Pre-Algebra: Week of May 11 – May 15, 2020...2020/05/07 · Pre-Algebra: Chapter 12 Statistics...

Pre-Algebra: Week of May 11 – May 15, 2020 Time Allotment: 40 minutes per day

Student Name: ________________________________ Teacher Name: ________________________________

Pre-Algebra: Chapter 12 Statistics May 11 – May 15

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Packet Overview Date Objective(s) Page Number

Monday, May 11 Construct a Stem-and-Leaf Plot from Analyzed Data 2

Tuesday, May 12 Construct a Box-and-Whisker Plot for a given set of Data 6

Wednesday, May 13 Construct a Histogram 10

Thursday, May 14 Calculating the Standard Deviation 13

Friday, May 15 Calculating the Standard Deviation ***Quiz***

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Notes: We included space for you to write your work directly in this packet on each day. Show your work where you can. All problems can be done in the packet, so all you need to submit at the end of this week is this completed packet! ***Checking/correcting your answers using a red pen is required! Options to turn in completed packets:

1. Upload it to your Google Classroom! This allows for quicker review, more in-depth feedback and easier communication.

2. Physically turn it back in. Guided Instruction via Zoom. For example, if you had Pre-Algebra with Mrs. Walters during Period 5, then on Tuesday and Thursday from 11-11:50 you can connect with her via Zoom.

Email Us: [email protected] or [email protected] Academic Honesty

I certify that I completed this assignment independently in accordance with the GHNO

Academy Honor Code. Student signature:

___________________________

I certify that my student completed this assignment independently in accordance with the

GHNO Academy Honor Code. Parent signature:

___________________________


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Monday, May 11, 2020 Pre-Algebra: Chapter 12 Lesson: Statistics Objective: Construct a Stem-and-Leaf Plot from Analyzed Data Lesson When we gather data, we often need to display the data so that it can be analyzed. Suppose the following test scores were made by a class of 30 students:

92, 75, 69, 56, 88, 62, 75, 82, 90, 74, 65, 70, 80, 96, 81, 87, 95, 58, 96, 89, 91, 81, 83, 94, 86, 88, 95, 74, 87, 85

We can find the mean (average) of the scores by adding the scores and dividing by 30. If we do this, we find that the mean score is 81.47.

Mean = !"#%&'()*+!,%-)!

*+= .///

*+= 81.47

To find the range, mode and median of the scores, we need to arrange the numbers in order from least to greatest. A graphing calculator will do this for us in one step. If a graphing calculator is not available, a stem-and-leaf plot can be used to organize the data quickly. We note that all the scores begin with 5, 6, 7, 8, and 9, so we list these numbers vertically as our stem, as shown in the leftmost column. STEM STEM LEAF 5 5 6 6 6 9 7 7 5 8 8 8 9 9 2 We call the second part of each number a leaf. The first five numbers in our list are 92, 75, 69, 56, and 88. So we use 2 as the first leaf on the 9 stem, 5 as the first leaf on the 7 stem, 9 as the first leaf on the 6 stem, 6 as the first leaf on the 5 stem, and 8 as the first leaf on the 8 stem, as we show in the figure on the right above. Now we place the rest of the leaves on the proper stem in the order that they appear, as shown.

STEM STEM LEAF 5 5 6, 8 6 6 9, 2, 5 7 7 5,5, 4, 0, 4 8 8 8, 2, 0, 1, 7, 9, 1, 3, 6, 8, 7, 5 9 9 2, 0, 6, 5, 6, 1, 4, 5 This method of displaying data is called a stem-and-leaf plot since the first number can be thought of as a branch on a tree and the numbers to the right represent leaves on the branch.


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Now, if we arrange the leaves on each stem in order from least to greatest, we get the figure below. STEM STEM LEAF 5 5 6, 8 6 6 2, 5, 9 7 7 0, 4, 4, 5, 5 8 8 0, 1, 1, 2, 3, 5, 6, 7, 7, 8, 8, 9 9 9 0, 1, 2, 4, 5, 5, 6, 6 The stem-and- leaf plot has the same shape as the one above, but we can use this plot to arrange the scores in order from least to greatest.

