resources
A A A A A D DD A A D A
11
B
efor
e yo
u be
gin
Cha
pter
11
•
Rea
d ea
ch s
tate
men
t.
•
Dec
ide
whe
ther
you
Agr
ee (A
) or
Dis
agre
e (D
) wit
h th
e st
atem
ent.
•
Wri
te A
or
D in
the
firs
t co
lum
n O
R if
you
are
not
sur
e w
heth
er y
ou a
gree
or
disa
gree
, wri
te N
S (N
ot S
ure)
.
A
fter
you
com
plet
e C
hapt
er 1
1
•
Rer
ead
each
sta
tem
ent
and
com
plet
e th
e la
st c
olum
n by
ent
erin
g an
A o
r a
D.
•
Did
any
of y
our
opin
ions
abo
ut t
he s
tate
men
ts c
hang
e fr
om t
he fi
rst
colu
mn?
•
For
thos
e st
atem
ents
tha
t yo
u m
ark
wit
h a
D, u
se a
pie
ce o
f pap
er t
o w
rite
an
exam
ple
of w
hy y
ou d
isag
ree.
ST
EP
1
A,
D,
or
NS
Sta
tem
en
tS
TE
P 2
A
or
D
1.
In a
neg
ativ
ely
skew
ed d
istr
ibut
ion,
the
mea
n is
less
tha
n th
e m
edia
n. 2
. Pe
rcen
tile
s di
vide
a d
istr
ibut
ion
into
100
equ
al g
roup
s.
3.
A c
onti
nuou
s ra
ndom
var
iabl
e ta
kes
on a
cou
ntab
le n
umbe
r of
pos
sibl
e va
lues
. 4
. In
a p
roba
bilit
y di
stri
buti
on, t
he s
um o
f P(X
) mus
t be
1.
5.
In a
nor
mal
dis
trib
utio
n, t
he m
ean,
med
ian,
and
mod
e ar
e eq
ual a
nd a
re lo
cate
d at
the
cen
ter
of t
he d
istr
ibut
ion.
6.
A z
-val
ue r
epre
sent
s th
e nu
mbe
r of
sta
ndar
d de
viat
ions
tha
t a
give
n da
ta v
alue
is fr
om t
he m
ean.
7.
If t
he s
ampl
e si
ze is
ade
quat
ely
larg
e, t
he d
istr
ibut
ion
of t
he
sam
ple
mea
ns w
ill a
hav
e a
mea
n an
d st
anda
rd d
evia
tion
eq
ual t
o th
e po
pula
tion
mea
n an
d st
anda
rd d
evia
tion
. 8
. A
s th
e pr
obab
ility
of s
ucce
ss in
crea
ses
to 0
.5, t
he s
hape
of t
he
bino
mia
l dis
trib
utio
n be
gins
to
rese
mbl
e th
e no
rmal
di
stri
buti
on.
9.
A c
onfid
ence
leve
l giv
es t
he p
roba
bilit
y th
at t
he in
terv
al
esti
mat
e w
ill in
clud
e a
give
n pa
ram
eter
.10
. Th
e nu
ll hy
poth
esis
sta
tes
that
the
re is
a d
iffer
ence
bet
wee
n th
e sa
mpl
e va
lue
and
the
popu
lati
on p
aram
eter
. 11
. E
xtra
pola
tion
use
s an
equ
atio
n to
mak
e pr
edic
tion
s ov
er t
he
rang
e of
the
dat
a.12
. A
cor
rela
tion
coe
ffici
ent
of -
0.95
sho
ws
a st
rong
neg
ativ
e co
rrel
atio
n be
twee
n tw
o va
riab
les.
Antic
ipat
ion
Guid
eIn
fere
nti
al
Sta
tist
ics
Step
2
Step
1
Ch
ap
ter
11
3
Gle
ncoe
Pre
calc
ulus
0ii_
004_
PC
CR
MC
11_8
9381
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ddS
ec1:
33/
25/0
912
:40:
05P
M
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e/M
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-Hill, a
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isio
n o
f Th
e M
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-Hill C
om
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ies, In
c.
Chapter 11 A2 Glencoe Precalculus
Answers (Lesson 11-1)
Pdf Pass
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Ch
ap
ter
11
6
Gle
ncoe
Pre
calc
ulus
11-1
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Desc
rip
tive
Sta
tist
ics
Mea
sure
s of
Pos
itio
n Th
e qu
arti
les
give
n by
the
five
-num
ber
sum
mar
y sp
ecify
the
pos
itio
ns o
f dat
a va
lues
wit
hin
a di
stri
buti
on. F
or t
his
reas
on, b
ox p
lots
are
mos
t us
eful
for
side
-by-
side
com
pari
sons
of t
wo
or m
ore
dist
ribu
tion
s.
T
he a
ges
of O
scar
-win
ning
act
ress
es f
rom
tw
o 20
-yea
r pe
riod
s ar
e sh
own.
Con
stru
ct s
ide-
by-s
ide
box
plot
s of
the
dat
a se
ts.
The
n us
e th
is d
ispl
ay t
o co
mpa
re t
he d
istr
ibut
ions
.
19
69
–1
98
8
61
35
34
34
26
37
42
41
35
31
41
33
30
74
33
49
38
61
21
41
19
89
–2
00
8
26
80
42
29
33
36
45
49
39
34
26
25
33
35
35
28
30
29
61
32
Inpu
t th
e da
ta in
to L
1 an
d L
2. T
urn
on P
lot1
and
Plo
t2 a
ndch
oose
a b
ox p
lot
wit
h ou
tlie
rs s
how
n as
the
typ
e of
gra
ph.
From
the
gra
ph, w
e ca
n se
e th
at t
he m
edia
n ag
e fo
r th
e fir
st20
yea
rs is
just
slig
htly
hig
her
than
the
med
ian
age
for
the
seco
nd20
yea
rs. T
he r
ange
of a
ges
for
the
mid
dle
50%
of t
he d
ata
sets
is g
reat
er fo
r th
e ac
tres
ses
from
the
sec
ond
20-y
ear
peri
od.
Bot
h da
ta s
ets
have
out
liers
gre
ater
tha
n th
e re
st o
f the
dat
a. I
gnor
ing
the
outl
iers
, the
dis
trib
utio
n fo
r th
e se
cond
20-
year
per
iod
is m
ore
sym
met
ric
than
the
firs
t an
d th
e ra
nge
of d
ata
for
the
seco
nd is
less
tha
n th
at o
f the
fir
st.
Exer
cise
The
num
ber
of c
redi
t ca
rds
owne
d by
stu
dent
s in
a s
tati
stic
scl
ass
is s
how
n. C
onst
ruct
sid
e-by
-sid
e bo
x pl
ots
of t
he d
ata
sets
.Th
en u
se t
his
disp
lay
to c
ompa
re t
he d
istr
ibut
ions
.
Mo
re t
ha
n o
ne
-fo
urt
h o
f th
e s
tud
en
ts i
n e
ac
h g
rou
p h
av
e n
o c
red
it c
ard
s.
Th
e m
ed
ian
nu
mb
er
of
cre
dit
ca
rds
fo
r th
e f
em
ale
s i
s h
igh
er
tha
n f
or
ma
les
. T
he
ra
ng
e a
nd
ma
xim
um
nu
mb
er
of
cre
dit
ca
rds
is
sli
gh
tly
hig
he
r fo
r th
e
fem
ale
s t
ha
n t
he
ma
les
.
Exam
ple
[ 15,
85]
scl:
5 b
y [ 0
, 0.5
] scl:
0.1
25
[-1,
10]
scl:
1 b
y [ 0
, 1] s
cl: 0
.5
Fe
ma
les
18
01
5
44
05
0
00
30
1
Ma
les
04
00
2
05
70
0
35
00
1
005_
042_
PC
CR
MC
11_8
9381
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:12
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 11-1
Ch
ap
ter
11
7
Gle
ncoe
Pre
calc
ulus
11-1
Prac
tice
Desc
rip
tive
Sta
tist
ics
1. W
EATH
ER T
he a
vera
ge w
ind
spee
ds r
ecor
ded
at v
ario
us w
eath
er s
tati
ons
in t
he U
nite
d St
ates
are
list
ed b
elow
.
Sta
tio
nS
pe
ed
(mp
h)
Sta
tio
nS
pe
ed
(mp
h)
Sta
tio
nS
pe
ed
(mp
h)
Alb
uq
ue
rqu
e8
.9A
nch
ora
ge
*7
.1A
tlan
ta*
9.1
Ba
ltim
ore
*9
.1B
ost
on
*1
2.5
Ch
ica
go
10
.4
Da
llas–
Ft.
Wo
rth
10
.8H
on
olu
lu*
11
.3In
dia
na
po
lis9
.6
Ka
nsa
s C
ity1
0.7
La
s V
eg
as
9.3
Litt
le R
ock
7.8
Lo
s A
ng
ele
s*6
.2M
em
ph
is8
.8M
iam
i*9
.2
Min
ne
ap
olis
– S
t. P
au
l1
0.5
Ne
w O
rle
an
s8
.1N
ew
Yo
rk C
ity*
9.4
Ph
ilad
elp
hia
*9
.5P
ho
en
ix6
.2S
ea
ttle
*9
.0
Sour
ce: N
atio
nal C
limat
ic D
ata
Cent
er
a. C
onst
ruct
a h
isto
gram
and
use
it t
o de
scri
be t
he s
hape
of t
he d
istr
ibut
ion.
Th
e d
istr
ibu
tio
n i
s s
ing
le-p
ea
ke
d a
nd
ro
ug
hly
sy
mm
etr
ic.
b. S
umm
ariz
e th
e ce
nter
and
spr
ead
of t
he d
ata
usin
g ei
ther
the
mea
n an
d st
anda
rd d
evia
tion
or
the
five-
num
ber
sum
mar
y. J
usti
fy y
our
choi
ce.
Be
ca
us
e t
he
da
ta a
re s
ym
me
tric
, u
se
th
e
me
an
an
d s
tan
da
rd d
ev
iati
on
. T
he
me
an
win
d s
pe
ed
is
9.2
1 m
ph
an
d
the
sta
nd
ard
de
via
tio
n i
s 1
.53
mp
h.
2. O
CEA
NS
The
ten
wea
ther
sta
tion
s w
ith
an a
ster
isk
have
rela
tive
ly c
lose
pro
xim
ity
to e
ithe
r th
e A
tlan
tic
Oce
an o
rPa
cific
Oce
an. C
onst
ruct
sid
e-by
-sid
e bo
x pl
ots
of t
he d
ata
sets
. The
n us
e th
is d
ispl
ay t
o co
mpa
re t
he d
istr
ibut
ions
.T
he
me
dia
n w
ind
sp
ee
d f
or
sta
tio
ns
ne
ar
an
oc
ea
n i
s a
bo
ut
the
sa
me
as
th
e o
the
r s
tati
on
s.
Th
e m
idd
le h
alf
of
the
sp
ee
ds
fo
r th
os
e n
ea
r a
n o
ce
an
va
rie
s l
es
s t
ha
n
the
mid
dle
ha
lf o
f th
e s
pe
ed
s a
t th
e o
the
r s
tati
on
s.
3. S
CORE
S Th
e ta
ble
give
s th
e fr
eque
ncy
dist
ribu
tion
of t
hesc
ores
on
a te
st.
a. C
onst
ruct
a p
erce
ntile
grap
h of
the
dat
a.
b. E
stim
ate
the
perc
enti
le r
ank
a sc
ore
of 6
2 w
ould
hav
e in
thi
sdi
stri
buti
on. I
nter
pret
its
mea
ning
. 2
5th
pe
rce
nti
le;
A s
tud
en
t w
ith
a s
co
reo
f 6
2 s
co
red
be
tte
r th
an
ab
ou
t 2
5%
of
the
stu
de
nts
wh
o t
oo
k t
he
te
st.
[ 6, 1
4] s
cl: 1
by
[ 0, 1
0] s
cl: 1
[ 5, 1
4] s
cl: 1
by
[ 0, 1
] scl:
0.5
10 0203040
Cumulative Percentage
5060708090100
Scor
e50
6070
8090
100
Cla
ss
Bo
un
da
rie
sf
55
.5–
60
.53
60
.5–
65
.58
65
.5–
70
.51
2
70
.5–
75
.55
75
.5–
80
.59
005_
042_
PC
CR
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11_8
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:51
PM
A01_A19_PCCRMC11_893812.indd 2A01_A19_PCCRMC11_893812.indd 2 11/19/09 8:00:21 PM11/19/09 8:00:21 PM
Copyright
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 A3 Glencoe Precalculus
Answers (Lesson 11-1)
An
swer
s
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
8
Gle
ncoe
Pre
calc
ulus
11-1
Wor
d Pr
oble
m P
ract
ice
Desc
rip
tive
Sta
tist
ics
1. S
HO
ES A
sho
e st
ore
empl
oyee
des
igns
a
disp
lay
by p
laci
ng s
hoe
boxe
s in
ten
st
acks
. The
num
ber
of b
oxes
in e
ach
stac
k ar
e 5,
7, 9
, 11,
13,
10,
9, 8
, 7, a
nd 5
.
a. C
onst
ruct
a b
ox p
lot
and
use
it t
o de
scri
be t
he s
hape
of t
he d
istr
ibut
ion.
[ 4
, 14]
scl:
1 b
y [ 0
, 1] s
cl: 0
.5
b. S
umm
ariz
e th
e ce
nter
and
spr
ead
of
the
data
usi
ng e
ithe
r th
e m
ean
and
stan
dard
dev
iati
on o
r th
e fiv
e-nu
mbe
r su
mm
ary.
Jus
tify
you
r ch
oice
.
th
e m
ea
n a
nd
sta
nd
ard
d
ev
iati
on
; m
ea
n:
8.4
, s
tan
da
rd
de
via
tio
n:
ab
ou
t 2
.5
2. M
EDIC
INE
A h
isto
gram
for
the
num
ber
of p
atie
nts
trea
ted
at 5
0 U
.S. c
ance
r ce
nter
s in
one
yea
r is
sho
wn.
Sour
ce: U
.S. N
ews
Onl
ine
a. W
hich
is g
reat
er, t
he m
ean
or t
he
med
ian
of t
he d
ata
set?
Exp
lain
.m
ean
; T
he
dat
a a
re p
osi
tive
ly
skew
ed.
b. S
ketc
h th
e ge
nera
l sha
pe o
f a b
ox p
lot
that
wou
ld r
epre
sent
the
his
togr
am.
Sa
mp
le a
ns
we
r:
3. E
XA
MS
The
tabl
e gi
ves
the
freq
uenc
ies
of t
he fi
nal e
xam
sco
res
of 5
0 st
uden
ts in
tw
o pr
ecal
culu
s cl
asse
s.
Cla
ss
Bo
un
da
rie
sF
req
ue
nc
y
f
42
.5–
52
.54
52
.5–
62
.51
0
62
.5–
72
.51
5
72
.5–
82
.51
3
82
.5–
92
.57
92
.5–
10
2.5
1
a. C
onst
ruct
a p
erce
ntile
gra
ph o
fth
e da
ta. 10 0203040 Cumulative Percentage 506070809010
0
Scor
e50
4060
7080
9010
0
b. E
stim
ate
the
perc
enti
le r
ank
a te
st
scor
e of
77
wou
ld h
ave
in t
his
dist
ribu
tion
. Int
erpr
et it
s m
eani
ng.
90
th p
erc
en
tile
; A
stu
de
nt
wit
h
a s
co
re o
f 7
7 h
as
a b
ett
er
sc
ore
th
an
ab
ou
t 9
0%
of
the
s
tud
en
ts t
ak
ing
th
e e
xa
m.
4. M
OV
IES
The
ages
of m
ovie
pat
rons
ina
thea
ter
are
25, 4
7, 1
6, 4
5, 5
4, 1
7, 1
4,
16, 1
6, 3
9, 4
8, 4
8, 1
8, 1
2, 1
3, 6
2, 5
1, 4
6,
and
18. S
umm
ariz
e th
e di
stri
buti
on o
f th
e da
ta.
bim
od
al;
tw
o c
lus
ters
th
at
are
e
ac
h r
ou
gh
ly s
ym
me
tric
Patie
nts i
n U.
S. C
ance
r Cen
ters
2030 10 0
Number or Cancer Centers
40
Num
ber o
f Pat
ient
s
3500
500
1000
1500
2000
2500
3000
Th
e d
ata
are
ro
ug
hly
s
ym
me
tric
.
500
1000
1500
2000
2500
3000
3500
005_
042_
PC
CR
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11_8
9381
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11/1
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12:0
1:18
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 11-1
Ch
ap
ter
11
9
Gle
ncoe
Pre
calc
ulus
11-1
Enri
chm
ent
Mean
s
The
aver
age
of a
dat
a se
t is
kno
wn
as t
he a
rith
met
ic m
ean.
The
re a
re o
ther
m
eans
.
A t
rim
med
mea
n is
the
ari
thm
etic
mea
n of
a d
ata
set
afte
r th
e to
p 10
% a
nd
bott
om 1
0% o
f the
dat
a va
lues
are
tri
mm
ed o
ff. T
here
fore
, if t
here
are
100
dat
a va
lues
, the
leas
t 10
val
ues
and
grea
test
10
valu
es a
re r
emov
ed.
A g
eom
etri
c m
ean
is t
he n
th r
oot
of t
he p
rodu
ct o
f n d
ata
valu
es. I
t is
oft
en
used
in e
cono
mic
s to
find
an
aver
age
rate
of g
row
th.
A h
arm
onic
mea
n H
is t
he n
umbe
r of
dat
a va
lues
div
ided
by
the
sum
of t
he
mul
tipl
icat
ive
inve
rses
of a
ll th
e da
ta v
alue
s. I
t is
oft
en u
sed
in a
vera
ging
spe
eds.
It
can
not
be u
sed
if 0
is a
dat
a va
lue.
H
=
n −
∑
1 −
x , whe
re n
is t
he n
umbe
r of
dat
a va
lues
and
x is
a d
ata
valu
e.
Exer
cise
s 1.
Fin
d th
e tr
imm
ed m
ean
of t
he d
ata
set.
32
, 24,
56,
102
, 54,
12,
27,
49,
35,
23,
44,
51,
66,
36,
52,
16,
63,
75,
21,
41
4
2.1
25
2. F
ind
the
arit
hmet
ic m
ean
and
med
ian
of t
he d
ata
set
in E
xerc
ise
1. W
hat
is t
he b
enef
it o
f usi
ng a
tri
mm
ed m
ean
inst
ead
of a
n ar
ithm
etic
mea
n?
a
rith
me
tic
me
an
: 4
3.9
5,
me
dia
n:
42
.5;
Th
e t
rim
me
d m
ea
n i
s
les
s s
en
sit
ive
to
ou
tlie
rs.
3. W
ATE
R Th
e E
PA r
ecom
men
ds t
hat
qual
ity
fres
hwat
er b
e m
onit
ored
for
e.co
li co
ncen
trat
ion.
The
geo
met
ric
mea
n of
the
e.c
oli c
once
ntra
tion
s in
5
or m
ore
sam
ples
tak
en o
ver
30 d
ays
shou
ld b
e le
ss t
han
126
per
100
mill
ilite
rs. D
o th
e da
ta {2
25, 1
81, 1
10, 1
18, 1
07} m
eet
this
cri
teri
on?
Exp
lain
. N
o,
the
ge
om
etr
ic m
ea
n i
s a
bo
ut
14
1.4
pe
r m
L.
4. T
RAV
EL I
f you
tra
vel a
t on
e sp
eed
for
half
the
dist
ance
of a
tri
p an
d yo
u tr
avel
at
a di
ffere
nt s
peed
for
the
othe
r ha
lf of
the
dis
tanc
e, t
hen
your
av
erag
e sp
eed
for
the
trip
is t
he h
arm
onic
mea
n of
the
spe
eds.
If G
reg
drov
e at
60
mph
for
half
a tr
ip a
nd 7
0 m
ph w
hen
the
spee
d lim
it in
crea
sed
for
the
seco
nd h
alf,
wha
t w
as h
is a
vera
ge s
peed
for
the
trip
? a
bo
ut
64
.6 m
ph
005_
042_
PC
CR
MC
11_8
9381
2.in
dd9
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/09
12:3
6:57
PM
A01_A19_PCCRMC11_893812.indd 3A01_A19_PCCRMC11_893812.indd 3 11/19/09 8:27:48 PM11/19/09 8:27:48 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
Chapter 11 A4 Glencoe Precalculus
Answers (Lesson 11-1 and Lesson 11-2)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
10
Gle
ncoe
Pre
calc
ulus
11-1
TI-N
spir
eTM A
ctiv
ity
Gra
ph
ing
Data
You
can
use
a T
I-N
spir
e to
gra
ph d
ata
sets
and
des
crib
e th
e sh
ape
of t
he d
istr
ibut
ion.
T
he le
ngth
s of
fis
h ca
ught
on
a on
e-da
yfi
shin
g tr
ip a
re s
how
n in
the
tab
le. M
ake
a hi
stog
ram
and
box
plot
of
the
data
.
Step
1:
Add
a L
ists
& S
prea
dshe
et p
age.
Lis
t th
e da
tain
col
umn
A a
nd t
itle
the
col
umn
leng
ths.
