SECTION 3.2 MEASURES OF SPREAD Copyright © The McGraw-Hill Companies, Inc. Permission required for...

Section 3.2Measures of Spread

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Section 3.2 - Objectives

1. Compute the range of a data set2. Compute the variance of a population and a

sample3. Compute the standard deviation of a population

and a sample4. Approximate the standard deviation using

grouped data5. Use the Empirical Rule to summarize data that

are unimodal and approximately symmetric6. Use Chebyshev’s Inequality to describe a data

set7. Compute the coefficient of variation


Compute the range of a data set

Objective 1


The Range

The range of a data set is a measure of spread. That is, it measure how spread out the data are.

The range of a data set is the difference between the largest and the smallest value.

Range = Largest Value – Smallest Value


Example 3.10

The following table presents the average monthly temperature, in degrees Fahrenheit, for the cities of San Francisco and St. Louis. Compute the range for each city.

Jan Feb Mar

Apr May

Jun Jul Aug

Sep

Oct Nov

Dec

San Francisco

51 54 55 56 58 60 60 61 63 62 58 52

St. Louis 30 35 44 57 66 75 79 78 70 59 45 35Source: National Weather Service


Solution

The largest value for San Francisco is 63 and the smallest is 51.

The range for San Francisco is 63 – 51 = 12.

The largest value for St. Louis is 79 and the smallest is 30.

The range for St. Louis is 79 – 30 = 49.


The Range is Not Used in Practice

Although the range is easy to compute, it is not often used in practice. The reason is that the range involves only two values from the data set; the largest and smallest.

The measures of spread that are most often used are the variance and the standard deviation, which use every value in the data set.


Compute the variance of a population and a sample

Objective 2


Variance

When a data set has a small amount of spread, like the San Francisco temperatures, most of the values will be close to the mean. When a data set has a larger amount of spread, more of the data values will be far from the mean.

The variance is a measure of how far the values in a data set are from the mean, on the average.

The variance is computed slightly differently for populations and samples. The population variance is presented first.


Definition: Population Variance

Let x1, x2, x3, … xN denote the values in a population of size N. Let μ denote the population mean. The population variance, denoted by σ2 , is

2

2 ( )ix

N


Procedure for Computing the Population Variance

Following is the procedure for computing the population variance of a data set:

Step 1:Compute the population mean μ.

Step 2:For each population value xi compute xi – μ. This is called the deviation for the value xi.

Step 3:Square the deviations to obtain the quantity (xi – μ)2.

Step 4:Sum the squared deviations to obtain the quantity Σ(xi – μ)2.

Step 5:Divide the sum obtained in Step 4 by the population size N to obtain the population variance σ2.


Example 3.11

Compute the population variance for the San Francisco temperatures.

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

San Francisco

51 54 55 56 58 60 60 61 63 62 58 52


Step 2:For each population value xi compute xi – μ. These values are shown in the second row below.

Xi 51 54 55 56 58 60 60 61 63 62 58 52

xi – μ –6.5 –3.5 –2.5 –1.5 0.5 2.5 2.5 3.5 5.5 4.5 0.5 -5.5

Solution

Step 1: Compute the population mean μ.

51 54 55 56 58 60 60 61 63 62 58 5212

57.5

ix

N


Solution

Step 3:Square the deviations to obtain the quantity (xi – μ)2. These values are shown in the third row.

xi 51 54 55 56 58 60 60 61 63 62 58 52

xi – μ –6.5 –3.5 –2.5 –1.5 0.5 2.5 2.5 3.5 5.5 4.5 0.5 -5.5

(xi – μ)2

42.25

12.25

6.25 2.25 0.25 6.25 6.25 12.25

30.25

20.25

0.25 30.25


Solution

Step 4: Sum the squared deviations to obtain the quantity Σ(xi – μ)2.

Step 5:Divide the sum obtained in Step 4 by the population size N to obtain the population variance σ2.