56, 58, 62, 65, 69, 70, 74, 74, 75, 75, 80, 81, 81, 82, 83, 85, 86, 87, 87, 88, 88, 89, 90, 91, 92, 94, 95, 95, 96, 96

We know that it is possible to compute the mean of the scores without arranging the scores in order from least to greatest. But now that the scores are in order from least to greatest, we can find that range, mean, mode, and median. The scores range from 56 to 96 and the range of the scores is the difference between these numbers which is 40.

Range = 96 – 56 = 40 From our list of scores, we can see that seven scores appear twice. But no score appears more than twice, so there are seven modes. They are 74, 75, 81, 87, 88, 95, and 96. There are 30 scores and so there are an even number of scores. Thus, the median score will be halfway between the fifteenth and sixteenth score. The fifteenth score is 83 and the sixteenth score is 85, so the median score is 84.

Median = 0*102.

= 84 Your Turn to try it!

Please write answer here:


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Please complete the class exercises on the space provided below.

Write your answers for the Class Exercises in the space below:

1. 2. 3. 4.

***Please check answer key before proceeding forward Exercises for Monday, May 4, 2020 *PLEASE circle answers here and not on an answer packet. Thanks!


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Please write the answers in the space provided here below:

1. 2. 3. 4.

5. 6. 7.


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Tuesday, May 12, 2020 Pre-Algebra: Chapter 12 Lesson: Statistics Objective: Construct a Box-and-Whisker Plot for a given set of Data Lesson Another graphical method used to display data is called the box-and-whisker plot. It uses five dots, two rectangles called boxes, and lines on the end of the boxes called whiskers to divide a set of data into four equal or approximately equal groups. We will show how to construct a box-and-whisker plot for a given set of data. We shall use the same data as presented in yesterday’s lesson. We were given test scores for a class of 30 students. We begin by rewriting the test scores in order from least to greatest.

56, 58, 62, 65, 69, 70, 74, 74, 75, 75, 80, 81, 81, 82, 83, 85, 86, 87, 87, 88, 88, 89, 90, 91, 92, 94, 95, 95, 96, 96

Now we draw a number line that covers the full range of our data. 56 84 96 Least score Median Greatest Score 50 60 70 80 90 100 The three dots above the number line represent the location of the least score 56, the median score 84, and the greatest score 96. Now we find the median score of those scores less than 84 and the median score of those scores greater than 84. There are fifteen scores less than 84. Fifteen is an odd number so the median score of these numbers, called the first quartile, is the eighth score on the ordered list. Therefore,

First quartile = 74 There are fifteen scores greater than 84. Fifteen is an odd number, so that median score between these numbers, called the third quartile, is the eighth score after the median of the ordered list. Therefore,

Third quartile = 90

Now we have five dots above the number line that represent the location of the least score 56. The first quartile score is 74, the median score 84, the third quartile score 90, and the greatest score 96, as shown below 56 74 84 90 96 Least score First quartile Median 3rd Greatest

Quartile Score 50 60 70 80 90 100


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Next, we make a box with the first quartile score and the third quartile score on the outer sides. Then we draw a line inside the box, through the median score. 56 74 84 90 96 Least score First quartile Median 3rd Greatest

Quartile Score 50 60 70 80 90 100 Finally, we draw two lines called whiskers from the sides of the box to the dots that mark the location of the least score and the greatest score, as shown below. 56 74 84 90 96 Least score First quartile Median 3rd Greatest

Quartile Score 50 60 70 80 90 100 The box-and-whisker plot shown above separates the scores into four approximately equal groups. This allows us to see how the scores are spread out. Approximately one fourth of the scores lie in the region denoted by the leftmost whisker, which extends from the least score to the first quartile. The left half of the box, which extends from the first quartile to the median, shows where the next approximate one fourth of the scores lie. The right half of the box, which extends from the median to the third quartile shows where the next approximate one fourth of the scores lie. The next approximate one fourth of the scores lie in the region denoted by the rightmost whisker, which extends from the third quartile to the greatest score. Thus, the box contains approximately one half of the data points, while each whisker contains approximately one fourth of the data points.