Step
2:
Wit
h th
e cu
rsor
in c
olum
n A
, pre
ss b
and
choo
se D
ata
> Q
uick
Gra
ph. P
ress
b a
ndch
oose
Plo
t T
ype
> H
isto
gram
. Pre
ss b
agai
n an
d ch
oose
Plo
t P
rope
rtie
s >
His
togr
am P
rope
rtie
s >
Bin
Set
ting
s to
cha
nge
the
wid
th o
f the
bar
s. P
ress
b a
nd c
hoos
eW
indo
w/Z
oom
> Z
oom
-Dat
a to
aut
omat
ical
ly s
ize
the
axes
to
see
all t
he d
ata.
You
can
cha
nge
the
vert
ical
cat
egor
y to
per
cent
by p
ress
ing b
and
cho
osin
g P
lot
Pro
pert
ies
>H
isto
gram
Pro
pert
ies
> H
isto
gram
Sca
le >
Per
cent
.
Step
3:
To c
hang
e th
e gr
aph
to a
box
plo
t, pr
ess b
and
choo
se P
lot
Typ
e >
Box
Plo
t. M
ove
the
curs
or o
ver
the
grap
h to
vie
w t
he fi
ve-n
umbe
r su
mm
ary.
Exer
cise
s
Ope
n a
new
Lis
ts &
Spr
eads
heet
pag
e. T
itle
col
umn
A a
s da
ta. U
se it
to
com
plet
e th
e fo
llow
ing.
1. M
ake
a hi
stog
ram
of t
he d
ata
set:
44.5
, 13.
7, 2
9.4,
22.
0, 3
2.5,
45.
8, 3
8.6,
24.
3,18
.1, a
nd 5
0.3.
Use
the
gra
ph t
o de
scri
be t
he s
hape
of t
he d
istr
ibut
ion.
s
ym
me
tric
2. C
hang
e th
e hi
stog
ram
in E
xerc
ise
1 to
a b
ox p
lot.
List
the
five
-num
ber
sum
mar
y.1
3.7
, 2
2,
30
.95
, 4
4.5
, 5
0.3
3. O
pen
a ne
w L
ists
& S
prea
dshe
et p
age.
Tit
le c
olum
n A
as
data
1 an
d co
lum
n B
as d
ata2
. Ent
er t
he d
ata
show
n be
low
. Use
Qui
ck G
raph
in e
ach
colu
mn
to m
ake
box
plot
s. D
escr
ibe
and
com
pare
the
dis
trib
utio
ns.
D
ata1
: 102
, 116
, 132
, 111
, 124
, 103
, 101
, 108
, 129
, 103
, 109
, 115
, 111
D
ata2
: 115
, 129
, 101
, 128
, 125
, 115
, 101
, 124
, 124
, 118
, 121
, 119
, 120
T
he
ra
ng
es
are
alm
os
t th
e s
am
e.
Ab
ou
t 7
5%
of
the
da
ta i
n d
ata
2 a
re
gre
ate
r th
an
th
e m
ed
ian
of
da
ta1.
Da
ta1 i
s p
os
itiv
ely
sk
ew
ed
an
d d
ata
2 i
s
ne
ga
tiv
ely
sk
ew
ed
.
Le
ng
ths
of
Fis
h (
in.)
14
16
18
18
14
21
26
23
35
15
32
30
92
21
2
Exam
ple
005_
042_
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11_8
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/09
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-2
NA
ME
DA
TE
PE
RIO
D
Lesson 11-2
Ch
ap
ter
11
11
Gle
ncoe
Pre
calc
ulus
Prob
abili
ty D
istr
ibut
ion
A d
iscr
ete
rand
om v
aria
ble
X c
an t
ake
on a
coun
tabl
e nu
mbe
r of
val
ues.
A p
roba
bili
ty d
istr
ibut
ion
links
eac
h po
ssib
leva
lue
of X
wit
h it
s pr
obab
ility
of o
ccur
ring
.
Mea
n of
pro
babi
lity
dist
ribu
tion
of X
: μ =
Σ
[ X �
P(X
)]
Stan
dard
dev
iati
on o
f pro
babi
lity
dist
ribu
tion
of X
: σ =
√
��
��
��
�
Σ [(X
- μ
)2 � P
(X) ]
T
he n
umbe
r of
boo
ks b
ough
t by
eac
h of
100
ran
dom
ly
sele
cted
boo
ksto
re c
usto
mer
s du
ring
one
wee
k is
sho
wn.
a. C
onst
ruct
a p
roba
bili
ty d
istr
ibut
ion
for
X.
Th
ere
are
100
cust
omer
s, s
o P(
0) =
0.4
5, P
(1) =
0.3
0,
P(2)
= o
r 0.
15, a
nd P
(3) =
0.1
0.
b. F
ind
and
inte
rpre
t th
e m
ean
in t
he c
onte
xt o
f th
e pr
oble
m s
itua
tion
.F
ind
the
vari
ance
and
sta
ndar
d de
viat
ion.
O
rgan
ize
your
cal
cula
tion
s in
a t
able
.
Th
e m
ean
of t
he d
istr
ibut
ion
is 0
.9. O
n av
erag
e, e
ach
cust
omer
bou
ght
one
book
.Th
e va
rian
ce =
0.9
9, s
o th
e st
anda
rd d
evia
tion
is √
��
0.
99 o
r ab
out
0.99
5.
Exer
cise
1. T
he t
able
sho
ws
the
num
ber
of m
edic
al t
ests
tha
t15
ran
dom
ly s
elec
ted
pati
ents
ent
erin
g a
part
icul
arho
spit
al r
ecei
ved
one
day.
a. C
onst
ruct
a p
roba
bilit
y di
stri
buti
on fo
r X
.
b. F
ind
and
inte
rpre
t th
e m
ean
in t
he c
onte
xt o
f the
pro
blem
sit
uati
on. F
ind
the
vari
ance
and
sta
ndar
d de
viat
ion.
T
he
me
an
is
ab
ou
t 0
.93
, s
o,
on
av
era
ge
, e
ac
h p
ati
en
t re
ce
ive
d o
ne
te
st;
va
ria
nc
e ≈
0.8
6;
sta
nd
ard
de
via
tio
n
≈0
.93
11-2
Stud
y Gu
ide
and
Inte
rven
tion
Pro
bab
ilit
y D
istr
ibu
tio
ns
Exam
ple
Bo
ok
s,
XF
req
ue
nc
y
04
5
13
0
21
5
31
0B
oo
ks
, X
01
23
Fre
qu
en
cy
0.4
50
.30
0.1
50
.10
Bo
ok
s,
XP
(X)
X �
P(X
)(X
- μ
)2(X
- μ
)2 �
P(X
)
00
.45
0 �
0.4
5 =
0(0
- 0
.9)2
= 0
.81
0.8
1 �
0.4
5 =
0.3
64
5
10
.30
1 �
0.3
0 =
0.3
0(1
- 0
.9)2
= 0
.01
0.0
1 �
0.3
0 =
0.0
03
20
.15
2 �
0.1
5 =
0.3
0(2
- 0
.9)2
= 1
.21
1.2
1 �
0.1
5 =
0.1
81
5
30
.10
3 �
0.1
0 =
0.3
0(3
- 0
.9)2
= 4
.41
4.4
1 �
0.1
0 =
0.4
41
μ =
0.9
σ2 =
0.9
9
Te
sts
, X
Fre
qu
en
cy
06
15
23
31
Te
sts
, X
01
23
Fre
qu
en
cy
2
−
5
1
−
3
1
−
5
1
−
15
005_
042_
PC
CR
MC
11_8
9381
2.in
dd11
3/25
/09
12:3
7:05
PM
A01_A19_PCCRMC11_893812.indd 4A01_A19_PCCRMC11_893812.indd 4 3/25/09 2:47:24 PM3/25/09 2:47:24 PM
Copyright
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 A5 Glencoe Precalculus
An
swer
s
Answers (Lesson 11-2)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
12
Gle
ncoe
Pre
calc
ulus
Bino
mia
l Dis
trib
utio
n In
a b
inom
ial e
xper
imen
t, t
he o
utco
mes
are
suc
cess
or
failu
re. T
here
are
a fi
xed
num
ber
of in
depe
nden
t tr
ials
n a
nd t
he r
ando
m v
aria
ble
X
repr
esen
ts t
he n
umbe
r of
suc
cess
es. T
he p
roba
bilit
y of
suc
cess
p a
nd t
he p
roba
bilit
y of
fa
ilure
q o
r 1
- p
rem
ain
cons
tant
. The
pro
babi
lity
of X
suc
cess
es in
n in
depe
nden
t tr
ials
is
P(X
) = nC
x px q
n -
x =
n!
−
(n -
x)!x
! px q
n -
x .
A
sur
vey
foun
d th
at 2
0% o
f A
mer
ican
s ha
ve v
isit
ed a
doc
tor
in t
he
past
six
mon
ths.
Fiv
e pe
ople
wil
l be
sele
cted
at
rand
om a
nd a
sked
if t
hey
visi
ted
a do
ctor
in t
he p
ast
six
mon
ths.
Con
stru
ct a
nd g
raph
a b
inom
ial d
istr
ibut
ion
for
the
rand
om v
aria
ble
X, r
epre
sent
ing
the
num
ber
of p
eopl
e w
ho s
ay y
es. T
hen
find
the
pr
obab
ilit
y th
at a
t le
ast
four
of
thes
e pe
ople
say
yes
.
For
this
bin
omia
l exp
erim
ent,
n =
5, p
= 0
.2, a
nd q
= 1
- 0
.2 o
r 0.
8. C
ompu
te e
ach
poss
ible
va
lue
of X
usi
ng t
he B
inom
ial P
roba
bilit
y Fo
rmul
a.P(
0) =
5C0 �
0.2
0 � 0
.85 ≈
0.3
28
P(3)
= 5C
3 � 0
.23 �
0.8
2 ≈ 0
.051
P(1)
= 5C
1 � 0
.21 �
0.8
4 ≈ 0
.410
P(
4) =
5C4 �
0.2
4 � 0
.81 ≈
0.0
06P(
2) =
5C2 �
0.2
2 � 0
.83 ≈
0.2
05
P(5)
= 5C
5 � 0
.25 �
0.8
0 ≈ 0
.000
To fi
nd t
he p
roba
bilit
y th
at a
t le
ast
four
peo
ple
said
yes
, fin
d th
e su
m o
f P(4
) and
P(5
).P(
X ≥
4) =
P(4
) + P
(5) =
0.0
06 +
0.0
00 o
r 0.
6%
Exer
cise
A s
urve
y fo
und
that
60%
of A
mer
ican
vic
tim
s of
hea
lth-
care
frau
d w
ere
seni
or c
itiz
ens.
Six
vi
ctim
s of
hea
lth-
care
frau
d w
ill b
e ch
osen
at
rand
om a
nd t
heir
age
s w
ill b
e re
cord
ed.
Con
stru
ct a
nd g
raph
a b
inom
ial d
istr
ibut
ion
for
the
rand
om v
aria
ble
X, r
epre
sent
ing
the
num
ber
of s
enio
r ci
tize
ns c
hose
n. T
hen
find
the
prob
abili
ty t
hat
at le
ast
thre
e of
the
vic
tim
s w
ill b
e se
nior
cit
izen
s.
P(X
≥ 3
) =
0.8
21
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Pro
bab
ilit
y D
istr
ibu
tio
ns
11-2
Exam
ple
XP
(X)
00
.32
8
10
.41
0
20
.20
5
30
.05
1
40
.00
6
50
.00
0
00
12
34
5
0.1
0.2
0.3
0.4
0.5
Peop
le
Probability
P(x)
x
XP
(X)
XP
(X)
00
.00
44
0.3
11
10
.03
75
0.1
87
20
.13
86
0.0
47
30
.27
6
01
23
45
6
0.050.1
0.150.2
0.250.3
0.35
Seni
or C
itize
ns
Probability
0
P(x)
x
005_
042_
PC
CR
MC
11_8
9381
2.in
dd12
3/27
/09
7:14
:30
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-2
NA
ME
DA
TE
PE
RIO
D
Lesson 11-2
Ch
ap
ter
11
13
Gle
ncoe
Pre
calc
ulus
Cla
ssif
y ea
ch r
ando
m v
aria
ble
X a
s di
scre
te o
r co
ntin
uous
. Exp
lain
you
r re
ason
ing.
1. X
rep
rese
nts
the
tim
e it
tak
es a
ran
dom
ly s
elec
ted
clas
sroo
m t
o re
ach
68°F
from
60°
F.
co
nti
nu
ou
s;
Th
e t
ime
ca
n b
e a
ny
nu
mb
er.
2. X
rep
rese
nts
the
num
ber
of p
hoto
grap
hs t
aken
by
a ph
otog
raph
er a
t a
rand
omly
sele
cted
wed
ding
. d
isc
rete
; T
he
nu
mb
er
of
ph
oto
gra
ph
s i
s c
ou
nta
ble
.
3. T
he t
able
sho
ws
the
num
ber
of c
ell p
hone
s ow
ned
by
100
rand
omly
sel
ecte
d ho
useh
olds
. Con
stru
ct a
nd g
raph
a
prob
abili
ty d
istr
ibut
ion
for
X. T
hen
find
and
inte
rpre
t th
e m
ean
in t
he c
onte
xt o
f the
pro
blem
sit
uati
on. F
ind
the
vari
ance
and
sta
ndar
d de
viat
ion.
4. R
ACE
A r
esor
t is
pla
nnin
g a
bicy
cle
race
. The
cos
t of
spo
nsor
ing
the
race
is$8
000.
The
res
ort
expe
cts
to m
ake
$15,
000
on t
he e
vent
. The
re is
a 3
0% c
hanc
eof
a h
urri
cane
arr
ivin
g th
e da
y of
the
rac
e. I
f thi
s ha
ppen
s, t
he r
ace
will
be
canc
elle
d an
d w
ill n
ot b
e re
sche
dule
d. W
hat
is t
he r
esor
t’s e
xpec
ted
prof
it?
$8
10
0
5. C
OM
MU
TE I
n a
rece
nt p
oll,
45%
of a
tow
n’s
citi
zens
sai
d th
ey u
se t
he b
us t
oge
t to
wor
k. F
ive
of t
hese
cit
izen
s w
ill b
e ra
ndom
ly c
hose
n an
d as
ked
if th
eyus
e th
e bu
s to
get
to
wor
k.
a. C
onst
ruct
a b
inom
ial d
istr
ibut
ion
for
the
rand
om v
aria
ble
X, r
epre
sent
ing
the
peop
le w
ho s
ay y
es.
b. F
ind
the
mea
n, v
aria
nce,
and
sta
ndar
d de
viat
ion
of t
his
dist
ribu
tion
. Int
erpr
etth
e m
ean
in t
he c
onte
xt o
f the
pro
blem
sit
uati
on.
m
ea
n:
2.2
5,
va
ria
nc
e:
1.2
37
5,
std
de
v:
ab
ou
t 1
.11
1;
On
av
era
ge
, 2
of
ev
ery
5 r
an
do
mly
se
lec
ted
pe
op
le i
n t
he
to
wn
wo
uld
sa
y t
he
y t
ak
e t
he
b
us
to
ge
t to
wo
rk.
11-2
Prac
tice
Pro
bab
ilit
y D
istr
ibu
tio
ns
Th
e m
ea
n i
s 1
.93
, s
o e
ac
h
ho
us
eh
old
ha
s a
bo
ut
two
c
ell
ph
on
es
; v
ari
an
ce
≈ 0
.79
; s
tan
da
rd d
ev
iati
on
≈ 0
.88
6
Ph
on
es
, X
01
23
4
Fre
qu
en
cy
0.0
20
.30
0.4
80
.13
0.0
7
Ph
on
es
, X
Fre
qu
en
cy
02
13
0
24
8
31
3
47
01
23
4
Cell
Phon
es
Probability
00.1
0.20.3
0.4
0.5
P(x)
x
X0
12
34
5
P(X
)0
.05
00
.20
60
.33
70
.27
60
.11
30
.01
8
005_
042_
PC
CR
MC
11_8
9381
2.in
dd13
11/1
4/09
4:53
:59
PM
A01_A19_PCCRMC11_893812.indd 5A01_A19_PCCRMC11_893812.indd 5 11/19/09 8:27:56 PM11/19/09 8:27:56 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 11 A6 Glencoe Precalculus
Answers (Lesson 11-2)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
14
Gle
ncoe
Pre
calc
ulus
1. R
ETA
IL A
sto
re m
anag
er m
ade
the
prob
abili
ty d
istr
ibut
ion
show
n be
low
. It
sho
ws
the
prob
abili
ty o
f sel
ling
Xsw
imsu
its
on a
ran
dom
ly s
elec
ted
day
in J
une.
Find
the
mea
n, v
aria
nce,
and
sta
ndar
d de
viat
ion
of t
he d
istr
ibut
ion.
2
0.8
; 1
.6;
1.2
2. I
NSU
RAN
CE A
n in
sura
nce
com
pany
in
sure
s a
pain
ting
wor
th $
20,0
00 a
gain
st
thef
t fo
r $3
00 p
er y
ear.
The
com
pany
has
as
sess
ed t
he p
roba
bilit
y of
the
pai
ntin
g be
ing
stol
en in
a g
iven
yea
r as
0.0
02.
Wha
t is
the
insu
ranc
e co
mpa
ny’s
expe
cted
ann
ual p
rofit
?
$
25
9.4
0
3. R
ESTA
URA
NT
A s
urve
y fo
und
that
25%
of
all
part
ies
at a
res
taur
ant
wer
e gr
oups
of
five
or
larg
er. E
ight
een
part
ies
are
rand
omly
sel
ecte
d.
a. F
ind
the
prob
abili
ty t
hat
exac
tly
five
part
ies
are
mad
e up
of f
ive
or
mor
e pe
ople
.
19
.9%
b. F
ind
the
prob
abili
ty t
hat
5, 6
, or
7 pa
rtie
s ar
e m
ade
up o
f fiv
e or
mor
e pe
ople
.
4
2.4
%
4. P
ETS
Acc
ordi
ng t
o on
e po
ll, a
bout
63%
of
Am
eric
an h
ouse
hold
s in
clud
e at
leas
t on
e pe
t. Si
x ne
w h
omes
are
bui
lt a
nd s
old.
a. C
onst
ruct
a b
inom
ial d
istr
ibut
ion
for
the
rand
om v
aria
ble
X, r
epre
sent
ing
the
num
ber
of t
hese
hom
es t
hat
will
ha
ve a
t le
ast
one
pet.
b. F
ind
the
mea
n, v
aria
nce,
and
st
anda
rd d
evia
tion
of t
his
dist
ribu
tion
. 3
.78
; 1
.39
86
; a
bo
ut
1.1
8
c. F
ind
the
prob
abili
ty t
hat
at le
ast
half
of t
he n
ew h
omes
hav
e pe
ts.
Ab
ou
t 8
6%
5. T
ESTI
NG
Mr.
Han
lon
dist
ribu
ted
a 5-
ques
tion
mul
tipl
e ch
oice
qui
z to
his
st
uden
ts. T
here
wer
e 5
choi
ces
for
each
qu
esti
on. A
shle
y gu
esse
s th
e an
swer
on
each
que
stio
n.
a. W
hat
is A
shle
y’s
prob
abili
ty o
f gu
essi
ng e
xact
ly 3
que
stio
ns
corr
ectl
y?
5.1
2%
b. W
hat
wou
ld b
e th
e pr
obab
ility
inpa
rt a
if t
here
wer
e 4
choi
ces
for
each
que
stio
n?
8.8
%
c.
Wha
t w
ould
be
the
prob
abili
ty in
part
a if
the
qui
z co
ntai
ned
only
true
/fals
e qu
esti
ons?
3
1.2
5%
11-2
Wor
d Pr
oble
m P
ract
ice
Pro
bab
ilit
y D
istr
ibu
tio
ns
Sw
ims
uit
s,
X1
92
02
12
22
3
P(X
)0
.20
0.2
00
.30
0.2
00
.10
X0
12
3
P(X
)0
.00
30
.02
60
.11
20
.25
3
X4
56
P(X
)0
.32
30
.22
00
.06
3
005_
042_
PC
CR
MC
11_8
9381
2.in
dd14
11/1
4/09
4:56
:17
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-2
NA
ME
DA
TE
PE
RIO
D
Lesson 11-2
Ch
ap
ter
11
15
Gle
ncoe
Pre
calc
ulus
The
Pois
son
dist
ribu
tion
is a
pro
babi
lity
dist
ribu
tion
for
an e
vent
occ
urri
ngx
tim
es o
ver
a gi
ven
inte
rval
. The
inte
rval
is u
sual
ly a
tim
e in
terv
al, s
uch
as
the
num
ber
of p
eopl
e en
teri
ng a
sto
re d
urin
g on
e ho
ur. T
he d
istr
ibut
ion
is a
di
scre
te d
istr
ibut
ion,
so
x m
ust
be a
who
le n
umbe
r.
The
prob
abili
ty o
f an
even
t oc
curr
ing
x ti
mes
ove
r a
give
n in
terv
al is
λx ·
e-
λ
−
x!
,
whe
re λ
is t
he a
vera
ge n
umbe
r of
tim
es t
he e
vent
occ
urs
duri
ng t
he in
terv
al.
A b
ank
man
ager
det
erm
ined
tha
t an
ave
rage
of 9
.2 c
usto
mer
s us
e a
cert
ain
ATM
eve
ry h
our.