(xi – μ)2

42.25

12.25

6.25 2.25 0.25 6.25 6.25 12.25

30.25

20.25

0.25 30.25

2( ) 42.25 12.25 6.25 2.25 0.25 6.25 6.25

12.25 30.25 20.25 0.25 30.25

169

ix

2

2 ( ) 16914.083

12ix

N


Sample Variance

When the data values come from a sample rather than a population, the variance is called the sample variance. The procedure for computing the sample variance is a bit different from the one used to compute a population variance.

In the formula, the mean μ is replaced by the sample mean and the denominator is n – 1 instead of N. The sample variance is denoted by s2.

22 ( )

1ix x

sn


Why Divide by n –1?

When computing the sample variance, we use the sample mean to compute the deviations. For the population variance we use the population mean for the deviations.

It turns out that the deviations using the sample mean tend to be a bit smaller than the deviations using the population mean. If we were to divide by n when computing a sample variance, the value would tend to be a bit smaller than the population variance.

It can be shown mathematically that the appropriate correction is to divide the sum of the squared deviations by n –1 rather than n.


Example 3.12

A company that manufactures batteries is testing a new type of battery designed for laptop computers. They measure the lifetimes, in hours, of six batteries, and the results are presented in the following table. Find the sample variance of the lifetimes.

Battery Lifetime

3 4 6 5 4 2


Solution

Step 1: Compute the sample mean.

3 4 6 5 4 24

6ix

xn


Solution

Solution (continued):

Step 2:For each population value xi compute xi – . These values are shown in the second row below.

Step 3:Square the deviations to obtain the quantity (xi – )2. These values are shown in the third row.

xi 3 4 6 5 4 2

xi – –1 0 2 1 0 –2

xi 3 4 6 5 4 2

xi – –1 0 2 1 0 –2

(xi – )2

1 0 4 1 0 4


Solution

Solution (continued):

Step 4: Sum the squared deviations to obtain the quantity Σ(xi – )2.

Step 5:Divide the sum obtained in Step 4 by the n –1 to obtain the sample variance s2.

2( ) 1 0 4 1 0 4 10ix x

22 ( ) 10 10

21 6 1 5

ix xs

n

(xi – )2

1 0 4 1 0 4


Compute the standard deviation of a population and a sample

Objective 3


Standard Deviation

Because the variance is computed using squared deviations, the units of the variance are the squared units of the data. For example, in Battery Lifetime example, the units of the data are hours, and the units of variance are squared hours. In most situations, it is better to use a measure of spread that has the same units as the data.

We do this simply by taking the square root of the variance. This quantity is called the standard deviation. The standard deviation of a sample is denoted s, and the standard deviation of a population is denoted by σ.

2 2s s


Example 3.12

Recall that in the Battery Lifetime example, the sample variance was computed as s2 = 2. Find the sample standard deviation.

Battery Lifetime

3 4 6 5 4 2


Solution

The sample standard deviation, s, is the square root of the sample variance.

2 2 1.414s s


Standard Deviation on the TI-84 PLUSThe following steps will compute the standard deviation for both sample data and population data on the TI-84 PLUS Calculator:

Enter the data into L1 in the data editor.

Run the 1-Var Stats command (the same command used for means and medians), selecting L1 as the location of the data.


Example – 3.14 (TI-84 PLUS Calculator)

Using the TI-84 PLUS Calculator, find the sample variance.

A company that manufactures batteries is testing a new type of battery designed for laptop computers. They measure the lifetimes, in hours, of six batteries, and the results are presented in the following table.

Solution:Running the 1-Var Stats command, we find the sample standard deviation to be s = 1.414213562. We square this quantity to find the sample variance: s2 = (1.414213562)2 = 2.

Battery Lifetime

3 4 6 5 4 2


Standard Deviation & Resistance

Recall that a statistic is resistant if its value is not affected much by extreme data values.

The standard deviation is not resistant.

That is, the standard deviation is affected by extreme data values.