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Try it! Use the space below…not extra paper needed!

***Please check the answer key before moving forward.

Exercises for Tuesday, May 12, 2020 ***Reminder fill in answers here.*** Written exercises page 460 #1 – 6

STEPS FOR EACH PROBLEM

1. Order the list from least to greatest 2. Draw a number line that covers the full range of the data 3. Put a dot on the number line for least score, median and greatest score (There should be three dots!) 4. Calculate 1st and 3rd Quartile (Remember 1st Quartile is the median between the least score and

overall median and the 3rd Quartile is the median between the greatest score and overall median.) 5. Put a date on the number line for the 1st Quartile and 3rd Quartile (There should now be five dots!) 6. Make a “Box” around 1st Quartile, Median and 3rd Quartiles 7. Make “Whiskers” by drawing a line from the box to both the least score and the greatest score


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Wednesday, May 13, 2020 Pre-Algebra: Chapter 12 Lesson: Statistical Measures Objective: Construct a Histogram Lesson The data from our stem-and-leaf plot of the test scores show us that there were two scores in the 50s, three scores in the 60s, five scores in the 70s, twelve scores in the 80s, and eight scores in the 90s. If we turn the stem-and-leaf plot on its side, we get the figure on the left below. 5 7 8 6 3 5 1 4 9 1 4 7 6 0 1 5 5 4 0 6 8 2 5 2 0 6 9 5 8 2 5 6 7 8 9

On the right above we show that same data displayed in a figure called a histogram, which is a bar graph of a frequency distribution. As we see, histograms do not give us precise information but do give us at a glance a good idea of how data is distributed. Try it!

Please read the examples on pages 461 – 462 for more examples in your textbook and answer the following questions:

(a) On page 462, what is the range of the scores? (b) How many students received scores less than 7?

(c) How many students took the quiz?

***Please check the answer key before moving forward.

Test Scores

[56, 64] (64, 72] (72, 80] (80, 88] (88, 96]

Num

ber o

f Tes

t Sco

res

0

2

4

6

8

10

12

Histogram


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Exercises for Wednesday, May 13, 2020

For Exercises 11 – 14, x represents an item of data and f represents the frequency of the item. Make a histogram from the data given.


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Make a histogram for each set of data

1.

2. 3. 4.

5.

6. 7. 8.


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Thursday, May 14, 2020 Pre-Algebra: Chapter 12 Lesson: Statistical Measures Objective: Calculating the Standard Deviation Lesson: Standard Deviation

The Standard Deviation is a measure of how spread out numbers are.

Its symbol is σ (the Greek letter sigma)

The formula is easy: it is the square root of the Variance.

So now you ask, "What is the Variance?"

Variance is defined as: the Average of the squared differences from the Mean.

To calculate the variance, follow these steps:

• Work out the Mean • Then, for each number: subtract the Mean and the square the result (the squared difference) • Then, work out the average of those squared differences.

Example:

You and your friends have just measured the heights of your dogs (in millimeters):

The heights (at the shoulders) are: 600mm, 470mm, 170mm, 430mm and 300mm.

Find out the Mean, the Variance, and the Standard Deviation.


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Your first step is to find the Mean:

Mean = !""#$%"#&%"#$'"#'""(

= &)%"(= 394

So, the mean height is 394mm. Let’s plot on the chart:

Now, we calculate each dog’s difference from the Mean:

To calculate the Variance, take each difference, square it, and then average the result:

Variance

𝜎* =206* + 76* + (−224)* + 36* + (−94)*

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= $*$'!#(%%!#("&%!#&*)!#++'!

(

= &"+(*"

(

= 21704 So, the variance is 21,704


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And, the Standard Deviation is just the square root of Variance, so: Standard Deviation

𝜎 = √21704 = 147.32… = 147(𝑡𝑜𝑡ℎ𝑒𝑛𝑒𝑎𝑟𝑒𝑠𝑡𝑚𝑚) The Standard Deviation shows us how far away each point is away from the Mean.

So, using the Standard Deviation we have a "standard" way of knowing what is normal, and what is extra-large or extra small.