You
can
cre
ate
a Po
isso
n di
stri
buti
on b
y us
ing
9.2
for
λ.
For
exam
ple,
the
pro
babi
lity
that
5 c
usto
mer
s us
e th
e A
TM in
a g
iven
hou
r
is g
iven
by
9.25 ·
e-
9.2
−
5!
. U
se y
our
calc
ulat
or t
o ve
rify
P(5
) ≈ 0
.055
or
abou
t 5.
5%.
On
a gr
aphi
ng c
alcu
lato
r, y
ou c
an a
lso
find
the
prob
abili
ty b
y us
ing
the
pois
sonp
df(
com
man
d fo
und
by p
ress
ing
[DIS
TR].
In t
he p
aren
thes
es, e
nter
λ
, x.
1. T
he b
ank
man
ager
det
erm
ined
tha
t an
ave
rage
of 1
.8 c
usto
mer
s in
quir
e ab
out
open
ing
a ne
w a
ccou
nt a
t a
cert
ain
bran
ch e
very
hou
r. C
ompl
ete
the
tabl
e.
2. N
otic
e th
at u
nlik
e a
bino
mia
l dis
trib
utio
n, x
has
no
uppe
r lim
it. T
he t
able
in
Exe
rcis
e 2
stop
ped
at x
= 5
, but
it is
pos
sibl
e fo
r 6
or m
ore
peop
le t
o in
quir
e ab
out
open
ing
a ne
w a
ccou
nt. H
ow c
an y
ou fi
nd P
(x >
2)?
S
ub
tra
ct
P(0
) +
P(1
) +
P(2
) fr
om
1.
3. O
n av
erag
e, t
he b
ank
rece
ives
4.2
onl
ine
loan
app
licat
ions
per
day
. Fin
d ea
ch p
roba
bilit
y.
a. T
he b
ank
rece
ives
3 o
nlin
e lo
an a
pplic
atio
ns o
ne d
ay.
18
.5%
b. T
he b
ank
rece
ives
4 o
nlin
e lo
an a
pplic
atio
ns o
ne d
ay.
19
.4%
c. T
he b
ank
rece
ives
5 o
nlin
e lo
an a
pplic
atio
ns o
ne d
ay.
16
.3%
d. T
he b
ank
rece
ives
6 o
nlin
e lo
an a
pplic
atio
ns o
ne d
ay.
11
.4%
e. T
he b
ank
rece
ives
7 o
nlin
e lo
an a
pplic
atio
ns o
ne d
ay.
6.9
%
4. M
ake
a co
njec
ture
abo
ut t
he g
ener
al s
hape
of a
ny P
oiss
on d
istr
ibut
ion.
It
is
po
sit
ive
ly s
ke
we
d.
Th
e p
ea
k i
s a
t λ
.
11-2
Enri
chm
ent
Th
e P
ois
son
Dis
trib
uti
on
X0
12
34
5
P(X
)0
.16
50
.29
80
.26
80
.16
10
.07
20
.02
6
005_
042_
PC
CR
MC
11_8
9381
2.in
dd15
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/09
12:3
7:28
PM
A01_A19_PCCRMC11_893812.indd 6A01_A19_PCCRMC11_893812.indd 6 11/17/09 8:19:06 AM11/17/09 8:19:06 AM
Copyright
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 A7 Glencoe Precalculus
An
swer
s
Answers (Lesson 11-3)
An
swer
s
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
16
Gle
ncoe
Pre
calc
ulus
The
Nor
mal
Dis
trib
utio
n A
nor
mal
dis
trib
utio
n is
a c
onti
nuou
s pr
obab
ility
dis
trib
utio
n. T
he g
raph
of a
nor
mal
dis
trib
utio
n is
sym
met
ric
and
bell-
shap
ed. I
t ap
proa
ches
but
nev
er t
ouch
es t
he x
-axi
s. I
t in
clud
es 1
00%
of
the
data
, so
the
area
und
er t
he c
urve
is 1
. The
z-v
alue
rep
rese
nts
the
num
ber
of s
tand
ard
devi
atio
ns t
hat
a gi
ven
data
val
ue is
from
the
mea
n.
z =
X -
μ
−
σ
, w
here
X is
the
dat
a va
lue,
μ is
the
mea
n, a
nd σ
is t
he s
tand
ard
devi
atio
n.
The
stan
dard
nor
mal
dis
trib
utio
n ha
s a
mea
n of
0 a
nd a
sta
ndar
d de
viat
ion
of 1
.
On
his
last
20
airl
ine
trip
s, a
n em
ploy
ee h
ad a
n av
erag
e la
yove
r of
82
min
utes
wit
h a
stan
dard
dev
iati
on o
f7.
5 m
inut
es. F
ind
the
num
ber
of la
yove
rs t
hat
wer
e le
ss t
han
75 m
inut
es.
Firs
t, fin
d th
e z-
valu
e.
z =
X -
μ
−
σ
F
orm
ula
fo
r z
-va
lue
s
=
75 -
82
−
7.5
or
abou
t -
0.93
X
= 7
5,
μ =
82
, a
nd
σ =
7.5
Use
a g
raph
ing
calc
ulat
or t
o fin
d th
e ar
ea u
nder
the
cur
ve t
hat
is t
o th
e le
ft o
f 75.
Pre
ss
2n
d [D
ISTR
] and
cho
ose
norm
alcd
f(.
Ent
er t
he lo
wer
val
ue (y
ou c
an u
se -
4 in
stea
d of
neg
ativ
e in
finit
y) a
nd t
he u
pper
val
ue a
s -
0.93
. The
res
ulti
ng a
rea
is 0
.176
. Th
is m
eans
tha
t ab
out
17.6
% o
f the
dat
a va
lues
are
less
tha
n-
0.93
sta
ndar
d de
viat
ions
from
the
mea
n.
Bec
ause
the
re a
re 2
0 fli
ghts
, abo
ut 2
0 � 0
.176
or
abou
t 4
fligh
ts h
ad la
yove
r ti
mes
tha
t w
ere
less
tha
n 75
min
utes
.
Exer
cise
s
1. A
t a
rest
aura
nt, t
he a
vera
ge t
ime
betw
een
whe
n an
ord
er is
pla
ced
and
whe
n th
e en
tree
is s
erve
d is
12.
5 m
inut
es w
ith
a st
anda
rd d
evia
tion
of
1.2
min
utes
. Out
of 1
00 r
ando
mly
sel
ecte
d cu
stom
ers,
how
man
y w
ill b
e se
rved
the
ir e
ntre
es w
ithi
n 14
min
utes
of o
rder
ing?
ap
pro
xim
ate
ly 8
9 c
us
tom
ers
2. M
rs. Q
uan,
a fu
ll pr
ofes
sor
at a
com
mun
ity
colle
ge, e
arns
a s
alar
y of
$4
8,60
0. T
he a
vera
ge s
alar
y fo
r a
full
prof
esso
r at
the
col
lege
is $
52,0
00
wit
h a
stan
dard
dev
iati
on o
f $36
00. H
ow m
any
of t
he 4
5 fu
ll pr
ofes
sors
ea
rn le
ss t
han
Mrs
. Qua
n?
ap
pro
xim
ate
ly 8
pro
fes
so
rs
3. D
urin
g on
e O
ctob
er, t
he a
vera
ge w
ater
tem
pera
ture
of a
pon
d w
as 5
3.2°
w
ith
a st
anda
rd d
evia
tion
of 2
.3°. H
ow m
any
days
was
the
tem
pera
ture
gr
eate
r th
an 5
0°?
ap
pro
xim
ate
ly 2
8 d
ay
s
Stud
y Gu
ide
and
Inte
rven
tion
Th
e N
orm
al
Dis
trib
uti
on
11-3
Exam
ple
005_
042_
PC
CR
MC
11_8
9381
2.in
dd16
3/25
/09
12:3
7:32
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-3
NA
ME
DA
TE
PE
RIO
D
Lesson 11-3
Ch
ap
ter
11
17
Gle
ncoe
Pre
calc
ulus
Prob
abili
ty a
nd t
he N
orm
al D
istr
ibut
ion
The
area
und
er t
he
norm
al c
urve
cor
resp
onds
to
the
prob
abili
ty o
f dat
a va
lues
falli
ng w
ithi
n a
give
n in
terv
al.
T
he a
vera
ge m
onth
ly t
empe
ratu
res
for
a ci
ty f
or o
ne
year
wer
e no
rmal
ly d
istr
ibut
ed w
ith
μ =
65°
and
σ =
5. F
ind
P(5
0 <
X <
70)
. Use
a g
raph
ing
calc
ulat
or t
o sk
etch
the
co
rres
pond
ing
area
und
er t
he c
urve
.
Find
the
z-v
alue
s fo
r X
= 5
0 an
d X
= 7
0.
z =
X
- μ
−
σ
z
= X
- μ
−
σ
=
50
- 6
5 −
5 =
-3
= 70
- 6
5 −
5 =
1
Find
the
are
a be
twee
n z
= -
3 an
d z
= 1
. On
your
cal
cula
tor,
pre
ss
2n
d
[DIS
TR] a
nd c
hoos
e Sh
adeN
orm
und
er t
he D
RA
W m
enu.
Ent
er t
he lo
wer
an
d up
per
z-va
lues
and
pre
ss e
nter
.
P(50
< X
< 7
0) ≈
84%
Ther
efor
e, a
ppro
xim
atel
y 84
% o
f the
tem
pera
ture
s w
ere
betw
een
50°
and
70°.
Exer
cise
s 1.
The
ave
rage
age
of t
he s
wim
mer
s on
a m
aste
r sw
im t
eam
is n
orm
ally
di
stri
bute
d w
ith
μ =
56
and
σ =
4. F
ind
each
pro
babi
lity.
Use
a g
raph
ing
calc
ulat
or t
o sk
etch
the
cor
resp
ondi
ng a
rea
unde
r th
e cu
rve.
a. P
(53
< X
< 5
9)
54
.7%
b. P
(X <
53)
2
2.7
%
2. S
tude
nts
who
sco
re in
the
bot
tom
5%
of a
phy
sica
l edu
cati
on t
est
will
be
enro
lled
in a
sup
plem
enta
l phy
sica
l edu
cati
on p
rogr
am. T
he s
core
s of
all
of t
he s
tude
nts
who
too
k th
e te
st a
re n
orm
ally
dis
trib
uted
wit
h μ
= 12
2.6
and
σ =
18.
Wha
t is
the
gre
ates
t sc
ore
that
a s
tude
nt w
ho e
nrol
led
in t
he
supp
lem
enta
l pro
gram
cou
ld h
ave
rece
ived
? 9
2
11-3
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Th
e N
orm
al
Dis
trib
uti
on
Exam
ple
[ –4,
4] s
cl: 1
by
[ 0, 0
.5] s
cl: 0
.125
[ –4,
4] s
cl: 1
by
[ 0, 0
.5] s
cl: 0
.125
005_
042_
PC
CR
MC
11_8
9381
2.in
dd17
11/1
9/09
6:40
:36
PM
A01_A19_PCCRMC11_893812.indd 7A01_A19_PCCRMC11_893812.indd 7 11/19/09 8:28:05 PM11/19/09 8:28:05 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 11 A8 Glencoe Precalculus
Answers (Lesson 11-3)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
18
Gle
ncoe
Pre
calc
ulus
1. T
REES
The
hei
ghts
of 2
00 t
rees
in a
nur
sery
are
nor
mal
ly d
istr
ibut
ed
wit
h a
mea
n of
120
inch
es a
nd a
sta
ndar
d de
viat
ion
of 1
6 in
ches
.
a. A
ppro
xim
atel
y ho
w m
any
tree
s ar
e m
ore
than
136
inch
es t
all?
3
2 t
ree
s
b. W
hat
perc
ent
of t
he t
rees
are
bet
wee
n 88
inch
es a
nd 1
04 in
ches
tal
l?
13
.6%
Fin
d ea
ch o
f th
e fo
llow
ing.
2. z
if X
= 6
5, μ
= 5
0, a
nd σ
= 1
0 3.
X if
z =
-0.
4, μ
= 4
0, a
nd σ
= 5
1
.5
3
8
Fin
d th
e in
terv
al o
f z-
valu
es a
ssoc
iate
d w
ith
each
are
a.
4. t
he m
iddl
e 60
% o
f the
dat
a -
0.8
4 <
z <
0.8
4
5. t
he o
utsi
de 3
0% o
f the
dat
a z
< -
1.0
4 a
nd
z >
1.0
4
6. D
OG
S Th
e w
eigh
ts o
f the
42
full-
grow
n G
erm
an s
heph
erds
at
a ke
nnel
are
nor
mal
ly
dist
ribu
ted.
The
mea
n w
eigh
t is
86
poun
ds a
nd t
he s
tand
ard
devi
atio
n is
3 p
ound
s.
a. D
eter
min
e th
e nu
mbe
r of
Ger
man
she
pher
ds t
hat
wei
gh m
ore
than
82
poun
ds.
ab
ou
t 3
8 d
og
s
b. H
ow m
any
Ger
man
she
pher
ds w
eigh
less
tha
n 88
pou
nds?
a
bo
ut
31
do
gs
7. H
OTE
LS T
he p
rice
s of
roo
ms
at h
otel
s ar
ound
an
airp
ort
are
norm
ally
di
stri
bute
d w
ith
μ =
$12
0 an
d σ
= $
20. F
ind
each
pro
babi
lity.
a. T
he c
ost
of a
roo
m is
gre
ater
tha
n $1
50.
6
.7%
b. T
he c
ost
of a
roo
m is
bet
wee
n $1
10 a
nd $
130.
3
8.3
%
c. T
he c
ost
of a
roo
m is
bet
wee
n $9
0 an
d $1
00.
9
.2%
d. I
f onl
y th
e m
ost
expe
nsiv
e 10
% o
f the
roo
ms
are
avai
labl
e, w
hat
is t
he
leas
t am
ount
you
will
pay
for
a ro
om?
$
14
6
8. E
XA
MS
A s
tude
nt s
core
d 65
on
a bi
olog
y ex
am w
ith
μ =
50
and
σ =
10.
Sh
e sc
ored
30
on a
lite
ratu
re e
xam
wit
h μ
= 2
5 an
d σ
= 5
. Com
pare
her
sc
ores
on
each
tes
t. A
ssum
e th
at b
oth
sets
of s
core
s w
ere
norm
ally
di
stri
bute
d. S
he
did
be
tte
r o
n h
er
bio
log
y e
xa
m.
Th
e z
-sc
ore
fo
r th
e b
iolo
gy
ex
am
wa
s 1
.5 a
nd
th
e z
-sc
ore
fo
r th
e
lite
ratu
re e
xa
m w
as
1.
11-3
Prac
tice
Th
e N
orm
al
Dis
trib
uti
on
005_
042_
PC
CR
MC
11_8
9381
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/09
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-3
NA
ME
DA
TE
PE
RIO
D
Lesson 11-3
Ch
ap
ter
11
19
Gle
ncoe
Pre
calc
ulus
1. R
EAL
ESTA
TE T
he a
vera
ge p
rice
of a
on
e-be
droo
m c
ondo
min
ium
list
ed b
y a
real
tor
is $
145,
500
wit
h a
stan
dard
de
viat
ion
of $
1500
. The
pri
ces
are
norm
ally
dis
trib
uted
. Det
erm
ine
the
prob
abili
ty t
hat
a ra
ndom
ly s
elec
ted
cond
omin
ium
cos
ts b
etw
een
$143
,580
an
d $1
47,4
20.
8
0%
2. C
OM
MU
TIN
G T
he a
vera
ge t
imes
spe
nt
com
mut
ing
to w
ork
in a
cer
tain
cit
y ar
e no
rmal
ly d
istr
ibut
ed w
ith
a m
ean
of25
.5 m
inut
es a
nd a
sta
ndar
d de
viat
ion
of 6
.1 m
inut
es. W
hat
is t
he p
roba
bilit
y th
at a
ran
dom
ly s
elec
ted
com
mut
e to
w
ork
take
s lo
nger
tha
n a
half
hour
?
2
3.0
%
3. S
IGH
TSEE
ING
The
tim
es p
eopl
e sp
end
view
ing
cert
ain
anci
ent
ruin
s ar
e no
rmal
ly d
istr
ibut
ed w
ith
a m
ean
of
96 m
inut
es w
ith
a st
anda
rd d
evia
tion
of
17 m
inut
es.
a. F
ind
the
prob
abili
ty t
hat
a si
ghts
eer
will
spe
nd a
t le
ast
two
hour
s at
the
ru
ins.
7
.9%
b. F
ind
the
prob
abili
ty t
hat
a si
ghts
eer
will
spe
nd a
t m
ost
80 m
inut
es a
t th
e ru
ins.
1
7.3
%
c.
If a
tou
r bu
s dr
ops
off a
gro
up o
f si
ghts
eers
at
9 A.M
., w
hat
tim
e sh
ould
th
e bu
s pi
ck u
p th
e si
ghts
eers
? E
xpla
in.
S
am
ple
an
sw
er:
Ab
ou
t 9
5%
of
the
sig
hts
ee
rs w
ill
sp
en
d l
es
s
tha
n 1
24
min
ute
s a
t th
e r
uin
s,
so
th
e b
us
sh
ou
ld r
etu
rn a
t 1
1:0
5 A
.M.
4. T
RAIN
ING
To
qual
ify fo
r a
secu
rity
po
siti
on, c
andi
date
s m
ust
take
a p
hysi
cal
fitne
ss t
est.
The
scor
es o
n th
e te
st a
re
norm
ally
dis
trib
uted
wit
h a
mea
n of
400
an
d a
stan
dard
dev
iati
on o
f 100
.
a. C
andi
date
s sc
orin
g in
the
top
3%
are
la
ter
recr
uite
d as
tra
iner
s in
the
pr
ogra
m. W
hat
is t
he m
inim
um s
core
a
cand
idat
e ne
eds
in o
rder
to
be
recr
uite
d la
ter
as a
tra
iner
?
5
89
b. C
andi
date
s sc
orin
g in
the
bot
tom
1.5
%
mus
t re
take
the
phy
sica
l tra
inin
g pr
ogra
m. W
hat
is t
he m
inim
um s
core
a
cand
idat
e w
ould
nee
d to
avo
id
reta
king
the
tra
inin
g pr
ogra
m?
1
82
5. J
UIC
E Th
e am
ount
of j
uice
pou
red
into
bot
tles
in a
fact
ory
is n
orm
ally
di
stri
bute
d w
ith
a m
ean
of 1
6 ou
nces
and
a st
anda
rd d
evia
tion
of 0
.3 o
unce
. A
shi
pmen
t co
ntai
ns 2
80 b
ottl
es.
a. H
ow m
any
bott
les
are
expe
cted
to
cont
ain
mor
e th
an 1
6.5
ounc
es o
f ju
ice?
a
bo
ut
13
bo
ttle
s
b. H
ow m
any
bott
les
are
expe
cted
to
cont
ain
less
tha
n 15
.75
ounc
es o
f ju
ice?
a
bo
ut
57
bo
ttle
s
11-3
Wor
d Pr
oble
m P
ract
ice
Th
e N
orm
al
Dis
trib
uti
on
005_
042_
PC
CR
MC
11_8
9381
2.in
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11/1
4/09
4:59
:36
PM
A01_A19_PCCRMC11_893812.indd 8A01_A19_PCCRMC11_893812.indd 8 11/17/09 8:49:18 AM11/17/09 8:49:18 AM
Copyright
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 A9 Glencoe Precalculus
An
swer
s
Answers (Lesson 11-3)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
20
Gle
ncoe
Pre
calc
ulus
Bef
ore
grap
hing
cal
cula
tors
, stu
dent
s ha
d to
find
the
are
a un
der
a no
rmal
dis
trib
utio
n cu
rve
by u
sing
a t
able
of v
alue
s. Y
ou c
an s
till
find
thes
e ta
bles
in m
any
stat
isti
cs b
ooks
.It
is g
ood
to k
now
how
to
use
them
, in
case
you
find
you
rsel
f in
a si
tuat
ion
wit
hout
a
grap
hing
cal
cula
tor.
Her
e is
par
t of
a s
tand
ard
norm
al p
roba
bilit
y ta
ble.
It
give
s th
e ar
ea t
hat
is b
etw
een
the
mea
n, 0
, and
the
z-s
core
. To
find
the
area
bet
wee
n a
z-sc
ore
of 0
and
a z
-sco
re o
f 1.3
5, fi
nd
1.3
in t
he v
erti
cal c
olum
n an
d 0.
05 in
the
top
row
. It
is 0
.411
5.
To fi
nd t
he a
rea
that
is t
o th
e le
ft o
f a c
erta
in p
osit
ive
z-sc
ore,
rem
embe
r to
add
the
are
ath
at is
to
the
left
of t
he m
ean,
0.5
.
Bec
ause
a n
orm
al d
istr
ibut
ion
is s
ymm
etri
c ab
out
the
mea
n, y
ou c
an u
se t
he s
ame
char
t to
find
the
are
a be
twee
n a
nega
tive
z-s
core
and
the
mea
n. F
or e
xam
ple,
the
area
bet
wee
n a
z-sc
ore
of -
1.35
and
0 is
the
sam
e as
bet
wee
n 0
and
1.35
: 0.4
115.
Use
the
tab
le t
o an
swer
eac
h qu
esti
on.
1. W
hat
is t
he a
rea
betw
een
a z-
scor
e of
0 a
nd a
z-s
core
of 1
.02?