Approximate the standard deviation using grouped data

Objective 4


Approximating the Standard Deviation

Sometimes we don’t have access to the raw data in a data set, but we are given a frequency distribution. In these cases we can approximate the standard deviation.


Approximating the Standard DeviationFollowing is the procedure for approximating the standard deviation:

Step 1: Compute the midpoint of each class and approximate the mean of the frequency distribution.

Step 2: For each class, subtract mean from the class midpoint to obtain (Midpoint – Mean).

Step 3: For each class, square the differences obtained in Step 2 to obtain (Midpoint – Mean)2, and multiply by the frequency to obtain (Midpoint – Mean)2 x (Frequency).

Step 4: Add the products (Midpoint – Mean)2 x (Frequency) over all classes.

Step 5: To compute the population variance, divide the sum obtained in Step 4 by n. To compute the sample variance, divide the sum obtained in Step 4 by n –1.

Step 6: Take the square root of the variance obtained in Step 5. The result is the standard deviation.Copyright © The McGraw-Hill Companies, Inc. Permission required for

reproduction or display.

Example 3.16

The following table presents the number of text messages sent via cell phone by a sample of 50 high school students. Approximate the sample standard deviation number of messages sent.Number of Text Messages

SentFrequen

cy

0 – 49 10

50 – 99 5

100 – 149 13

150 – 199 11

200 – 249 7

250 – 299 4


Solution

Step 1: Compute the midpoint of each class. Recall from Section 3.1 that the sample mean was computed as 137.

Number of Text Messages Sent

Class Midpoin

t

0 – 49 25

50 – 99 75

100 – 149 125

150 – 199 175

200 – 249 225

250 – 299 275


Solution

Step 2: For each class, subtract mean from the class midpoint to obtain (Midpoint – Mean).


Class Midpoin

t

(Midpoint – Mean)

0 – 49 25 –112

50 – 99 75 –62

100 – 149 125 –12

150 – 199 175 38

200 – 249 225 88

250 – 299 275 138


Solution

Step 3: For each class, square the differences obtained in Step 2 to obtain (Midpoint – Mean)2, and multiply by the frequency to obtain (Midpoint – Mean)2 x (Frequency).


Frequency

(Midpoint – Mean)

(Midpoint – Mean)2

x (Frequen

cy)

0 – 49 10 –112 125,440

50 – 99 5 –62 19,220

100 – 149 13 –12 1,872

150 – 199 11 38 15,884

200 – 249 7 88 54,208

250 – 299 4 138 76,176Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Solution

Step 4: Add the products (Midpoint – Mean)2 x (Frequency) over all classes.

(Midpoint – Mean)2 x

(Frequency)

125,440

19,220

1,872

15,884

54,208

76,176

2(Midpoint-Mean) Frequency

= 125,440+19,220+1,872+15,884+54,208+76,176

292,800


Solution

Step 5: Since we are computing the sample variance, we divide the sum obtained in Step 4 by n –1.

Step 6: Take the square root of the variance to obtain the standard deviation.

2

2(Midpoint-Mean) Frequency 292,800

1 50 1s

n

5975.51020

2 5975.51020 77.30142s s


Grouped Data on the TI-84 PLUS The same procedure used to compute the mean for grouped data in a frequency distribution may be used to compute the standard deviation.

Enter the midpoint for each class into L1 and the corresponding frequencies in L2. Next, select the 1-Var stats command and enter L1 in the List field and L2 in the FreqList field, if using Stats Wizards. If you are not using Stats Wizards, you may rund1-Var Stats command followed by L1, comma, L2.


Example 3.16 using TI-84 PLUS

The output for Example 3.16 on the TI-84 PLUS Calculator is presented below.

The value of s represents the approximate sample standard deviation. In this example s = 77.30142. Therefore the approximate standard deviation is 77.30142.

Class Midpoint

Frequency

25 10

75 5

125 13

175 11

225 7

275 4


Use the Empirical Rule to summarize data that are unimodal and approximately symmetric

Objective 5


Bell-Shaped Histogram

Many histograms have a single mode near the center of the data, and are approximately symmetric. Such histograms are often referred to as bell-shaped.