Rottweilers are tall dogs. And Dachshunds are a bit short, right? YES!

Exercises for Thursday May 14, 2020


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Question 1: Calculate the mean score for each boy:

NOTE: The mean provides a measure of the center of a statistical variable, but not information about how spread out the scores may be. To measure the ‘spread’ of the data, we will look at how much each individual score deviates (is different) from the mean score. Question 2: Variance

Rudi’s

Question 3: Calculate the Standard Deviation for each boy. Question 4:


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Friday, May 15, 2020 Pre-Algebra: Chapter 12 Lesson: Statistical Measures Objective: Calculating Standard Deviation

***Please take the quiz on pages 21-22 after completing today’s lesson*** Lesson: Take a couple of minutes to review Thursday’s Lesson to include definitions and the formula. Try it!

#1 The standard deviation of a student's grades is zero. What do you know about the grades of this student? Why do you say so? A. The student scored low grades. B. The student scored the same grade every time. C. The student's grade is unpredictable. D. The student got a B on all tests. #2

Variance:

Standard Deviation:

***Make sure you check your answers to Try It! There is a similar problem to #2 on the quiz!

***Also, No Friday exercises. Just review the packet for the week and take the quiz


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ANSWER KEY Answers to Exercises for Monday, May 11, 2020 Try It!

Solution for the Class Exercises:

Answers to Exercise problem set:

Answers to Exercises for Tuesday, May 12, 2020

Try it! class exercises

Written Exercises

Answers to Exercises for Wednesday, May 13, 2020

Try it! a)8 b)15 c)40


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Exercises pages 463 – 464 #11-16 in your textbook:

Review Exercises:

Answers to Exercises for Thursday, May 14, 2020 Answers to Jamal and Rudi Cricket Club Scores: Question 1: a. Jamal’s mean batting score = *)#'(#*)#*%

$= &*"

$= 30

Rudi’s mean batting score = %#&*#&+#+'$

= &*"$= 30

b. The mean tells us that Jamal and Rudi have the same ability. The mean does NOT tell us that Jamal is the more consistent player.

Question 2: The Variance

Jamal’s mean squared deviation = +$&#&**(#+$&#%*)$

= '!'!$= 909

Rudi’s mean batting score = $)#&$$#'*$#!++)$

= %$"!$= 1851.5

Question 3: The Standard Deviation

Jamal = √909 = 30.15 and Rudi = √1851.5 = 43.03

Question 4:

You notice Rudi has a bigger standard deviation which means Rudi is susceptible for having much lower and much higher scores, and thus it is unpredictable what type of game he will have. Whereas, Jamal as a much lower standard deviation which means Jamal can be counted on to consistently have around the same score.


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Answers to Try It! for Friday, May 15, 2020

#1: The answer is B. If the student scored the same grade every time, then the student's grade did not deviate from the mean at all. With no deviation, the standard deviation is 0

#2:


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Week 7 Quiz

Questions 1 and 2 Refer to the bar graph to the right. (Note, graph also found on page 471 of textbook.)

1. What is the approximate area of Delaware?

_________________________________

2. About how much smaller than New Jersey is Rhode Island? (Hint: Pay close attention to units!) _________________________________

Questions 3 and 4 Refer to the line graph to the right. (Note, graph also found on page 471 of textbook.)

3. During which time period was there no increase in area? _________________________________

4. During which 80 years was there the greatest increase? _________________________________

5. What was the approximate area of the United States in 1840? ____________________________

For Questions 6-8 refer to this data: 28, 15, 28, 15, 40, 40, 21, 28, 28, 47, 40

6. Find the:

a. Range: b. Mean: c. Median: d. Mode:

7. Make a steam-and-left plot of the given data

8. Make a box-and-whisker plot of the given data

***Quiz Continued on Back****


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9. Find the variance and standard deviation of the average temperatures recorded over a five-day period last winter:

18, 22, 19, 25, 12

Variance:

Standard Deviation:

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Pre-Algebra: Week of May 11 – May 15, 2020...2020/05/07 · Pre-Algebra: Chapter 12 Statistics...

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