0
.34
61
2. W
hat
is t
he a
rea
betw
een
a z-
scor
e of
1.1
8 an
d a
z-sc
ore
of 1
.43?
Exp
lain
how
you
foun
d it
. 0
.04
26
: S
ub
tra
ct
the
are
a b
etw
ee
n 0
an
d 1
.18
fro
m t
he
are
a b
etw
ee
n 0
an
d 1
.43
: 0
.42
36
- 0
.38
10
= 0
.04
26
.
3. F
ind
the
area
to
the
righ
t of
a z
-sco
re o
f 1.1
7. E
xpla
in h
ow y
ou fo
und
it.
0.1
21
; S
ub
tra
ct
the
are
a i
n t
he
ch
art
fro
m 0
.5:
0.5
- 0
.37
90
= 0
.12
1.
4. W
hat
is t
he a
rea
to t
he le
ft o
f a z
-sco
re o
f 1.4
1?
0.9
20
7
5. W
hat
is t
he a
rea
betw
een
a z-
scor
e of
-1.
25 a
nd a
z-s
core
of 0
? 0
.39
44
6. W
hat
is t
he a
rea
to t
he le
ft o
f a z
-sco
re o
f -1.
33?
0.0
91
8
7. W
hat
is t
he a
rea
betw
een
a z-
scor
e of
-1.
05 a
nd a
z-s
core
of 1
.38?
0
.76
93
11-3
Enri
chm
ent
Fin
din
g A
reas
by
Usi
ng
a T
ab
le
z.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9
1.0
.34
13
.34
38
.34
61
.34
85
.35
08
.35
31
.35
54
.35
77
.35
99
.36
21
1.1
.36
43
.36
65
.36
86
.37
08
.37
29
.37
49
.37
70
.37
90
.38
10
.38
30
1.2
.38
49
.38
69
.38
88
.39
07
.39
25
.39
44
.39
62
.39
80
.39
97
.40
15
1.3
.40
32
.40
49
.40
66
.40
82
.40
99
.41
15
.41
31
.41
47
.41
62
.41
77
1.4
.41
92
.42
07
.42
22
.42
36
.42
51
.42
65
.42
79
.42
92
.43
06
.43
19
005_
042_
PC
CR
MC
11_8
9381
2.in
dd20
11/1
9/09
6:41
:03
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-3
NA
ME
DA
TE
PE
RIO
D
Lesson 11-3
Ch
ap
ter
11
21
Gle
ncoe
Pre
calc
ulus
The
grap
h sh
own
at t
he r
ight
is k
now
n as
the
sta
ndar
d no
rmal
curv
e. T
he s
tand
ard
norm
al c
urve
is t
he g
raph
of f
(x) =
1
−
√
��
2π
e
-
x2 −
2 .
You
can
use
a g
raph
ing
calc
ulat
or t
o in
vest
igat
e pr
oper
ties
of
this
func
tion
and
its
grap
h. E
nter
the
func
tion
for
the
norm
al
curv
e in
the
Y =
list
of a
gra
phin
g ca
lcul
ator
.
1. T
he s
tand
ard
norm
al c
urve
mod
els
a pr
obab
ility
dis
trib
utio
n. A
s a
resu
lt, p
roba
bilit
ies
for
inte
rval
s of
x-v
alue
s ar
e eq
ual t
o ar
eas
of
regi
ons
boun
ded
by t
he c
urve
, the
x-a
xis,
and
the
ver
tica
l lin
es
thro
ugh
the
endp
oint
s of
the
inte
rval
s. T
he c
alcu
lato
r ca
n ap
prox
imat
e th
e ar
eas
of s
uch
regi
ons.
To
find
the
area
of t
he r
egio
n bo
unde
d by
the
cur
ve, t
he x
-axi
s, a
nd t
he v
erti
cal l
ines
x =
-1
and
x =
1, g
o to
the
CA
LC
men
u an
d se
lect
7: ∫
f(x)
dx. M
ove
the
curs
or t
o
the
poin
t w
here
x =
-1.
Pre
ss E
NT
ER
. The
n m
ove
the
curs
or t
o th
e
poin
t w
here
x =
1 a
nd p
ress
EN
TE
R. T
he c
alcu
lato
r w
ill s
hade
the
re
gion
and
dis
play
its
appr
oxim
ate
area
. Wha
t nu
mbe
r do
es t
he
calc
ulat
or d
ispl
ay fo
r th
e ar
ea o
f the
sha
ded
regi
on?
0.6
82
68
94
9
2. E
nter
2
nd
[DR
AW
] 1. T
his
caus
es t
he c
alcu
lato
r to
cle
ar t
he s
hadi
ng a
nd
redi
spla
y th
e gr
aph.
Fin
d th
e ar
ea o
f the
reg
ion
boun
ded
by t
he c
urve
, th
e x-
axis
, and
the
ver
tica
l lin
es x
= -
2 an
d x
= 2
. 0.9
5449974
3. F
ind
the
area
of t
he r
egio
n bo
unde
d by
the
cur
ve, t
he x
-axi
s, a
nd t
he v
erti
cal
lines
x =
-3
and
x =
3.
0.9
97
30
02
4. W
itho
ut u
sing
a c
alcu
lato
r, e
stim
ate
the
area
of t
he r
egio
n bo
unde
d by
the
cu
rve,
the
x-a
xis,
and
the
ver
tica
l lin
es x
= -
4 an
d x
= 4
to
four
dec
imal
pl
aces
. Ver
ify y
our
conj
ectu
re.
Sa
mp
le a
ns
we
r: 1
; s
ee
stu
de
nts
’ w
ork
.
11-3
Grap
hing
Cal
cula
tor
Activ
ityTh
e S
tan
dard
No
rmal
Cu
rve
[ -4.
7, 4
.7] s
cl: 1
by
[ -0.
2, 0
.5] s
cl: 0
.1
005_
042_
PC
CR
MC
11_8
9381
2.in
dd21
11/1
4/09
5:03
:38
PM
A01_A19_PCCRMC11_893812.indd 9A01_A19_PCCRMC11_893812.indd 9 11/19/09 8:00:54 PM11/19/09 8:00:54 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 11 A10 Glencoe Precalculus
Answers (Lesson 11-4)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
22
Gle
ncoe
Pre
calc
ulus
11-4
Stud
y Gu
ide
and
Inte
rven
tion
Th
e C
en
tral
Lim
it T
heo
rem
The
Cent
ral L
imit
The
orem
The
Cen
tral
Lim
it T
heor
em s
tate
s th
at a
s th
e sa
mpl
ing
size
n in
crea
ses:
•
the
shap
e of
the
dis
trib
utio
n of
the
sam
ple
mea
ns o
f a p
opul
atio
n w
ith
mea
n μ
and
stan
dard
dev
iati
on σ
will
app
roac
h a
norm
al d
istr
ibut
ion
and
•
the
corr
espo
ndin
g di
stri
buti
on w
ill h
ave
a m
ean
μ a
nd s
tand
ard
devi
atio
n σ
−
x =
σ
−
√
�
n
.
The
z-va
lue
for
a sa
mpl
e m
ean
in a
pop
ulat
ion
is g
iven
by
z =
−
x -
μ
−
σ −
x , w
here
−
x is
the
sam
ple
mea
n, μ
is t
he m
ean
of t
he p
opul
atio
n, a
nd σ
−
x is
the
stan
dard
err
or.
A
stu
dy w
as d
one
in w
hich
par
ents
rep
orte
d th
e nu
mbe
r of
hou
rs p
er d
ay t
hat
thei
r ch
ildr
en, b
etw
een
the
ages
of
2.0
and
5.9,
wat
ched
tel
evis
ion
or v
ideo
s. T
he m
ean
age
of t
he c
hild
ren
was
3.3
yea
rs w
ith
a st
anda
rd d
evia
tion
of
0.9
year
. Ass
ume
that
the
va
riab
le is
nor
mal
ly d
istr
ibut
ed. I
f a
rand
om s
ampl
e of
30
chil
dren
in
thi
s st
udy
is s
elec
ted,
fin
d th
e pr
obab
ilit
y th
at t
he m
ean
age
is
less
tha
n 3
year
s.
The
dist
ribu
tion
of s
ampl
e m
eans
will
be
appr
oxim
atel
y no
rmal
wit
h μ
= 3
.3 a
nd
σ
−
x =
0.9
−
√
��
30
or
abou
t 0.
164.
Fin
d th
e z-
valu
e.
z =
−
x -
μ
−
σ −
x
z-va
lue
fo
r a
sa
mp
le m
ea
n
z =
3 -
3.3
−
0.16
4 or
abo
ut -
1.83
−
x =
3,
μ =
3.3
, σ
−
x =
0.1
64
The
area
to
the
left
of a
z-v
alue
of -
1.83
is 0
.033
59.
The
prob
abili
ty t
hat
the
mea
n ag
e of
the
sam
ple
is le
ss t
han
3 ye
ars
or P
( −
x <
3) i
s ab
out
3.36
%.
Exer
cise
s 1.
The
mea
n pi
tch
of t
he e
xper
t sl
opes
at
the
ski r
esor
ts in
a c
erta
in r
egio
n is
25˚
wit
h a
stan
dard
dev
iati
on o
f 3˚. A
ssum
e th
at t
he v
aria
ble
is
norm
ally
dis
trib
uted
. If 2
0 ex
pert
slo
pes
are
chos
en a
t ra
ndom
, wha
t is
th
e pr
obab
ility
tha
t th
e m
ean
of t
he s
lope
s w
ill b
e gr
eate
r th
an
26.3
˚?
2.6
2%
2. A
vet
erin
aria
n re
port
s th
at t
he a
vera
ge a
ge o
f the
cat
s th
at s
he t
reat
s is
96
mon
ths
wit
h a
stan
dard
dev
iati
on o
f 16
mon
ths.
If a
ran
dom
sam
ple
of
36 o
f her
cat
pat
ient
s is
sel
ecte
d, fi
nd t
he p
roba
bilit
y th
at t
he m
ean
age
is
betw
een
90 a
nd 1
00 m
onth
s. 9
2.1
%
Exam
ple
005_
042_
PC
CR
MC
11_8
9381
2.in
dd22
11/1
4/09
5:06
:47
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-4
NA
ME
DA
TE
PE
RIO
D
Lesson 11-4
Ch
ap
ter
11
23
Gle
ncoe
Pre
calc
ulus
11-4
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Th
e C
en
tral
Lim
it T
heo
rem
The
Nor
mal
App
roxi
mat
ion
The
norm
al d
istr
ibut
ion
can
be u
sed
to a
ppro
xim
ate
a bi
nom
ial d
istr
ibut
ion
if:
•
the
orig
inal
var
iabl
e is
nor
mal
ly d
istr
ibut
ed o
r n
≥ 3
0 an
d
•
np ≥
5 a
nd n
q ≥
5,
whe
re n
is t
he n
umbe
r of
tri
als,
p is
the
pro
babi
lity
of s
ucce
ss, a
nd q
is t
hepr
obab
ility
of f
ailu
re.
Whe
n us
ing
the
norm
al d
istr
ibut
ion
to a
ppro
xim
ate
a bi
nom
ial d
istr
ibut
ion,
the
cont
inui
ty c
orre
ctio
n fa
ctor
mus
t be
use
d. T
his
invo
lves
add
ing
0.5
unit
to
or
subt
ract
ing
0.5
unit
from
a g
iven
dis
cret
e bo
unda
ry.
T
en p
erce
nt o
f th
e m
embe
rs o
f a
golf
leag
ue a
re y
oung
erth
an 3
0. I
f 20
0 m
embe
rs a
re r
ando
mly
sel
ecte
d, f
ind
the
prob
abil
ity
that
mor
e th
an 1
0 w
ill b
e yo
unge
r th
an 3
0.
Step
1
Find
the
mea
n μ
and
sta
ndar
d de
viat
ion
σ o
f the
bin
omia
l di
stri
buti
on.
μ =
np
σ =
√
��
np
q
=
200
· 0.
10 o
r 20
= √
��
��
��
�
200
· 0.
10 ·
0.90
or
abou
t 4.
24
Si
nce
np =
20
and
nq =
180
, the
nor
mal
dis
trib
utio
n ca
n be
use
d.
Step
2
Wri
te t
he p
robl
em in
pro
babi
lity
nota
tion
: P ( X
> 1
0) .
Step
3
Rew
rite
the
pro
blem
wit
h th
e co
ntin
uity
fact
or in
clud
ed. S
ince
the
qu
esti
on is
ask
ing
for
the
prob
abili
ty t
hat
mor
e th
an 1
0 m
embe
rs
will
be
youn
ger
than
30,
add
0.5
to
10.
P (
X >
10)
= P
( X >
10
+ 0
.5)
or
P (X
> 1
0.5)
Step
4
Find
the
cor
resp
ondi
ng z
-val
ue fo
r X
= 1
0.5.
z =
X -
μ
−
σ
= 10
.5 -
20
−
4.24
≈ -
2.24
Step
5
Find
the
cor
resp
ondi
ng a
rea
abov
e z
= -
2.24
. It
isab
out
0.98
7. T
he p
roba
bilit
y is
abo
ut 9
8.7%
.
Exer
cise
s 1.
A s
tudy
foun
d th
at 4
0% o
f all
of a
tow
n’s
citi
zens
app
rove
of a
new
tra
in s
tati
on.
If 5
0 ci
tize
ns a
re r
ando
mly
sel
ecte
d, fi
nd t
he p
roba
bilit
y th
at fe
wer
tha
n 20
cit
izen
s ap
prov
e of
a n
ew t
rain
sta
tion
. 4
4.2
%
2. A
bas
ebal
l pla
yer
gets
a h
it 3
2% o
f the
tim
e. F
ind
the
prob
abili
ty t
hat
the
play
er w
ill g
et a
t le
ast
26 h
its
in h
is n
ext
100
tim
es a
t ba
t. 9
1.8
%
Exam
ple
005_
042_
PC
CR
MC
11_8
9381
2.in
dd23
11/1
4/09
5:07
:35
PM
A01_A19_PCCRMC11_893812.indd 10A01_A19_PCCRMC11_893812.indd 10 11/17/09 9:21:50 AM11/17/09 9:21:50 AM
Copyright
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 A11 Glencoe Precalculus
An
swer
s
Answers (Lesson 11-4)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
24
Gle
ncoe
Pre
calc
ulus
11-4
Prac
tice
Th
e C
en
tral
Lim
it T
heo
rem
1. N
URS
ING
The
mea
n sa
lary
for
nurs
es in
a c
ity
is $
52,1
29 w
ith
a st
anda
rd d
evia
tion
of $
1800
. Wha
t is
the
pro
babi
lity
that
the
mea
n sa
lary
for
a ra
ndom
ly s
elec
ted
grou
p of
50
nurs
es in
the
cit
y is
gre
ater
tha
n $5
2,50
0?
7
.21
%
2. U
TILI
TIES
The
ave
rage
ele
ctri
c bi
ll in
a r
esid
enti
al a
rea
in J
une
is $
72. A
ssum
eth
is v
aria
ble
is n
orm
ally
dis
trib
uted
wit
h a
stan
dard
dev
iati
on o
f $6.
Fin
d th
epr
obab
ility
tha
t th
e m
ean
elec
tric
bill
for
a ra
ndom
ly s
elec
ted
grou
p of
15
resi
dent
sis
less
tha
n $7
5.
9
7.3
8%
3. S
HO
ES T
he p
rice
s of
sho
es in
a s
tore
are
nor
mal
ly d
istr
ibut
ed w
ith
a m
ean
of$9
3 an
d a
stan
dard
dev
iati
on o
f $18
. If n
ine
pair
s of
sho
es a
re r
ando
mly
sel
ecte
d,fin
d th
e pr
obab
ility
tha
t th
e m
ean
cost
is b
etw
een
$100
and
$11
0.
1
1.8
6%
4. C
OLL
EGE
Of t
he t
otal
pop
ulat
ion
at a
sm
all c
olle
ge, 2
0% a
re fr
om t
he M
id-A
tlan
tic
stat
es. I
f 200
stu
dent
s ar
e ra
ndom
ly s
elec
ted,
find
the
pro
babi
lity
that
at
leas
t50
are
from
the
Mid
-Atl
anti
c st
ates
.
4
.64
%
5. W
EATH
ER K
yle
has
rese
arch
ed t
he a
vera
ge a
nnua
l pre
cipi
tati
on in
his
cit
y fo
rse
vera
l yea
rs a
nd c
alcu
late
d th
e m
ean
as 1
9.32
inch
es. A
ssum
e th
e av
erag
epr
ecip
itat
ion
is n
orm
ally
dis
trib
uted
and
the
sta
ndar
d de
viat
ion
is 2
.44
inch
es.
a. I
f one
yea
r fr
om t
he t
ime
peri
od t
hat
Kyl
e re
sear
ched
is r
ando
mly
sel
ecte
d,w
hat
is t
he p
roba
bilit
y th
at t
he p
reci
pita
tion
is m
ore
than
18
inch
es?
7
0.5
4%
b. I
f fiv
e ye
ars
from
the
tim
e pe
riod
tha
t K
yle
rese
arch
ed a
re r
ando
mly
sel
ecte
d,w
hat
is t
he p
roba
bilit
y th
at t
he m
ean
prec
ipit
atio
n is
mor
e th
an 1
8 in
ches
?
8
8.6
9%
6. G
ROCE
RY E
ight
y-fiv
e pe
rcen
t of
the
sho
pper
s at
a g
roce
ry s
tore
hav
e a
freq
uent
-buy
er c
ard.
Thi
rty-
five
shop
pers
are
ran
dom
ly s
elec
ted
for
a ta
ste
test
. Wha
t is
the
pro
babi
lity
that
at
leas
t 25
and
at
mos
t 30
of t
he t
aste
test
ers
have
a fr
eque
nt-b
uyer
car
d?
6
3.4
%
Fin
d th
e m
inim
um s
ampl
e si
ze n
eede
d fo
r ea
ch p
roba
bili
ty s
o th
at t
he n
orm
al
dist
ribu
tion
can
be
used
to
appr
oxim
ate
the
bino
mia
l dis
trib
utio
n.
7. p
= 0
.6
8. p
= 0
.15
1
3
3
4
Ch
ap
ter
11
24
Gle
ncoe
Pre
calc
ulus
005_
042_
PC
CR
MC
11_8
9381
2.in
dd24
11/1
4/09
5:10
:57
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-4
NA
ME
DA
TE
PE
RIO
D
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson 11-4
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
25
Gle
ncoe
Pre
calc
ulus
Ch
ap
ter
11
25
Gle
ncoe
Pre
calc
ulus
Wor
d Pr
oble
m P
ract
ice
Th
e C
en
tral
Lim
it T
heo
rem
11-4
1. H
EATI
NG
Wor
kers
at
a pu
blic
uti
litie
s co
mpa
ny s
urve
yed
200
of t
heir
cus
tom
ers
and
foun
d th
at t
he m
ean
tem
pera
ture
at
whi
ch t
he c
usto
mer
s’ th
erm
osta
ts w
ere
set
in J
anua
ry w
as 7
0˚ w
ith
a st
anda
rd
devi
atio
n of
1.8
˚. I
f 30
cust
omer
s ar
e ra
ndom
ly s
elec
ted
for
a fo
llow
-up
surv
ey, f
ind
the
prob
abili
ty t
hat
the
mea
n te
mpe
ratu
re a
t w
hich
the
y se
t th
eir
ther
mos
tats
in J
anua
ry is
less
than
69.
5˚.
6
.30
%
2. M
ILK
A fo
od s
tudy
in a
cit
y fo
und
that
th
e pr
ice
of a
gal
lon
of w
hole
milk
in it
s st
ores
was
$3.
72. T
his
vari
able
is
norm
ally
dis
trib
uted
wit
h a
stan
dard
de
viat
ion
of $
0.08
.
a. I
f a g
allo
n of
who
le m
ilk is
ran
dom
ly
sele
cted
from
one
sto
re, f
ind
the
prob
abili
ty t
hat
it c
osts
less
tha
n $3
.70.
4
0.1
3%
b. I
f a g
allo
n of
milk
is r
ando
mly
se
lect
ed fr
om e
ach
of 4
0 di
ffere
nt
stor
es, f
ind
the
prob
abili
ty t
hat
the
mea
n co
st o
f the
sam
ple
is le
ss t
han
$3.7
0.
6
.17
%
3. B
ICY
CLES
Thi
rty-
seve
n pe
rcen
t of
the
re
side
nts
in a
nei
ghbo
rhoo
d ow
n bi
cycl
es.
If a
gro
up o
f 40
resi
dent
s ar
e ra
ndom
ly
sele
cted
, wha
t is
the
pro
babi
lity
that
at
leas
t ha
lf ow
n bi
cycl
es?
6
.17
%
4. T
ESTI
NG
The
ave
rage
tim
e it
tak
es a
gr
oup
of s
tude
nts
to c
ompl
ete
a re
adin
g te
st is
46.
2 m
inut
es w
ith
a st
anda
rd
devi
atio
n of
8 m
inut
es. T
he t
imes
are
no
rmal
ly d
istr
ibut
ed.
a. A
gro
up o
f 10
stud
ents
is r
ando
mly
se
lect
ed. F
ind
the
prob
abili
ty t
hat
the
mea
n ti
me
to c
ompl
ete
the
test
is
mor
e th
an 4
5 m
inut
es.
6
8.2
2%
b. H
ow d
oes
your
ans
wer
to
part
ach
ange
if t
he g
roup
con
sist
s of
15 s
tude
nts?
T
he
pro
ba
bil
ity
in
cre
as
es
by
3.7
%.
5. C
AFE
TERI
A C
olle
ge fr
eshm
en w
ere
aske
d if
they
ate
bre
akfa
st o
n Su
nday
m
orni
ng in
the
caf
eter
ia. T
he g
raph
sh
ows
the
perc
ent
of m
ales
and
fem
ales
w
ho s
aid
yes.
Fem
aleM
ale
4060 20 0
Percent
80
a. O
ut o
f 30
rand
omly
sel
ecte
d fe
mal
e fr
eshm
en, w
hat
is t
he a
ppro
xim
ate
prob
abili
ty t
hat
at le
ast
25 o
f the
m
ate
brea
kfas
t in
the
caf
eter
ia o
n Su
nday
mor
ning
?
a
bo
ut
46
%
b. O
ut o
f 30
rand
omly
sel
ecte
d m
ale
fres
hmen
, wha
t is
the
app
roxi
mat
e pr
obab
ility
tha
t at
leas
t 25
of t
hem
at
e br
eakf
ast
in t
he c
afet
eria
on
Sund
ay m
orni
ng?
a
bo
ut
3%
005_
042_
PC
CR
MC
11_8
9381
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Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
Chapter 11 A12 Glencoe Precalculus
Answers (Lesson 11-4 and Lesson 11-5)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
26
Gle
ncoe
Pre
calc
ulus
11-4
Exp
eri
en
ce t
he C
en
tral
Lim
it T
heo
rem
The
Cen
tral
Lim
it T
heor
em is
bes
t ap
prec
iate
d w
hen
expe
rien
ced.
Thi
s ac
tivi
tyw
ill a
llow
you
thi
s pr
ivile
ge. Y
ou m
ay w
ish
to w
ork
wit
h a
part
ner.
Si
mul
ate
spin
ning
a s
pinn
er w
ith
equa
l sec
tion
s la
bele
d 1–
10tw
o ti
mes
by
sele
ctin
g th
e ra
ndIn
t( c
omm
and
from
the
PR
B m
enu
afte
r pr
essi
ng M
AT
H. T
he s
cree
n sh
own
indi
cate
sth
e sp
inne
r la
nded
on
7 on
the
firs
t sp
in a
nd 1
on
the
seco
nd s
pin.
The
mea
n of
the
se s
pins
is 4
.
1.
Com
plet
e 25
sim
ulat
ions
and
list
the
mea
ns o
f the
spi
ns.
An
sw
ers
wil
l v
ary
.
2.
Find
the
mea
n an
d st
anda
rd d
evia
tion
of t
he m
eans
in E
xerc
ise
1.
An
sw
ers
wil
l v
ary
.
Now
sim
ulat
e sp
inni
ng t
he s
ame
spin
ner
four
tim
es.
3.
Com
plet
e 25
sim
ulat
ions
and
list
the
mea
ns o
f the
spi
ns.
An
sw
ers
wil
l v
ary
.
4.
Find
the
mea
n an
d st
anda
rd d
evia
tion
of t
he m
eans
in E
xerc
ise
3.
An
sw
ers
wil
l v
ary
.
Now
sim
ulat
e sp
inni
ng t
he s
ame
spin
ner
six
tim
es.
5.
Com
plet
e 25
sim
ulat
ions
and
list
the
mea
ns o
f the
spi
ns.
An
sw
ers
wil
l v
ary
.
6.
Find
the
mea
n an
d st
anda
rd d
evia
tion
of t
he m
eans
in E
xerc
ise
5.
An
sw
ers
wil
l v
ary
.
7. W
hat
do y
ou n
otic
e ab
out
the
mea
ns in
Exe
rcis
es 2
, 4, a
nd 6
?
Th
ey
ap
pro
ac
h t
he
me
an
of
the
da
ta s
et,
5.5
.
8.
Wha
t do
you
not
ice
abou
t th
e st
anda
rd d
evia
tion
s in
Exe
rcis
es 2
, 4, a
nd 6
?
Th
ey
ge
t s
ma
lle
r a
s t
he
nu
mb
er
of
sp
ins
in
cre
as
es
.
9.
Find
σ
−
√
�
n
for
n =
2, 4
, and
6 g
iven
σ =
2.8
7. W
hat
do y
ou n
otic
e?
Th
ey
are
clo
se
to
th
e s
am
ple
sta
nd
ard
de
via
tio
ns
fo
r 2
, 4
, a
nd
6 s
pin
s.
10. I
f n =
1, t
hat
is t
he s
pinn
er is
spu
n on
ce, w
hat
is t
he d
istr
ibut
ion
of t
he s
ampl
es?
How
doe
s it
cha
nge
as n
incr
ease
s?
Th
e d
istr
ibu
tio
n i
s u
nif
orm
; it
be
co
me
s n
orm
al
as
n i
nc
rea
se
s.
Enri
chm
ent
005_
042_
PC
CR
MC
11_8
9381
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dd26
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/09
12:3
8:18
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-5
NA
ME
DA
TE
PE
RIO
D
Lesson 11-5
Ch
ap
ter
11
27
Gle
ncoe
Pre
calc
ulus
Nor
mal
Dis
trib
utio
n Th
e co
nfid
ence
leve
l c is
the
pro
babi
lity
that
an
inte
rval
es
tim
ate
cont
ains
a g
iven
par
amet
er, s
uch
as a
pop
ulat
ion
mea
n. T
he m
axim
um e
rror
of
esti
mat
e E
is t
he m
axim
um d
iffer
ence
bet
wee
n th
e po
int
esti
mat
e an
d th
e ac
tual
val
ue o
f th
e pa
ram
eter
. A c
onfi
denc
e in
terv
al C
I is
a s
peci
fic in
terv
al e
stim
ate
of a
par
amet
er.
For
a po
pula
tion
mea
n:
CI
= −
x ±
E o
r −
x ±
z ·
σ
−
√
�
n
.
z is
the
cri
tica
l val
ue t
hat
corr
espo
nds
to a
cer
tain
con
fiden
ce le
vel.
The
mos
t co
mm
on a
re
show
n be
low
.
Co
nfi
de
nc
e L
ev
el
90
%9
5%
99
%
z-V
alu
e1
.64
51
.96
02
.57
6
Whe
n σ
is u
nkno
wn,
the
sam
ple
stan
dard
dev
iati
on s
can
be
subs
titu
ted
for
σ, p
rovi
ded
the
sam
ple
size
n is
gre
ater
tha
n or
equ
al t
o 30
.
In
a s
ampl
e of
50
peop
le w
ho b
uy m
agaz
ines
, a r
esea
rche
r fi
nds
the
mea
n am
ount
spe
nt p
er m
onth
to
be $
12. A
ssum
e a
stan
dard
dev
iati
on o
f $4
.50.
Fin
d th
e 95
% c
onfi
denc
e in
terv
al f
or t
he m
ean
amou
nt s
pent
for
m
agaz
ines
eac
h m
onth
.C
I =
−
x ±
z ·
σ
−
√
�
n
C
on
fide
nce
In
terv
al f
or
the
Me
an
= 1
2 ±
1.9
6 · 4.
50
−
√
��
50
−
x =
12
, z
= 1
.96
, σ
= 4
.50
, a
nd
n =
50
≈ 1
2 ±
1.2
5 S
imp
lify.
Add
and
sub
trac
t th
e m
argi
n of
err
or.
Left
Bou
ndar
y: 1
2 -
1.2
5 =
10.
75
Rig
ht B
ound
ary:
12
+ 1
.25
= 1
3.25
The
95%
con
fiden
ce in
terv
al is
10.
75 <
μ <
13.
25. T
here
fore
, wit
h 95
% c
onfid
ence
, th
e m
ean
amou
nt s
pent
on
mag
azin
es p
er m
onth
is b
etw
een
$10.
75 a
nd $
13.2
5.
Exer
cise
s
The
num
ber
of d
ays
wit
h te
mpe
ratu
res
abov
e fr
eezi
ng f
or a
sam
ple
of 3
5 ci
ties
ha
d a
mea
n of
190
.7 d
ays
and
a sa
mpl
e st
anda
rd d
evia
tion
of
54.2
day
s.
1. F
ind
the
95%
con
fiden
ce in
terv
al fo
r th
e m
ean
num
ber
of d
ays
wit
h te
mpe
ratu
res
abov
e fr
eezi
ng.
1
72
.74
< μ
< 2
08
.66
2. F
ind
the
90%
con
fiden
ce in
terv
al fo
r th
e m
ean
num
ber
of d
ays
wit
h te
mpe
ratu
res
abov
e fr
eezi
ng.
1
75
.63
< μ
< 2
05
.77
Stud
y Gu
ide
and
Inte
rven
tion
Co
nfi
den
ce I
nte
rvals
Exam
ple
11-5
005_
042_
PC
CR
MC
11_8
9381
2.in
dd27
3/25
/09
12:3
8:25
PM
A01_A19_PCCRMC11_893812.indd 12A01_A19_PCCRMC11_893812.indd 12 3/25/09 2:47:52 PM3/25/09 2:47:52 PM
Copyright
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 A13 Glencoe Precalculus
An
swer
s
Answers (Lesson 11-5)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
28
Gle
ncoe
Pre
calc
ulus
t-D
istr
ibut
ion
To fi
nd a
con
fiden
ce in
terv
al w
hen
σ is
unk
now
n an
d th
e sa
mpl
e si
ze
n is
less
tha
n 30
, use
the
t-d
istr
ibut
ion,
pro
vide
d th
e va
riab
le is
app
roxi
mat
ely
norm
ally
di
stri
bute
d. T
he fo
rmul
a fo
r us
ing
the
t-di
stri
buti
on t
o co
nstr
uct
a co
nfid
ence
inte
rval
is
CI
= −
x ±
t ·
s −
√
�
n
,
whe
re −
x is
the
sam
ple
mea
n, t
is a
cri
tica
l val
ue w
ith
n -
1 d
egre
es o
f fre
edom
, s is
the
sa
mpl
e st
anda
rd d
evia
tion
, and
n is
the
sam
ple
size
.
T
he m
ean
pric
e of
9 r
ando
mly
sel
ecte
d te
levi
sion
s at
an
elec
tron
ics
stor
e is
$78
3 w
ith
a st
anda
rd d
evia
tion
of
$116
. Fin
d th
e 90
% c
onfi
denc
e in
terv
al
of t
he m
ean
pric
e of
all
of
the
tele
visi
ons
at t
he s
tore
. Ass
ume
that
the
var
iabl
e is
no
rmal
ly d
istr
ibut
ed.
Bec
ause
the
pop
ulat
ion
stan
dard
dev
iati
on is
unk
now
n an
d n
< 3
0, u
se t
he t-
dist
ribu
tion
. B
ecau
se n
= 9
, the
re a
re 9
- 1
or
8 de
gree
s of
free
dom
.
To fi
nd t
he t-
valu
e, d
eter
min
e th
e ar
ea in
the
low
er t
ail o
f the
di
stri
buti
on. S
ince
100
% -
90%
or
10%
is in
the
tai
ls, t
hen
5%
is in
eac
h ta
il. P
ress
2
nd
DIS
TR
and
cho
ose
invT
(. Ty
pe t
hear
ea t
o th
e le
ft o
f eac
h cr
itic
al v
alue
, as
a de
cim
al, f
ollo
wed
by
the
degr
ees
of fr
eedo
m.
CI
= −
x
± t
·
s −
√
�
n
C
on
fide
nce
inte
rva
l usi
ng
t-d
istr
ibu
tion
≈
783
± 1
.860
· 11
6 −
√
�
9 −
x =
78
3,
t ≈ 1
.86
0,
s =
11
6,
an
d n
= 9
≈
783
± 7
1.92
S
imp
lify.
Ther
efor
e, t
he 9
0% c
onfid
ence
inte
rval
is $
711.
08 <
μ <
$85
4.92
.
Exer
cise
s
1. A
t a
man
ufac
turi
ng p
lant
, the
thr
eads
on
ten
rand
omly
sel
ecte
d sc
rew
s ha
ve a
m
ean
dept
h of
0.3
2 in
ch a
nd a
sta
ndar
d de
viat
ion
of 0
.03
inch
. Fin
d th
e 95
%
conf
iden
ce in
terv
al o
f the
mea
n de
pth
of a
ll th
e sc
rew
s, a
ssum
ing
that
the
var
iabl
e is
no
rmal
ly d
istr
ibut
ed.
0
.30
< μ
< 0
.34
2. A
n en
viro
nmen
tal s
tudy
invo
lves
cou
ntin
g th
e nu
mbe
r of
ligh
t bu
lbs
in r
ando
mly
sel
ecte
d ro
oms
of a
bui
ldin
g. A
bui
ldin
g m
anag
er c
ount
s th
e bu
lbs
in 1
6 ro
oms
and
finds
the
mea
n nu
mbe
r of
bul
bs t
o be
21
wit
h a
stan
dard
dev
iati
on o
f 4.8
. Fin
d th
e 99
% c
onfid
ence
in
terv
al o
f the
mea
n nu
mbe
r of
bul
bs in
all
the
room
s of
the
bui
ldin
g, a
ssum
ing
the
vari
able
is n
orm
ally
dis
trib
uted
.
1
7.4
6 <
μ <
24
.54
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Co
nfi
den
ce I
nte
rvals
Exam
ple
11-5
005_
042_
PC
CR
MC
11_8
9381
2.in
dd28
3/25
/09
12:3
8:29
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-5
NA
ME
DA
TE
PE
RIO
D
Lesson 11-5
Ch
ap
ter
11
29
Gle
ncoe
Pre
calc
ulus
1. C
HIL
D C
ARE
A r
ando
m s
ampl
e of
50
pare
nts
of y
oung
chi
ldre
n fo
und
that
the
y sp
end
a m
ean
of $
7648
eac
h ye
ar fo
r ch
ild c
are.
The
sta
ndar
d de
viat
ion
of t
he s
ampl
e w
as $
630.
Fi
nd t
he 9
0% c
onfid
ence
inte
rval
of t
he m
ean
annu
al c
ost
of c
hild
car
e.
$7
50
1 <
μ <
$7
79
5
2. F
ITN
ESS
Twen
ty-e
ight
peo
ple
who
enr
olle
d in
a fi
tnes
s pr
ogra
m lo
st a
mea
n of
14
.3 p
ound
s w
ith
a st
anda
rd d
evia
tion
of 2
pou
nds.
Fin
d th
e 95
% c
onfid
ence
inte
rval
of
the
mea
n w
eigh
t lo
ss in
pou
nds
of a
ll of
the
mem
bers
enr
olle
d in
the
pro
gram
. 1
3.5
< μ
< 1
5.1
3. C
OM
MU
TE T
he a
vera
ge n
umbe
r of
min
utes
it t
akes
8 p
eopl
e to
com
mut
e to
and
from
w
ork
duri
ng r
ush
hour
is s
how
n. A
ssum
e th
at t
he t
imes
are
nor
mal
ly d
istr
ibut
ed.
Co
mm
uti
ng
Tim
e (
min
ute
s)
55
70
60
55
60
56
55
60
a. D
ecid
e th
e ty
pe o
f dis
trib
utio
n th
at c
an b
e us
ed t
o es
tim
ate
the
com
mut
ing
tim
e m
ean.
Exp
lain
you
r re
ason
ing.
Us
e a
t-d
istr
ibu
tio
n b
ec
au
se
σ i
s u
nk
no
wn
an
d n
< 3
0.
b. C
alcu
late
the
mea
n an
d st
anda
rd d
evia
tion
to
the
near
est
hund
redt
h.
58
.88
min
; 5
.08
min
c. C
onst
ruct
a 9
0% c
onfid
ence
inte
rval
for
the
aver
age
com
mut
ing
tim
e in
min
utes
for
a co
mm
uter
from
thi
s ci
ty.
55
.48
< μ
< 6
2.2
8
4. T
RAIN
ING
In
a su
rvey
of 4
42 e
mpl
oyee
s at
a c
all c
ente
r, t
he m
ean
tim
e th
at e
mpl
oyee
s fe
lt w
as n
eede
d fo
r ad
equa
te t
rain
ing
for
thei
r jo
bs w
as 7
day
s. T
he s
ampl
e st
anda
rd
devi
atio
n w
as 1
.5 d
ays.
Con
stru
ct a
98%
con
fiden
ce in
terv
al fo
r th
e am
ount
of t
rain
ing
tim
e th
at e
mpl
oyee
s fe
lt w
as a
dequ
ate
to b
egin
the
ir jo
bs.
6.8
< μ
< 7
.2
Det
erm
ine
whe
ther
the
nor
mal
dis
trib
utio
n or
t-d
istr
ibut
ion
shou
ld b
e us
ed fo
r ea
ch q
uest
ion.
The
n fi
nd e
ach
conf
iden
ce in
terv
al g
iven
the
follo
win
g in
form
atio
n.
5. 9
5%;
−
x =
115
, s =
6, n
= 6
t-
dis
trib
uti
on
; 1
08
.7 <
μ <
12
1.3
6. 9
6%;
−
x =
18.
5, s
= 1
.2, n
= 4
0 n
orm
al
dis
trib
uti
on
; 1
8.1
1 <
μ <
18
.89
7. 9
9%;
−
x =
236
, σ =
8, n
= 4
5 n
orm
al
dis
trib
uti
on
; 2
32
.9 <
μ <
23
9.1
8. F
OO
D T
he o
wne
rs o
f a s
andw
ich
shop
wan
t to
find
the
95%
con
fiden
ce in
terv
al o
f the
tr
ue m
ean
cost
of a
ham
burg
er in
the
ir c
ity.
How
larg
e sh
ould
the
ir s
ampl
e be
if t
hey
wan
t to
be
accu
rate
to
wit
hin
15 c
ents
? In
an
earl
ier
surv
ey, t
he s
tand
ard
devi
atio
n of
th
e pr
ice
was
26
cent
s.
at
lea
st
12
ob
se
rva
tio
ns
9. S
PORT
S A
tea
cher
wan
ts t
o es
tim
ate
the
aver
age
num
ber
of h
ours
per
wee
k th
at h
er
stud
ents
are
at
prac
tice
or
at g
ames
for
a sp
orts
tea
m. T
he s
tand
ard
devi
atio
n fr
om a
pr
evio
us y
ear
was
6.2
hou
rs. H
ow la
rge
of a
sam
ple
mus
t sh
e se
lect
if s
he w
ants
to
be
99%
con
fiden
t of
find
ing
the
aver
age
amou
nt o
f tim
e st
uden
ts s
pent
par
tici
pati
ng in
sp
orts
wit
hin
1.5
hour
s?
at
lea
st
11
4 o
bs
erv
ati
on
s
Prac
tice
Co
nfi
den
ce I
nte
rvals
11-5
005_
042_
PC
CR
MC
11_8
9381
2.in
dd29
11/1
4/09
5:24
:25
PM
A01_A19_PCCRMC11_893812.indd 13A01_A19_PCCRMC11_893812.indd 13 11/17/09 9:06:13 AM11/17/09 9:06:13 AM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 11 A14 Glencoe Precalculus
Answers (Lesson 11-5)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
30
Gle
ncoe
Pre
calc
ulus
1. R
EAD
ING
A s
ampl
e of
nor
mal
ly
dist
ribu
ted
read
ing
scor
es o
f for
ty
eigh
th-g
rade
stu
dent
s ha
s a
mea
n of
82
and
a s
tand
ard
devi
atio
n of
15.
Fin
d th
e 95
% c
onfid
ence
inte
rval
for
the
mea
n of
all
of t
he r
eadi
ng s
core
s.
7
7.4
< μ
< 8
6.6
2. C
HO
LEST
ERO
L Th
e se
rum
cho
lest
erol
le
vel w
as c
olle
cted
for
a gr
oup
of52
5 co
llege
wom
en. T
he m
ean
of t
he
sam
ple
was
191
.7
mg
−
100
mL w
ith
a
stan
dard
dev
iati
on o
f 41.
0.
a. C
onst
ruct
a 9
0% c
onfid
ence
leve
l for
th
e m
ean
seru
m c
hole
ster
ol le
vel.
1
88
.8 <
μ <
19
4.6
b. C
onst
ruct
a 9
5% c
onfid
ence
leve
l for
th
e m
ean
seru
m c
hole
ster
ol le
vel.
1
88
.2 <
μ <
19
5.2
c. S
uppo
se y
ou h
ear
a cl
aim
tha
t th
e m
ean
seru
m c
hole
ster
ol le
vel f
or
wom
en in
col
lege
is 2
00. W
hat
wou
ld
be y
our
reac
tion
bas
ed o
n yo
ur
answ
ers
to p
arts
a a
nd b
? W
hy?
S
am
ple
an
sw
er:
Th
e c
laim
is
in
co
rre
ct;
ne
ith
er
of
the
in
terv
als
co
nta
ins
20
0,
so
a
lev
el
of
20
0 i
s n
ot
lik
ely
.
3. C
AR
POLL
UTI
ON
Sev
en c
ars
wer
e te
sted
for
nitr
ogen
-oxi
de e
mis
sion
s.
The
resu
lts
are
show
n in
the
tab
le.
Em
iss
ion
s (
gra
ms
pe
r m
ile
)
0.0
50
.12
0.1
60
.15
0.1
40
.19
0.1
4
a. F
ind
the
mea
n of
the
dat
a.
0.1
36
b. F
ind
the
stan
dard
dev
iati
on o
f th
e da
ta.
0.0
44
c. F
ind
the
99%
con
fiden
ce in
terv
al o
f th
e m
ean
nitr
ogen
-oxi
de e
mis
sion
s.
0.0
74
7 <
μ <
0.1
96
7
4. F
UEL
CO
NSU
MPT
ION
The
mea
n an
d st
anda
rd d
evia
tion
for
city
and
hig
hway
fu
el c
onsu
mpt
ion
in m
iles
per
gallo
n fo
r 33
ran
dom
ly s
elec
ted
pre-
owne
d ca
rs o
n a
deal
er’s
lot
is s
how
n. A
ssum
e th
e va
riab
les
are
norm
ally
dis
trib
uted
.
−
x s
City
21
.35
4.1
3
Hig
hw
ay
29
.65
3.6
5
a. F
ind
the
98%
con
fiden
ce in
terv
al
for
the
mea
n fu
el c
onsu
mpt
ion
in
the
city
. 1
9.6
77
< μ
< 2
3.0
23
b. F
ind
the
98%
con
fiden
ce in
terv
al fo
r th
e m
ean
fuel
con
sum
ptio
n on
the
hi
ghw
ay.
28
.17
2 <
μ <
31
.12
8
c.
Com
pare
the
con
fiden
ce in
terv
als
in
part
s a
and
b.
S
am
ple
an
sw
er:
Th
e h
igh
wa
y
mil
ea
ge
is
co
ns
ide
rab
ly h
igh
er
tha
n t
he
cit
y m
ile
ag
e,
alt
ho
ug
h
the
ra
ng
e o
f v
alu
es
is
sm
all
er
for
hig
hw
ay
th
an
cit
y,
as
wo
uld
b
e e
xp
ec
ted
giv
en
th
at
the
s
tan
da
rd d
ev
iati
on
sh
ow
s l
es
s
va
ria
bil
ity
fo
r h
igh
wa
y m
ile
ag
e.
5. I
NTE
LLIG
ENCE
QU
OTI
ENT
Supp
ose
man
ager
s of
a c
orpo
rati
on w
ant
to
esti
mat
e th
e IQ
sco
re fo
r th
eir
empl
oyee
s. H
ow m
any
empl
oyee
s m
ust
be r
ando
mly
sel
ecte
d fo
r IQ
tes
ts if
the
m
anag
ers
wan
t to
be
95%
con
fiden
t th
at
the
mea
n is
wit
hin
2 IQ
poi
nts
of t
he
popu
lati
on m
ean?
The
y kn
ow fr
om
prev
ious
stu
dies
tha
t th
e st
anda
rd
devi
atio
n is
15
poin
ts.
a
t le
as
t 2
17
em
plo
ye
es
Wor
d Pr
oble
m P
ract
ice
Co
nfi
den
ce I
nte
rvals
11-5
005_
042_
PC
CR
MC
11_8
9381
2.in
dd30
11/1
4/09
5:22
:49
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-5
NA
ME
DA
TE
PE
RIO
D
Lesson 11-5
Ch
ap
ter
11
31
Gle
ncoe
Pre
calc
ulus
Th
e P
op
ula
tio
n C
orr
ecti
on
Facto
rTh
e fo
rmul
a fo
r th
e m
axim
um e
rror
of e
stim
ate,
E =
z ·
σ
−
√
�
n
, as
sum
es a
ver
y la
rge
popu
lati
on o
r sa
mpl
ing
wit
h re
plac
emen
t. If
the
pop
ulat
ion
size
is N
and
the
sam
ple
num
ber
is n
, the
form
ula
shou
ld b
e m
odifi
ed w
ith
a co
rrec
tion
fact
or if
n >
0.0
5N.
The
popu
lati
on c
orre
ctio
n fa
ctor
is √
��
�
N
- n
−
N -
1 .
Ther
efor
e, if
n >
0.0
5N, t
he m
axim
um
erro
r of
est
imat
e is
E =
z ·
σ
−
√
�
n
· √
��
�
N
- n
−
N -
1 .
Det
erm
ine
if t
he p
opul
atio
n co
rrec
tion
fac
tor
wou
ld b
e us
ed w
hen
dete
rmin
ing
a co
nfid
ence
inte
rval
. Jus
tify
you
r an
swer
.
1. 2
9 of
the
140
em
ploy
ees
in a
com
pany
are
sur
veye
d
y
es
; n
= 2
9,
0.0
5N
= 7
, a
nd
29
> 7
2. 7
5 of
the
tow
n’s
resi
dent
s ar
e su
rvey
ed a
nd t
he t
own
popu
lati
on is
299
8
n
o;
n =
75
, 0
.05N
= 1
49
.9,
an
d 7
5 <
14
9.9
3. 1
0 of
the
80
page
s pr
inte
d by
a n
ew p
rint
er a
re e
xam
ined
y
es
; n
= 1
0,
0.0
5N
= 4
, a
nd
10
> 4
4. T
here
are
250
stu
dent
s in
a s
choo
l. A
sam
ple
of 3
5 ra
ndom
ly s
elec
ted
stud
ents
fin
ds t
hat
thei
r m
ean
daily
stu
dy t
ime
is 5
2 m
inut
es w
ith
a st
anda
rd d
evia
tion
of 3
.3 m
inut
es. F
ind
the
95%
con
fiden
ce in
terv
al fo
r th
e m
ean
daily
stu
dy t
ime
of a
ll of
the
stu
dent
s. R
ound
E t
o th
e ne
ares
t hu
ndre
dth.
5
0.9
8 <
μ <
53
.02
5. W
hat
wou
ld b
e th
e co
nfid
ence
inte
rval
in E
xerc
ise
4 if
ther
e w
ere
750
stud
ents
in
the
sch
ool?
5
0.9
1 <
μ <
53
.09
6. F
orty
of t
he 1
60 p
otat
oes
in a
bin
are
ran
dom
ly s
elec
ted
for
wei
ght.
The
mea
nw
eigh
t of
the
sam
ple
is 1
50 g
ram
s w
ith
a st
anda
rd d
evia
tion
of 6
.5 g
ram
s. F
ind
the
90%
con
fiden
ce in
terv
al fo
r th
e m
ean
wei
ght
of a
ll of
the
pot
atoe
s. R
ound
Eto
the
nea
rest
hun
dred
th. A
ssum
e th
e va
riab
le is
nor
mal
ly d
istr
ibut
ed.
1
48
.53
< μ
< 1
51
.47
Enri
chm
ent
11-5
005_
042_
PC
CR
MC
11_8
9381
2.in
dd31
3/25
/09
12:3
8:43
PM
A01_A19_PCCRMC11_893812.indd 14A01_A19_PCCRMC11_893812.indd 14 11/17/09 9:08:01 AM11/17/09 9:08:01 AM
Copyright
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 A15 Glencoe Precalculus
An
swer
s
Answers (Lesson 11-6)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
32
Gle
ncoe
Pre
calc
ulus
Hyp
othe
ses
A h
ypot
hesi
s te
st a
llow
s yo
u to
eva
luat
e a
clai
m a
bout
a p
opul
atio
n.
The
null
hyp
othe
sis
or H
0 sta
tes
that
the
re is
not
a s
igni
fican
t di
ffere
nce
betw
een
the
sam
ple
valu
e an
d th
e po
pula
tion
par
amet
er a
nd it
alw
ays
cont
ains
a s
tate
men
t of
equ
alit
y.Th
e al
tern
ativ
e hy
poth
esis
or
Ha s
tate
s th
at t
here
is a
sig
nific
ant
diffe
renc
e be
twee
nth
e sa
mpl
e va
lue
and
the
popu
lati
on p
aram
eter
and
it a
lway
s co
ntai
ns a
sta
tem
ent
of
ineq
ualit
y.
For
eac
h st
atem
ent,
wri
te t
he n
ull a
nd a
lter
nati
ve h
ypot
hese
s.
Stat
e w
hich
hyp
othe
sis
repr
esen
ts t
he c
laim
.
a. A
doc
tor
clai
ms
that
the
pul
se r
ate
of a
pat
ient
cha
nges
fro
m 8
2 be
ats
per
min
ute
afte
r ta
king
a n
ew m
edic
atio
n.
Th
is c
laim
bec
omes
μ ≠
82
and
is t
he a
lter
nati
ve h
ypot
hesi
s si
nce
it in
clud
es a
n in
equa
lity
sym
bol.
The
com
plem
ent
is μ
= 8
2.
H0:
μ =
82
Ha:
μ ≠
82
(cla
im)
b. A
tra
ck m
embe
r cl
aim
s th
at h
e ca
n ru
n at
leas
t 10
mil
es t
hat
day.
Th
is c
laim
bec
omes
μ ≥
10
and
is t
he n
ull h
ypot
hesi
s si
nce
it in
clud
es a
n eq
ualit
y sy
mbo
l. Th
e co
mpl
emen
t is
μ <
10.
H0:
μ ≥
10
(cla
im)
Ha:
μ <
10
c. A
con
trac
tor
clai
ms
that
inst
alli
ng a
par
ticu
lar
kind
of
insu
lati
on w
ill
low
er t
he a
vera
ge J
uly
cool
ing
bill
of
$68
in h
er a
rea.
Th
is c
laim
bec
omes
μ <
$68
and
is t
he a
lter
nati
ve h
ypot
hesi
s si
nce
it in
clud
esan
ineq
ualit
y sy
mbo
l. Th
e co
mpl
emen
t is
μ ≥
$68
.H
0: μ
≥ $
68
Ha:
μ <
$68
(cla
im)
Exer
cise
s
Wri
te t
he n
ull a
nd a
lter
nati
ve h
ypot
hese
s fo
r ea
ch s
tate
men
t. S
tate
whi
ch
hypo
thes
is r
epre
sent
s th
e cl
aim
.
1. O
n a
scal
e of
1 t
o 10
, pat
ient
s de
scri
be t
heir
anx
iety
leve
ls a
s 8
duri
ng d
enta
lpr
oced
ures
. A d
enta
l ass
ista
nt t
hink
s th
is le
vel i
s lo
wer
whe
n so
ft m
usic
ispl
ayed
dur
ing
the
proc
edur
es.
H0:
μ ≥
8;
Ha:
μ <
8 (
cla
im)
2. A
rea
l est
ate
agen
t cl
aim
s th
at t
he a
vera
ge h
ome
pric
e in
an
area
is $
250,
000.
H
0:
μ =
25
0,0
00
(c
laim
); H
a:
μ ≠
25
0,0
00
3. A
hik
er c
laim
s th
at t
he a
vera
ge t
rail
leng
th in
a p
ark
is a
t m
ost
10 m
iles.
H
0:
μ ≤
10
(c
laim
); H
a:
μ >
10
4. A
res
taur
ant
owne
r cl
aim
s th
at t
he a
vera
ge a
ge o
f din
ers
in a
cer
tain
are
a is
gr
eate
r th
an 4
0.
H0:
μ ≤
40
; H
a:
μ >
40
(c
laim
)
Stud
y Gu
ide
and
Inte
rven
tion
Hyp
oth
esi
s Test
ing
Exam
ple
11-6
005_
042_
PC
CR
MC
11_8
9381
2.in
dd32
3/25
/09
12:3
8:46
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-6
NA
ME
DA
TE
PE
RIO
D
Lesson 11-6
Ch
ap
ter
11
33
Gle
ncoe
Pre
calc
ulus
Sign
ific
ance
and
Tes
ts T
o va
lidat
e a
clai
m, t
he n
ull h
ypot
hesi
s is
al
way
s te
sted
at
a ch
osen
leve
l of
sign
ific
ance
, α, w
hich
def
ines
the
are
a of
the
cri
tica
l reg
ion.
If t
he t
est
stat
isti
c z-
or
t-va
lue
is in
the
cri
tica
l reg
ion,
th
en H
0 sho
uld
be r
ejec
ted.
Oth
erw
ise,
H0 s
houl
d no
t be
rej
ecte
d.
A
n em
ploy
ee a
t a
spor
ting
goo
ds s
tore
cla
ims
that
the
ave
rage
pr
ice
of a
pai
r of
bas
ebal
l cle
ats
is le
ss t
han
$80.
Ano
ther
em
ploy
ee r
ando
mly
se
lect
s a
sam
ple
of 3
6 pa
irs
and
find
s −
x =
75
and
s =
19.
2. D
eter
min
e if
the
resu
lts
are
stat
isti
call
y si
gnif
ican
t at
α =
0.1
0.St
ep 1
Sta
te t
he n
ull a
nd a
lter
nati
ve h
ypot
hese
s an
d id
enti
fy t
he c
laim
.
H0:
μ ≥
$80
Ha:
μ <
80
(cla
im)
Step
2 D
eter
min
e th
e cr
itic
al v
alue
and
reg
ion.
Bec
ause
n ≥
30,
use
the
z-v
alue
. The
tes
t is
left
-tai
led
sinc
e μ
< 8
0.
All
of t
he c
riti
cal r
egio
n, w
ith
an a
rea
of 0
.10,
is t
o th
e le
ft o
f the
cr
itic
al v
alue
. Use
2
nd
DIS
TR
invN
orm
to
see
that
the
cri
tica
lva
lue
is -
1.28
. The
cri
tica
l reg
ion
is t
he a
rea
to t
he le
ft o
f z =
-1.
28.
Step
3 C
alcu
late
the
tes
t st
atis
tic.
Use
σ −
x = 19
.2
−
√
��
36
= 3
.2.
z =
−
x -
μ
−
σ −
x
Fo
rmu
la f
or
z-st
atis
tic
=
75 -
80
−
3.2
or
-1.
5625
−
x =
75
, μ
= 8
0,
an
d σ
−
x =
3.2
Step
4 A
ccep
t or
rej
ect
the
null
hypo
thes
is. H
0 is
reje
cted
sin
ce t
he t
est
stat
isti
c of
-1.
56 fa
lls w
ithi
n th
e cr
itic
al r
egio
n. T
here
is e
noug
h ev
iden
ce t
o su
ppor
t th
e cl
aim
tha
t th
e av
erag
e co
st o
f bas
ebal
l cle
ats
is le
ss t
han
$80.
Exer
cise
s
1. M
anag
ers
of a
larg
e de
part
men
t st
ore
clai
m t
hat
the
aver
age
sala
ry fo
r th
eir
part
-tim
e em
ploy
ees
is $
24,0
00. A
sam
ple
of 1
0 pa
rt-t
ime
empl
oyee
s ha
s a
mea
nsa
lary
of $
23,4
50 a
nd a
sta
ndar
d de
viat
ion
of $
1400
. Det
erm
ine
whe
ther
the
reis
eno
ugh
evid
ence
to
supp
ort
the
clai
m a
t α
= 0
.05.
t
= -
1.2
42
3,
nu
ll h
yp
oth
es
is i
s n
ot
reje
cte
d.
Th
e e
vid
en
ce
do
es
no
t re
jec
t th
e c
laim
th
at
the
me
an
is
$2
4,0
00
.
2. I
n E
xerc
ise
1, s
uppo
se t
hat
the
mea
n of
$23
,450
was
tak
en fr
om a
sam
ple
of50
par
t-ti
me
empl
oyee
s. W
ould
the
re b
e en
ough
evi
denc
e to
sup
port
the
cla
imat
α =
0.0
5?
z
= -
2.7
8,
nu
ll h
yp
oth
es
is i
s r
eje
cte
d.
Th
ere
wo
uld
be
en
ou
gh
ev
ide
nc
e
to r
eje
ct
the
cla
im.
11-6
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Hyp
oth
esi
s Test
ing
Exam
ple
-1.
28
Criti
cal r
egio
n
005_
042_
PC
CR
MC
11_8
9381
2.in
dd33
3/25
/09
12:3
8:51
PM
A01_A19_PCCRMC11_893812.indd 15A01_A19_PCCRMC11_893812.indd 15 11/17/09 9:08:40 AM11/17/09 9:08:40 AM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 11 A16 Glencoe Precalculus
Answers (Lesson 11-6)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
34
Gle
ncoe
Pre
calc
ulus
Wri
te t
he n
ull a
nd a
lter
nati
ve h
ypot
hese
s fo
r ea
ch s
tate
men
t. S
tate
w
hich
hyp
othe
sis
repr
esen
ts t
he c
laim
.
1. A
flor
ist
clai
ms
that
a c
erta
in t
ype
of fl
ower
has
a v
ase
life
of a
t le
ast
7 da
ys.
H0:
μ ≥
7 (
cla
im);
Ha:
μ <
7
2. A
bra
nd o
f cer
eal c
laim
s th
at a
ser
ving
con
tain
s le
ss t
han
2 gr
ams
of
suga
r.H
0:
μ ≥
2;
Ha:
μ <
2 (
cla
im)
3. R
ober
t cl
aim
s th
at h
e sw
ims
100
laps
eac
h w
eek.
H0:
μ =
10
0 (
cla
im);
Ha:
μ ≠
10
0
For
eac
h cl
aim
k, u
se t
he s
peci
fied
info
rmat
ion
to c
alcu
late
the
tes
t st
atis
tic
and
dete
rmin
e w
heth
er t
here
is e
noug
h ev
iden
ce t
o re
ject
th
e nu
ll h
ypot
hesi
s. T
hen
mak
e a
stat
emen
t re
gard
ing
the
orig
inal
cl
aim
.
4. k
: μ ≥
60,
α =
0.1
0, −
x =
58.
88, s
= 5
.08,
n =
8
t =
-0
.62
4;
Th
e n
ull
h
yp
oth
es
is i
s n
ot
reje
cte
d;
the
cla
im i
s n
ot
reje
cte
d.
5. k
: μ =
8, α
= 0
.05,
−
x =
8.2
, s =
0.6
, n =
32
z =
1.8
9;
Th
e n
ull
h
yp
oth
es
is i
s n
ot
reje
cte
d;
the
cla
im i
s n
ot
reje
cte
d.
6. M
AIL
A m
ail c
arri
er c
laim
s th
at t
he a
vera
ge n
umbe
r of
pie
ces
of m
ail
rece
ived
dai
ly b
y ho
useh
olds
in a
cer
tain
nei
ghbo
rhoo
d is
7. S
ampl
e da
ta
for
15 h
ouse
hold
s is
col
lect
ed. T
he a
vera
ge n
umbe
r of
mai
l pie
ces
was
6.5
w
ith
a st
anda
rd d
evia
tion
of 0
.8.
a. I
s th
ere
enou
gh e
vide
nce
to r
ejec
t the
mai
l car
rier
’s cl
aim
at α
= 0
.05?
t
= -
2.4
2, n
ull
hyp
oth
esis
is r
eje
cte
d. T
here
is e
no
ug
h
evid
en
ce t
o r
eje
ct
the c
laim
.
b. I
s th
ere
enou
gh e
vide
nce
to r
ejec
t the
mai
l car
rier
’s cl
aim
at α
= 0
.01?
n
o;
P =
0.0
30
7. W
EDD
ING
S A
wed
ding
pla
nner
wan
ts t
o te
st t
he c
laim
tha
t th
e av
erag
e w
eddi
ng h
as 1
25 g
uest
s. I
n a
rand
om s
ampl
e of
35
wed
ding
s, h
e fo
und
the
mea
n to
be
110
gues
ts w
ith
a st
anda
rd d
evia
tion
of 3
0 gu
ests
. Fin
d th
e P-
valu
e an
d de
term
ine
whe
ther
the
re is
eno
ugh
evid
ence
to
supp
ort
the
clai
m a
t α
= 0
.01.
P
-va
lue
= 0
.00
3;
Th
ere
is
no
t e
no
ug
h e
vid
en
ce
to
su
pp
ort
th
e
cla
im.
Prac
tice
Hyp
oth
esi
s Test
ing
11-6
005_
042_
PC
CR
MC
11_8
9381
2.in
dd34
11/1
4/09
5:31
:54
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-6
NA
ME
DA
TE
PE
RIO
D
Lesson 11-6
Ch
ap
ter
11
35
Gle
ncoe
Pre
calc
ulus
1. B
OTT
LED
WA
TER
The
aver
age
volu
me
in o
unce
s of
a r
ando
m s
ampl
e of
36
bot
tles
of w
ater
at
a pa
ckag
ing
plan
t w
as fo
und
to b
e 12
.19
ounc
es w
ith
a st
anda
rd d
evia
tion
of 0
.11
ounc
e. T
he
floor
sup
ervi
sor
mad
e th
e cl
aim
tha
t th
e m
ean
volu
me
was
gre
ater
tha
n 12
oun
ces.
Tes
t he
r cl
aim
at
α =
0.0
1.
a. W
rite
the
nul
l and
alt
erna
tive
hy
poth
eses
and
sta
te w
hich
hy
poth
esis
rep
rese
nts
the
clai
m.
H0:
μ ≤
12
; H
a:
μ >
12
(c
laim
)
b. I
s th
ere
enou
gh e
vide
nce
to r
ejec
t th
e nu
ll hy
poth
esis
? W
hy?
ye
s;
z =
10
.36
, w
hic
h i
s g
rea
ter
tha
n t
he
cri
tic
al
va
lue
of
2.3
3.
c. M
ake
a st
atem
ent
rega
rdin
g th
e or
igin
al c
laim
.
Th
ere
is e
no
ug
h e
vid
en
ce t
o
su
pp
ort
th
e c
laim
th
at
the m
ean
vo
lum
e o
f w
ate
r in
th
e b
ott
les is
g
reate
r th
an
12 o
un
ces.
2. T
EMPE
RATU
RE I
t is
a lo
ng-e
stab
lishe
d cl
aim
tha
t th
e m
ean
body
tem
pera
ture
fo
r hu
man
s is
98.
6°F.
In
a ra
ndom
sa
mpl
e of
10
peop
le, t
he m
ean
body
te
mpe
ratu
re w
as 9
8.3°
F w
ith
a st
anda
rd
devi
atio
n of
0.2
3°F.
Tes
t th
e cl
aim
at
α =
0.0
2.
a. W
hat
are
the
crit
ical
val
ues?
-
2.3
3 a
nd
2.3
3
b. W
hat
is t
he t
est
stat
isti
c?
-
4.1
2
c. I
s th
is e
vide
nce
enou
gh t
o su
ppor
tth
e cl
aim
tha
t th
e m
ean
body
te
mpe
ratu
re fo
r hu
man
s is
98.
6°?
n
o
3. P
ULS
E RA
TES
An
aero
bics
inst
ruct
or
had
a pu
lse
rate
of 1
10 b
eats
per
min
ute
afte
r a
war
m-u
p ro
utin
e w
ith
her
clas
s.
Stud
ents
rec
orde
d th
eir
puls
e ra
tes
at
the
sam
e ti
me.
The
ir r
ates
are
rec
orde
d in
the
tab
le b
elow
. The
inst
ruct
or c
laim
s th
at t
heir
ave
rage
pul
se r
ate
is lo
wer
th
an h
ers.
Tes
t he
r cl
aim
at
α =
0.0
5.
80
70
90
75
11
0
10
51
20
11
08
51
15
95
95
10
59
07
0
10
59
51
00
10
59
0
a. W
rite
the
nul
l and
alt
erna
tive
hy
poth
eses
and
sta
te w
hich
hy
poth
esis
rep
rese
nts
the
clai
m.
H0:
μ ≥
11
0;
Ha:
μ <
11
0 (
cla
im)
b. W
hat
are
the
crit
ical
val
ues
and
test
sta
tist
ic?
cri
tic
al
va
lue
: -
1.7
3;
tes
t s
tati
sti
c:
-4
.5
c.
Doe
s th
e da
ta s
uppo
rt t
he in
stru
ctor
’s cl
aim
s?
y
es
4. M
OU
NTA
IN C
LIM
BIN
G A
mou
ntai
n cl
imbi
ng in
stru
ctor
cla
ims
that
his
st
uden
ts t
ake
long
er t
han
5 m
inut
es
to p
ack
thei
r ba
ckpa
cks.
In
a ra
ndom
sa
mpl
e of
80
stud
ents
, the
ave
rage
ti
me
it t
ook
to p
ack
a ba
ckpa
ck w
as
5.3
min
utes
wit
h a
stan
dard
dev
iati
onof
1.7
min
utes
. Fin
d th
e P-
valu
e an
d de
term
ine
whe
ther
the
re is
eno
ugh
evid
ence
to
supp
ort
the
clai
m a
tα
= 0
.05.
P
-va
lue
= 0
.05
7;
Th
ere
is
no
t e
no
ug
h e
vid
en
ce
to
su
pp
ort
th
e
cla
im.
11-6
Wor
d Pr
oble
m P
ract
ice
Hyp
oth
esi
s Test
ing
005_
042_
PC
CR
MC
11_8
9381
2.in
dd35
11/1
4/09
5:32
:52
PM
A01_A19_PCCRMC11_893812.indd 16A01_A19_PCCRMC11_893812.indd 16 11/17/09 9:10:59 AM11/17/09 9:10:59 AM
Copyright
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 A17 Glencoe Precalculus
An
swer
s
Answers (Lesson 11-6 and Lesson 11-7)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
36
Gle
ncoe
Pre
calc
ulus
Hyp
oth
esi
s Test
ing
You
hav
e te
sted
hyp
othe
ses
by fi
ndin
g a
test
sta
tist
ic a
nd s
eein
g if
it fa
lls w
ithi
n th
e cr
itic
al
regi
on. A
lter
nati
vely
, you
can
tes
t a
hypo
thes
is b
y fin
ding
a c
onfid
ence
inte
rval
and
see
ing
if th
e cl
aim
falls
wit
hin
that
inte
rval
.
A c
ondo
min
ium
sup
erin
tend
ent
buys
5-p
ound
bag
s of
sal
t fo
r m
elti
ng ic
e.H
e w
eigh
s th
e sa
lt in
50
rand
omly
sel
ecte
d ba
gs a
nd f
inds
the
mea
n to
be
4.6
poun
ds w
ith
a st
anda
rd d
evia
tion
of
0.7
poun
d. U
se a
con
fide
nce
leve
lto
tes
t th
e su
peri
nten
dent
’s c
laim
tha
t th
e m
ean
wei
ght
is n
ot 5
pou
nds
at α
= 0
.05.
Wri
te t
he h
ypot
hese
s: H
0: μ
= 5
H
a: μ
≠ 5
Use
the
nor
mal
dis
trib
utio
n si
nce
n >
30.
Sin
ce α
= 0
.05,
the
con
fiden
ce le
vel i
s 1
- α
= 0
.05,
or
0.95
. The
z-v
alue
ass
ocia
ted
wit
h a
95%
con
fiden
ce le
vel i
s 1.
96.
Find
the
95%
con
fiden
ce in
terv
al fo
r th
e m
ean
wei
ght
of t
he s
alt
in t
he b
ags.
CI
= −
x ±
z ·
σ
−
√
�
n
C
on
fide
nce
inte
rva
l fo
r th
e m
ea
n
= 4
.6 ±
0.1
94
−
x =
4.6
, a
nd
z ·
σ
−
√
�
n = 1
.96
·
0.7
−
√
��
50
or
ab
ou
t 0
.19
4
So, t
he 9
5% c
onfid
ence
inte
rval
is 4
.406
< μ
< 4
.794
.
The
conf
iden
ce in
terv
al fo
r th
e po
pula
tion
doe
s no
t co
ntai
n th
e cl
aim
of 5
, so
we
reje
ct t
he
null
hypo
thes
is. E
vide
nce
supp
orts
the
cla
im t
hat
the
mea
n is
not
5 p
ound
s.
Exer
cise
s
1. S
OCC
ER A
soc
cer
coac
h cl
aim
s th
at t
he m
ean
wei
ght
of t
he p
laye
rs o
n th
e op
posi
ng
team
s is
200
pou
nds.
A r
ando
m s
ampl
e of
10
play
ers
from
opp
osin
g te
ams
has
a m
ean
of
198.
2 po
unds
and
a s
tand
ard
devi
atio
n of
3.3
pou
nds.
At
α =
0.0
5, c
an t
he c
laim
be
supp
orte
d? S
uppo
rt y
our
answ
er w
ith
a co
nfid
ence
inte
rval
.
T
he
da
ta s
up
po
rts
th
e c
laim
; 1
95
.84
< μ
< 2
00
.56
.
2. C
ALO
RIES
A t
each
er w
alke
d by
a v
endi
ng m
achi
ne a
nd c
laim
ed t
hat
the
aver
age
num
ber
of C
alor
ies
in t
he s
nack
s w
as 2
50. S
tude
nts
who
ove
rhea
rd h
er w
ante
d to
pr
ove
her
inco
rrec
t. Th
ey b
ough
t a
rand
om s
ampl
e of
8 s
nack
s an
d fo
und
the
mea
n nu
mbe
r of
Cal
orie
s of
tho
se s
nack
s to
be
210
wit
h a
stan
dard
dev
iati
on o
f 24.
At
α =
0.1
0, d
oes
the
data
sup
port
the
tea
cher
or
the
stud
ents
? Su
ppor
t yo
ur a
nsw
er
wit
h a
conf
iden
ce in
terv
al.
T
he
da
ta s
up
po
rts
th
e s
tud
en
ts;
19
3.9
2 <
μ <
22
6.0
8.
3. H
OM
EWO
RK A
tea
cher
cla
ims
her
stud
ents
spe
nd a
n av
erag
e of
45
min
utes
on
her
hom
ewor
k ea
ch n
ight
. To
test
the
cla
im, s
he a
sks
35 s
tude
nts
how
muc
h ti
me,
on
aver
age,
the
y sp
end
on h
er h
omew
ork
each
nig
ht a
nd t
he r
esul
ts s
how
a m
ean
of
40 m
inut
es w
ith
a st
anda
rd d
evia
tion
of 9
min
utes
. At
α =
0.0
5, d
oes
the
data
sup
port
he
r cl
aim
? Su
ppor
t yo
ur a
nsw
er w
ith
a co
nfid
ence
inte
rval
.
T
he
da
ta d
oe
s n
ot
su
pp
ort
he
r c
laim
; 3
7 <
μ <
43
.
Enri
chm
ent
11-6
005_
042_
PC
CR
MC
11_8
9381
2.in
dd36
11/1
9/09
7:40
:52
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-7
NA
ME
DA
TE
PE
RIO
D
Lesson 11-7
Ch
ap
ter
11
37
Gle
ncoe
Pre
calc
ulus
11-7
Corr
elat
ion
To d
eter
min
e if
the
corr
elat
ion
coef
fici
ent
r is
sig
nific
ant,
perf
orm
a h
ypot
hesi
s te
st w
ith
H0 :
ρ =
0, a
nd t
as t
he t
est
stat
isti
c, t
= r
√
��
�
n
- 2
−
1 -
r2
, w
here
the
deg
rees
of f
reed
om is
n -
2.
The
tab
le s
how
s th
e nu
mbe
r of
abs
ence
s of
7 s
tude
nts
from
a p
sych
olog
y cl
ass
and
thei
r fi
nal g
rade
s.
Ab
se
nc
es
91
01
52
82
6
Gra
de
s7
05
84
39
07
88
77
9
a. M
ake
a sc
atte
r pl
ot o
f th
e da
ta a
nd id
enti
fy t
he r
elat
ions
hip.
T
hen
calc
ulat
e an
d in
terp
ret
the
corr
elat
ion
coef
fici
ent.
E
nter
the
dat
a in
to L
1 an
d L
2. T
urn
on P
lot1
and
cho
ose
a sc
atte
r pl
ot. T
he d
ata
appe
ars
to h
ave
a ne
gati
ve li
near
cor
rela
tion
.
Pr
ess
STA
T a
nd s
elec
t L
inR
eg(a
x +
b)
unde
r th
e C
alc
men
u.
The
corr
elat
ion
coef
ficie
nt r
is a
bout
-0.
9609
. Bec
ause
r is
clo
se
to -
1, t
his
sugg
ests
a s
tron
g ne
gati
ve c
orre
lati
on.
b. T
est
the
sign
ific
ance
of
the
corr
elat
ion
coef
fici
ent
from
pa
rt a
at
the
5% le
vel.
St
ep 1
Sta
te t
he h
ypot
hese
s.
H0:
ρ =
0
H
a: ρ ≠
0
St
ep 2
Det
erm
ine
the
crit
ical
val
ues
usin
g n
- 2
or
5 de
gree
s of
fr
eedo
m. T
he c
riti
cal v
alue
s ar
e ±
2.6.
St
ep 3
Cal
cula
te t
he t
est
stat
isti
c.
t = -
0.96
09 √
��
��
��
5
−
1 -
(-0.
9609
)2 ≈
-7.
7597
St
ep 4
Sin
ce t
< -
2.6,
rej
ect
the
null
hypo
thes
is. T
he e
vide
nce
supp
orts
a s
igni
fican
t co
rrel
atio
n be
twee
n th
e nu
mbe
r of
abs
ence
s an
d a
stud
ent’s
gra
de.
Exer
cise
s
The
tab
le s
how
s da
ta a
bout
gey
sers
at
Yel
low
ston
e N
atio
nal P
ark.
Du
rati
on
(m
in)
3.2
53
57
.51
7.5
3.5
63
51
06
.51
01
20
4
He
igh
t (f
t)1
84
25
15
02
07
57
51
01
00
16
06
05
25
75
5
Sour
ce: N
atio
nal P
ark
Serv
ice
1. M
ake
a sc
atte
r pl
ot o
f the
dat
a an
d id
enti
fy t
he r
elat
ions
hip.
Th
en c
alcu
late
and
inte
rpre
t th
e co
rrel
atio
n co
effic
ient
. It
ap
pe
ars
to
ha
ve
no
re
lati
on
sh
ip o
r a
we
ak
po
sit
ive
li
ne
ar
co
rre
lati
on
; r
= 0
.15
9;
litt
le t
o n
o c
orr
ela
tio
n
2. T
est
the
sign
ifica
nce
of t
he c
orre
lati
on c
oeffi
cien
t fr
om E
xerc
ise
1 at
the
5%
leve
l.
t =
0.5
58
; S
inc
e t
< 2
.2,
do
no
t re
jec
t n
ull
hy
po
the
sis
; n
o c
orr
ela
tio
n.
Stud
y Gu
ide
and
Inte
rven
tion
Co
rrela
tio
n a
nd
Lin
ear
Reg
ress
ion
Exam
ple
[ 1, 1
7] s
cl: 1
by
[ 35,
100
] scl:
5
[ –1,
40]
scl:
1 b
y [ –
1, 2
15] s
cl: 5
005_
042_
PC
CR
MC
11_8
9381
2.in
dd37
11/1
4/09
5:42
:10
PM
A01_A19_PCCRMC11_893812.indd 17A01_A19_PCCRMC11_893812.indd 17 11/19/09 8:01:16 PM11/19/09 8:01:16 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
Chapter 11 A18 Glencoe Precalculus
Answers (Lesson 11-7)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
38
Gle
ncoe
Pre
calc
ulus
Line
ar R
egre
ssio
n If
the
cor
rela
tion
bet
wee
n tw
o va
riab
les
is s
igni
fican
t, yo
u ca
n de
term
ine
the
leas
t-sq
uare
s re
gres
sion
line
, whi
ch is
a li
ne o
f be
st f
it. T
hen
you
can
use
the
equa
tion
of t
he li
ne t
o m
ake
pred
icti
ons
over
the
ran
ge o
f dat
a.
The
tab
le s
how
s th
e da
ta f
rom
the
pre
viou
s pa
ge, w
hich
sho
wed
a
sign
ific
ant
corr
elat
ion.
Fin
d th
e eq
uati
on o
f th
e re
gres
sion
line
for
the
dat
a.
Inte
rpre
t th
e sl
ope
and
inte
rcep
t in
con
text
. Use
the
equ
atio
n to
pre
dict
the
ex
pect
ed g
rade
for
a s
tude
nt w
ith
4 ab
senc
es a
nd s
tate
whe
ther
thi
s pr
edic
tion
is
rea
sona
ble.
Exp
lain
. Ab
se
nc
es
91
01
52
82
6
Gra
de
s7
05
84
39
07
88
77
9
Pres
s S
TA
T a
nd s
elec
t L
inR
eg(a
x +
b)
unde
r th
e C
alc
men
u.
The
equa
tion
is a
ppro
xim
atel
y y
= -
3.48
x +
97.
993.
The
slop
e in
dica
tes
that
for
ever
y ad
diti
onal
abs
ence
a s
tude
nt
has,
his
or
her
grad
e w
ill d
rop
by a
bout
3.5
poi
nts.
The
y-in
terc
ept
indi
cate
s th
at a
stu
dent
wit
h no
abs
ence
s w
ill h
ave
a gr
ade
of a
bout
98.
Eva
luat
e th
e re
gres
sion
equ
atio
n fo
r x
= 4
and
cal
cula
te y
.y
= -
3.48
x +
97.
993
Re
gre
ssio
n e
qu
atio
n
=
-3.
48(4
) + 9
7.99
3 x
= 4
=
84.
073
Sim
plif
y.
We
expe
ct t
hat
a st
uden
t w
ith
4 da
ys o
f abs
ence
s w
ould
hav
e a
final
gra
de o
f abo
ut 8
4.
This
is r
easo
nabl
e be
caus
e 4
is in
the
ran
ge o
f the
giv
en x
-val
ues
and
84 is
in t
he r
ange
of
the
giv
en y
-val
ues.
Exer
cise
s
A s
port
s w
rite
r ha
s al
read
y de
term
ined
tha
t th
ere
is a
neg
ativ
e co
rrel
atio
n be
twee
n th
e ba
ttin
g av
erag
e an
d th
e nu
mbe
r of
hom
e ru
ns f
or t
hird
bas
emen
w
hose
sta
tist
ics
are
show
n in
the
tab
le.
Av
era
ge
0.3
07
0.3
28
0.3
05
0.2
94
0.3
06
0.3
11
0.2
71
0.2
67
0.2
67
0.3
20
Ho
me
Ru
ns
96
11
83
17
65
78
10
35
12
26
85
48
58
1. F
ind
the
equa
tion
of t
he r
egre
ssio
n lin
e fo
r th
e da
ta.
y
= -
64
93
.7x
+ 2
14
8.8
2
2. U
se t
he e
quat
ion
to p
redi
ct t
he n
umbe
r of
hom
e ru
ns fo
r a
thir
d ba
sem
an w
ith
a ba
ttin
g av
erag
e of
0.3
50. S
tate
whe
ther
thi
s pr
edic
tion
is r
easo
nabl
e. E
xpla
in.
a
bo
ut
-1
24
ho
me
ru
ns
; n
ot
rea
so
na
ble
; T
he
nu
mb
er
of
ho
me
ru
ns
m
us
t b
e p
os
itiv
e.
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Co
rrela
tio
n a
nd
Lin
ear
Reg
ress
ion
11-7
Exam
ple
005_
042_
PC
CR
MC
11_8
9381
2.in
dd38
3/25
/09
12:3
9:18
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-7
NA
ME
DA
TE
PE
RIO
D
Lesson 11-7
Ch
ap
ter
11
39
Gle
ncoe
Pre
calc
ulus
A s
uper
viso
r of
a c
lean
ing
busi
ness
has
the
dat
a in
the
tab
le t
hat
show
s th
e ag
e of
her
wor
kers
and
the
num
ber
of s
ick
days
the
y ta
ke e
ach
year
. She
won
ders
if
ther
e is
a s
igni
fica
nt li
near
rel
atio
nshi
p be
twee
n th
e ag
e of
an
empl
oyee
and
the
nu
mbe
r of
sic
k da
ys h
e or
she
tak
es e
ach
year
.
Ag
e3
82
56
01
84
55
41
92
24
33
4
Da
ys
91
22
16
45
15
17
31
1. M
ake
a sc
atte
r pl
ot o
f the
dat
a an
d id
enti
fy t
he r
elat
ions
hip.
Th
en c
alcu
late
and
inte
rpre
t th
e co
rrel
atio
n co
effic
ient
. T
he
da
ta a
pp
ea
r to
ha
ve
a n
eg
ati
ve
lin
ea
r c
orr
ela
tio
n,
r ≈
-0
.83
24
; T
he
co
rre
lati
on
co
eff
icie
nt
ind
ica
tes
a
fair
ly s
tro
ng
ne
ga
tiv
e l
ine
ar
co
rre
lati
on
.
2. D
eter
min
e if
the
corr
elat
ion
coef
ficie
nt is
sig
nific
ant
at t
he 1
%, 5
%, a
nd 1
0% le
vels
. E
xpla
in y
our
reas
onin
g.
t ≈
-4
.25
an
d t
< -
3.3
6,
t <
-2
.31
, t
< -
1.8
6,
so
th
e
nu
ll h
yp
oth
es
is i
s r
eje
cte
d a
nd
th
ere
is
a s
ign
ific
an
t c
orr
ela
tio
n.
3. I
f the
cor
rela
tion
is s
igni
fican
t at
the
10%
leve
l, fin
d th
e le
ast-
squa
res
regr
essi
on
equa
tion
and
inte
rpre
t th
e sl
ope
and
y-in
terc
ept
in c
onte
xt.
y =
-0
.34
8x
+ 2
0.8
67
; T
he
slo
pe
in
dic
ate
s t
ha
t fo
r e
ve
ry a
dd
itio
na
l y
ea
r o
lde
r, t
he
nu
mb
er
of
sic
k d
ay
s d
ec
rea
se
s b
y a
bo
ut
on
e-t
hir
d d
ay
; th
e y
-in
terc
ep
t is
no
t m
ea
nin
gfu
l. A
pe
rso
n 0
ye
ars
old
wo
uld
us
e 2
1 s
ick
da
ys
pe
r y
ea
r.
4. G
raph
and
ana
lyze
the
res
idua
l plo
t. T
he
re
sid
ua
ls a
pp
ea
r to
be
fa
irly
ra
nd
om
ly s
ca
tte
red
an
d m
os
tly
ce
nte
red
a
bo
ut
the
re
gre
ss
ion
lin
e y
= 0
. T
his
su
pp
ort
s t
he
c
laim
th
at
the
us
e o
f a
lin
ea
r m
od
el
is a
pp
rop
ria
te.
5. I
dent
ify a
ny in
fluen
tial
out
liers
. Des
crib
e th
e ef
fect
the
out
lier
has
on t
he s
tren
gth
of t
he c
orre
lati
on a
nd o
n th
e sl
ope
and
inte
rcep
t of
the
ori
gina
l re
gres
sion
line
. T
he
po
int
(34
, 1
) is
an
ou
tlie
r in
th
e y
-dir
ec
tio
n.
Th
e
co
rre
lati
on
is
str
on
ge
r w
ith
ou
t it
(-
0.9
38
). T
he
slo
pe
ch
an
ge
s v
ery
lit
tle
w
ith
ou
t it
, b
ut
the
y-i
nte
rce
pt
ch
an
ge
s s
om
e.
6. I
f any
dat
a w
ere
rem
oved
, rea
sses
s th
e si
gnifi
canc
e of
the
cor
rela
tion
at
the
10%
leve
l an
d, if
sti
ll ap
prop
riat
e, r
ecal
cula
te t
he r
egre
ssio
n eq
uati
on.
Wit
h (
34
, 1
) re
mo
ve
d,
t ≈
-7
.2 a
nd
t <
-1
.89
, s
o t
he
co
rre
lati
on
is
sti
ll
sig
nif
ica
nt
at
the
10
% l
ev
el.
^ y =
-0
.35
6x
+ 2
2.0
53
7. I
f app
ropr
iate
, pre
dict
the
num
ber
of s
ick
days
for
empl
oyee
s w
ho a
re 3
0, 5
0, a
nd 7
0 ye
ars
old.
Int
erpr
et y
our
resu
lts
and
stat
e w
heth
er t
he p
redi
ctio
ns a
re r
easo
nabl
e.
Exp
lain
you
r re
ason
ing.
1
1,
4,
an
d -
3 d
ay
s;
Th
e f
irs
t tw
o p
red
icti
on
s a
re
rea
so
na
ble
be
ca
us
e b
oth
va
ria
ble
s a
re i
n t
he
ra
ng
e o
f g
ive
n d
ata
; th
e l
as
t p
red
icti
on
is
no
t re
as
on
ab
le b
ec
au
se
70
is
ou
t o
f th
e g
ive
n x
-va
lue
s a
nd
th
e n
um
be
r o
f s
ick
da
ys
ca
nn
ot
be
ne
ga
tiv
e.
Prac
tice
Co
rrela
tio
n a
nd
Lin
ear
Reg
ress
ion
11-7
[ 10,
65]
scl:
1 b
y [ 0
, 20]
scl:
1
[ 10,
65]
scl:
1 b
y [ –
10, 1
0] s
cl: 1
005_
042_
PC
CR
MC
11_8
9381
2.in
dd39
3/25
/09
12:3
9:24
PM
A01_A19_PCCRMC11_893812.indd 18A01_A19_PCCRMC11_893812.indd 18 3/25/09 2:48:13 PM3/25/09 2:48:13 PM
Copyright
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 A19 Glencoe Precalculus
An
swer
s
Answers (Lesson 11-7)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
11
40
Gle
ncoe
Pre
calc
ulus
1. C
OLL
EGE
The
data
in t
he t
able
re
pres
ent
the
Am
eric
an C
olle
ge T
est
(AC
T) c
ompo
site
sco
res
and
grad
e po
int
aver
ages
(GPA
) of 2
0 ra
ndom
ly s
elec
ted
stud
ents
aft
er t
heir
firs
t se
mes
ter
in
colle
ge. A
col
lege
cou
nsel
or w
ants
to
dete
rmin
e if
ther
e is
a c
orre
lati
on
betw
een
AC
T sc
ores
and
firs
t se
mes
ter
GPA
s.
AC
T2
71
81
71
5
GP
A3
.92
.93
.33
.0
AC
T2
22
01
72
1
GP
A3
.62
.72
.93
.4
AC
T2
51
72
51
8
GP
A3
.53
.14
.03
.0
AC
T2
31
92
02
9
GP
A3
.62
.63
.03
.4
AC
T2
32
82
22
0
GP
A1
.84
.03
.04
.0
a. C
alcu
late
and
inte
rpre
t th
e co
rrel
atio
n co
effic
ient
.
r ≈
0.4
52;
it s
ug
gests
th
at
the d
ata
h
as a
weak p
osit
ive c
orr
ela
tio
n
b. D
eter
min
e if
the
corr
elat
ion
coef
ficie
nt
is s
igni
fican
t at
the
1%
, 5%
, and
10%
le
vels
.
t ≈
2.1
5;
Becau
se t
< 2
.88,
t >
2.1
0,
an
d t
> 1
.73,
the c
orr
ela
tio
n
co
eff
icie
nt
is s
ign
ific
an
t at
the 5
%
an
d 1
0%
levels
.
c. G
raph
and
ana
lyze
the
res
idua
l plo
t.
T
he r
esid
uals
ap
pear
to b
e c
en
tere
d
ab
ou
t th
e r
eg
ressio
n lin
e y
= 0
, w
hic
h
mean
s a
lin
ear
reg
ressio
n m
od
el is
ap
pro
pri
ate
fo
r th
is d
ata
. T
here
ap
pears
to
be o
ne o
utl
ier
at
x =
23.
2.
SALE
S A
sal
es a
ssoc
iate
wan
ts t
o kn
ow
if th
ere
is a
rel
atio
nshi
p be
twee
n th
e av
erag
e nu
mbe
r of
tim
es h
is c
owor
kers
co
ntac
t cl
ient
s ea
ch m
onth
and
the
av
erag
e m
onth
ly s
ales
vol
ume
in
thou
sand
s of
dol
lars
. He
colle
cted
the
da
ta s
how
n in
the
tab
le.
Cli
en
ts2
12
34
85
04
6
Sa
les
30
30
95
11
08
0
Cli
en
ts1
25
51
45
01
6
Sa
les
15
13
02
59
03
0
a. M
ake
a sc
atte
r pl
ot o
f the
dat
a an
d id
enti
fy t
he r
elat
ions
hip.
The
n ca
lcul
ate
and
inte
rpre
t th
e co
rrel
atio
n co
effic
ient
.
T
here
ap
pears
to
be a
po
sit
ive l
inear
co
rrela
tio
n;
r ≈
0.9
73,
su
gg
esti
ng
a
str
on
g p
osit
ive l
inear
co
rrela
tio
n
b. D
eter
min
e if
the
corr
elat
ion
coef
ficie
nt is
sig
nific
ant
at t
he 1
%,
5%, a
nd 1
0% le
vels
. Exp
lain
you
r re
ason
ing.
t
≈ 1
1.9
; B
ecau
se t
> 3
.36,
t >
2.3
1,
an
d t
> 1
.86,
the c
orr
ela
tio
n
co
eff
icie
nt
is s
ign
ific
an
t at
all t
hre
e
levels
.
c. I
f the
cor
rela
tion
is s
igni
fican
t at
the
10
% le
vel,
find
the
leas
t-sq
uare
s re
gres
sion
equ
atio
n an
d in
terp
ret
the
slop
e an
d y-
inte
rcep
t in
con
text
.
y
= 2
.312x
- 1
3.9
58;
Th
e s
lop
e
su
gg
ests
th
at
every
clie
nt
co
nta
ct
per
mo
nth
co
rresp
on
ds t
o a
sale
s
incre
ase o
f 2.3
12 t
ho
usan
d d
olla
rs.
Th
e y
-in
terc
ep
t su
gg
ests
th
at
no
clie
nt
co
nta
ct
co
rresp
on
ds t
o a
sale
s
decre
ase o
f 13.9
58 t
ho
usan
d d
olla
rs.
11-7
Wor
d Pr
oble
m P
ract
ice
Co
rrela
tio
n a
nd
Lin
ear
Reg
ress
ion
[ 10,
34]
scl:
2 b
y [ –
2, 2
] scl:
1
[ 8, 6
0] s
cl: 2
by
[ –5,
150
] scl:
1
005_
042_
PC
CR
MC
11_8
9381
2.in
dd40
3/25
/09
12:3
9:30
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson X-7
NA
ME
DA
TE
PE
RIO
D
Lesson 11-7
Ch
ap
ter
11
41
Gle
ncoe
Pre
calc
ulus
Mu
ltip
le R
eg
ress
ion
In m
ulti
ple
regr
essi
on, t
here
are
tw
o or
mor
e in
depe
nden
t va
riab
les
and
one
depe
nden
t va
riab
le. T
he m
ulti
ple
regr
essi
on e
quat
ion
is y
= a
+ b
1x1 +
b2x
2 + …
+ b
kxk,
whe
re
x 1, x 2,
…, x
k are
the
inde
pend
ent
vari
able
s.
The
stre
ngth
of t
he r
elat
ions
hip
betw
een
the
inde
pend
ent
vari
able
s an
d th
e de
pend
ent
vari
able
is m
easu
red
by t
he m
ulti
ple
corr
elat
ion
coef
fici
ent
R. T
his
valu
e ca
n ra
nge
from
0 t
o 1,
whe
re v
alue
s cl
oser
to
1 in
dica
te a
str
onge
r re
lati
onsh
ip t
han
thos
e cl
oser
to
0.
The
form
ula
for
a m
ulti
ple
corr
elat
ion
coef
ficie
nt w
ith
two
inde
pend
ent
vari
able
s is
R =
√
��
��
��
��
��
��
�
(r
y x 1 )
2 + (
r yx
2 ) 2 - 2
r yx1 ( r
yx2 ) (
r x 1x2 )
−−
√
��
��
1
- (
r x 1x2 ) 2
,
whe
re r
y x 1 is
the
cor
rela
tion
coe
ffici
ent
for
y an
d x 1;
r y x 2 is
the
cor
rela
tion
coe
ffici
ent
for
y an
d x 2
; and
r x 1x
2 is t
he c
orre
lati
on c
oeffi
cien
t fo
r x 1 a
nd x
2.
Hos
pita
l adm
inis
trat
ors
wis
h to
see
if a
nur
sing
ap
plic
ant’s
GP
A a
nd a
ge a
re r
elat
ed t
o th
e nu
rse’
s sc
ore
on t
he s
tate
nur
sing
boa
rd e
xam
. The
tab
le
show
s th
ese
stat
isti
cs f
or f
ive
nurs
es.
1. F
ind
the
valu
es o
f ry x
1 , r y x
2 , and
r x 1x
2 .
0.8
63
; 0
.69
3;
0.3
58
2. F
ind
and
inte
rpre
t R
. a
bo
ut
0.9
56
; th
ere
is
a s
tro
ng
lin
ea
r re
lati
on
sh
ip
am
on
g t
he
ap
pli
ca
nt’
s g
rad
e p
oin
t a
ve
rag
e,
ag
e,
an
d s
co
re o
n t
he
sta
te
bo
ard
ex
am
.
3. A
res
earc
her
is c
ondu
ctin
g a
stud
y to
see
if a
per
son’
s ag
e x 1 a
nd c
hole
ster
ol le
vel x
2 are
re
late
d to
his
or
her
syst
olic
blo
od p
ress
ure
y. F
ind
and
inte
rpre
t R
if r
y x 1 =
0.6
81,
r y x 2 =
0.8
72, a
nd r
x 1x2 =
0.7
46.
0.8
73
; S
inc
e R
is
fa
irly
clo
se
to
1,
we
ca
n s
ay
tha
t th
ere
is
a l
ine
ar
rela
tio
ns
hip
am
on
g a
ge
, c
ho
les
tero
l le
ve
l, a
nd
blo
od
p
res
su
re.
A t
rack
coa
ch w
ants
to
see
if t
he 5
k ti
mes
for
the
fi
rst
two
mee
ts o
f th
e se
ason
are
rel
ated
to
the
5k t
imes
at
the
stat
e m
eet.
The
tab
le s
how
s th
ese
stat
isti
cs, i
n m
inut
es, f
or f
ive
runn
ers.
4. F
ind
the
valu
es o
f r yx
1 , r yx
2 , and
r x 1x
2 .
0.8
16
; 0
.56
5;
0.8
00
5. F
ind
and
inte
rpre
t R
. a
bo
ut
0.8
29
; S
inc
e R
is
fa
irly
clo
se
to
1,
we
ca
n s
ay
th
at
the
re i
s a
lin
ea
r re
lati
on
sh
ip a
mo
ng
th
e t
hre
e r
ac
es
.
Enri
chm
ent
11-7
GP
A
x 1
Ag
e
x 2
Sta
te B
oa
rd
Ex
am
y
3.5
28
67
5
2.7
28
57
0
3.2
22
56
0
2.2
23
49
0
2.4
24
55
0
Me
et
1
x 1
Me
et
2
x 2
Sta
te M
ee
t
y
17
.41
6.5
16
.5
15
.41
6.1
15
.8
18
18
.21
6.9
16
.51
6.3
16
.5
17
.21
7.5
16
.1
005_
042_
PC
CR
MC
11_8
9381
2.in
dd41
3/25
/09
12:3
9:36
PM
A01_A19_PCCRMC11_893812.indd 19A01_A19_PCCRMC11_893812.indd 19 11/14/09 5:45:28 PM11/14/09 5:45:28 PM
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cG
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-Hill, a
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isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 11 Assessment Answer Key
Pdf Pass
Chapter 11 A20 Glencoe Precalculus
Quiz 1 (Lessons 11-1 and 11-2) Quiz 3 (Lessons 11-5 and 11-6) Mid-Chapter TestPage 43 Page 44 Page 45
Quiz 2 (Lesson 11-3 and 11-4)
Page 43
Quiz 4 (Lesson 11-7)
Page 44
1.
2.
3.
4.
5.
D
-0.84 < z < 0.84
6.3%
35.8%
P(X < 15.5)
1.
2.
3.
4.
5.
0.607
9.61 < μ < 10.83
at least 35 boys
B
yes, t = 4.02, and 4.02 > 1.83.
1.
2.
3.
B
F
A
1.
2.
3.
4.
5.
[50, 100] scl: 10 by [0, 10] scl: 1
1.
2.
3.
4.
5.
4.
5.
6.
7.
[0, 15] scl: 1 by [0, 1] scl: 0.5
reasonably symmetric
2.57
C
use mean and std. dev. because data is symmetric;
mean: 76.9; std.dev.: 10.5
r ≈ 0.869; fairly strong positive linear correlationyes at all levels; t = 4.65 and t > 1.89, t > 2.36, and t > 3.5.
15
D
y = 1.07x - 1.05^
The median and spread is much greater for those who received the placebo.
z < -0.93 or z > 0.93
2.51 lb
A20_A28_PCCRMC11_893812.indd 20A20_A28_PCCRMC11_893812.indd 20 11/14/09 5:49:24 PM11/14/09 5:49:24 PM
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cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 Assessment Answer Key
An
swer
s
12.
13.
14.
15.
16.
17.
18.
G
D
G
D
J
B
G
B:
19.
20.
99.8%
F
D
Pdf Pass
Chapter 11 A21 Glencoe Precalculus
Vocabulary Test Form 1Page 46 Page 47 Page 48
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
normal distribution
random variable
hypothesis test
confidence interval
t-distribution
inferential statistics
z-value
As n increases, the distribution approaches a normal distribution, −
x approaches μ, and
σ −
x approaches σ −
√ � n .
A distribution that connects all possible values of X with their corresponding probabilities.
correlation coefficent
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
A
H
B
H
A
H
C
F
A
J
A
A20_A28_PCCRMC11_893812.indd 21A20_A28_PCCRMC11_893812.indd 21 11/14/09 6:03:54 PM11/14/09 6:03:54 PM
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-Hill, a
div
isio
n o
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e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
Chapter 11 Assessment Answer Key
12.
13.
14.
15.
16.
17.
18.
J
A
H
B
H
B
J
1.36
J
D
B:
19.
20.
1.22 B:
18.
19.
20. F
D
F
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
D
H
C
G
A
G
D
F
A
G
A
Pdf Pass
Chapter 11 A22 Glencoe Precalculus
Form 2A Form 2BPage 49 Page 50 Page 51 Page 52
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
C
J
A
H
B
G
D
G
A
G
B
12.
13.
14.
15.
16.
17.
H
B
F
D
H
D
A20_A28_PCCRMC11_893812.indd 22A20_A28_PCCRMC11_893812.indd 22 3/25/09 1:05:27 PM3/25/09 1:05:27 PM
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-Hill
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isio
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he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 11 Assessment Answer Key
An
swer
s
Pdf Pass
Chapter 11 A23 Glencoe Precalculus
Form 2CPage 53 Page 54
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
positively skewed
mean: 31.78; median: 21
6.1
2.40
-$4
201.125
about 29.7%
about 89
14.7%
[10, 90] scl: 5 by [0, 25] scl: 1
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
33.7%
0.14
5.06–5.34
28 clocks
H0: μ ≤ 0.910;
Ha: μ > 0.910
(claim)
crit. value: 1.70 test stat: 2.52
yes; 2.52 > 1.65
-0.824
0.8
yes; t = -4.6 and -4.6 < -2.23
y = -0.09x + 106.39^
A20_A28_PCCRMC11_893812.indd 23A20_A28_PCCRMC11_893812.indd 23 11/14/09 6:05:10 PM11/14/09 6:05:10 PM
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-Hill C
om
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ies, In
c.
Chapter 11 Assessment Answer Key
Pdf Pass
Chapter 11 A24 Glencoe Precalculus
Form 2DPage 55 Page 56
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
negatively skewed
mean: 41.9; median: 44
3.95
1.85
-$1.75
134.075
24.8%
about 6 days
69.2%
[0, 100] scl: 10 by [0, 15] scl: 1
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
72.5%
0.06
2.54–2.66
44 people
H0: μ = 0.045 (claim);
Ha: μ ≠ 0.045
crit. value: -2.2 test stat: -0.94
no; -0.94 > -2.2
-0.896
1.41
yes; t = -6.4 and -6.4 < -2.23
y = -2.18x + 9.17^
A20_A28_PCCRMC11_893812.indd 24A20_A28_PCCRMC11_893812.indd 24 11/17/09 9:24:35 AM11/17/09 9:24:35 AM
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-Hill
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McG
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-Hill
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an
ies,
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Chapter 11 Assessment Answer Key
An
swer
s
Pdf Pass
Chapter 11 A25 Glencoe Precalculus
Form 3Page 57 Page 58
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Top 50% of NH speeds are greater than top 75% of MA speeds. NH has greater range and maximumsomewhat negatively skewed
47
621.4; 25
$105
12.9225
12%
about 4 sofas
42.1%
[55, 95] scl: 5 by [0, 1] scl: 0.5
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
40.8%
5.1
444.9–455.1
26 batteries
H0: μ ≤ 10 (claim);
Ha: μ > 10
crit. value: -1.73; test stat: -0.895; P-value: 0.81
no; -0.895 > -1.73 and 0.81 > 0.05.
0.764; slight positive linear
correlation
7 people
0.6395
Signif at 1% only; t = 3.348 and
t > 1.86 and 2.306, but t < 3.355
A20_A28_PCCRMC11_893812.indd 25A20_A28_PCCRMC11_893812.indd 25 11/14/09 6:09:59 PM11/14/09 6:09:59 PM
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ies, In
c.
Chapter 11 Assessment Answer Key
Pdf Pass
Chapter 11 A26 Glencoe Precalculus
Page 59, Extended-Response Test Sample Answers
1a. negatively skewed
1b. Use the five-number summary since the data are skewed: 64, 80, 87, 93, 101. While weights range from 64 to 101 pounds, the median weight is 87 pounds. The middle half of the data varies by 13 pounds.
2a. about 13 students; 94 is 1.5 standard deviations above the mean. The area to the right of 1.5 is about 6.7% and 6.7% of 200 is 13.4.
2b. 88; First, find the z-value, which has 70% of the distribution below it. This is about 0.524. Then solve the formula for z-values to find X.
2c. 13.8%; Using the Central Limit Theorem to adjust the standard deviation, 83 is 1.972 standard deviations below the mean and 84 is 0.986 standard deviations below the mean. The area between -1.972 and -0.986 is about 13.8% of the distribution.
3. H0: μ = 14; Ha: μ ≠ 14; The critical values are -2.14 and 2.14. The test statistic is t = -2.3. Because -2.3 < -2.14, it falls in the critical region. The evidence supports rejecting the company’s claim of a mean of 14 ounces.
4a. 0.965; There is a strong positive linear correlation.
4b. It is significant at all three levels because it is significant at 1%: t = 11.12 and the positive critical t-value for 1%, 9 degrees of freedom is 3.25.
4c. y^ = 0.6x + 2.53; The slope means that for every increase of one ounce in volume, the cost increases by $0.60. The intercept means 0 ounces of food cost $2.53, which does not make sense for this situation.
[60, 105] scl: 5 by [0, 12] scl: 1
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Copyright
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-Hill
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an
ies,
Inc.
Chapter 11 Assessment Answer Key
An
swer
s
Pdf Pass
Chapter 11 A27 Glencoe Precalculus
Standardized Test PracticePage 60 Page 61
1.
2.
3.
4.
5.
6.
7.
8.
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11.
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A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
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Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 11 Assessment Answer Key
Pdf Pass
Chapter 11 A28 Glencoe Precalculus
Standardized Test Practice (continued)Page 62
19.
20.
21.
22.
23.
24.
25a.
25b.
25c.
25d.
25, 34, 43
1; 2π; 1 −
2π
; - π
−
2 ; 2
0; yes
(3, -1), (6, -1), y = 0, x = 0
0, π, 3π
−
4 ,
7π
−
4
H0: μ ≤ 20,000;
Ha: u > 20,000
(claim)
2.326
4.941
yes; 4.941 > 2.326
⎡
⎢
⎣
-15
26
-17
29 ⎤
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