The Empirical Rule

When a data set has a bell-shaped histogram, it is often possible to use the standard deviation to provide an approximate description of the data using a rule known as The Empirical Rule.

When a population has a histogram that is approximately bell-shaped, then:Approximately 68% of the data will be within one standard deviation of the mean. In other words, approximately 68% of the data will be in the interval µ - σ to µ + σ.Approximately 95% of the data will be within two standard deviations of the mean. In other words, approximately 95% of the data will be in the interval µ - 2σ to µ + 2σ .All, or almost all, of the data will be within three standard deviations of the mean. In other words, all, or almost all, of the data will be in the interval µ - 3σ to µ + 3σ.


The Empirical Rule


Example 3.17

The following table presents the U.S. Census Bureau projection for the percentage of the population aged 65 and over for each state and the District of Columbia. Compute the population mean and standard deviation and use The Empirical Rule to describe the data.Alabama 14.1 Rhode Island 14.1 Nevada 12.3 Kentucky 13.1

Arkansas 14.3 Tennessee 13.3 New Mexico 14.1 Maryland 12.2Connecticut 14.4 Vermont 14.3 North Dakota 15.3 Minnesota 12.4

Florida 17.8 West Virginia 16 Oregon 13 Montana 15

Idaho 12 Alaska 8.1South

Carolina 13.6New

Hampshire 12.6Iowa 14.9 California 11.5 Texas 10.5 New York 13.6

Louisiana 12.6 Delaware 14.1 Virginia 12.4 Ohio 13.7Massachusett

s 13.7 Georgia 10.2 Wisconsin 13.5 Pennsylvania 15.5Mississippi 12.8 Illinois 12.4 Arizona 13.9 South Dakota 14.6Nebraska 13.8 Kansas 13.4 Colorado 10.7 Utah 9

New Jersey 13.7 Maine 15.6 D.C. 11.5 Washington 12.2North

Carolina 12.4 Michigan 12.8 Hawaii 14.3 Wyoming 14Oklahoma 13.8 Missouri 13.9 Indiana 12.7


Solution

We first note that the histogram is approximately bell-shaped.

We may use the TI-84 PLUS Calculator – or other technology – to compute the population mean and population standard deviation.

Mean: µ = 13.24901961

Standard Deviation: σ = 1.682711694


Solution

Next, we compute the quantities:

3 13.24901961 3(1.682711694) 8.20

3 13.24901961 3(1.682711694) 18.30

Approximately 68% of the data values are between these.




Almost all of the data values are between these.

Almost all of the data values are between these.

2 13.24901961 2(1.682711694) 9.88

2 13.24901961 2(1.682711694) 16.61

13.24901961 1.682711694 11.57

13.24901961 1.682711694 14.93


Solution

8.20 9.88 11.57 14.93 16.61 18.30


Use Chebyshev’s Inequality to describe a data set

Objective 6


Any Data Set

When a distribution is bell-shaped, we use The Empirical Rule to approximate the proportion of data within one or two standard deviations. Another rule called Chebyshev’s Inequality holds for any data set.


Chebyshev’s Inequality

In any data set, the proportion of the data that is within K standard deviations of the mean is at least 1– 1/K2. Specifically, by setting K = 2 or K = 3, we obtain the following results.

At least 3/4 (75%) of the data are within two standard deviations of the mean.At least 8/9 (89%) of the data are within three standard deviations of the mean.


Example 3.20

As part of a public health study, systolic blood pressure was measured for a large group of people. The mean was 120 and the standard deviation was 10. What information does Chebyshev’s Inequality provide about these data?


Solution

We compute the following:

We conclude:At least 3/4 (75%) of the people had systolic blood pressures between 100 and 140. At least 8/9 (89%) of the people had systolic blood pressures between 90 and 150.

2 120 2(10) 100

2 120 2(10) 140

x s

x s

3 120 3(10) 90

3 120 3(10) 150

x s

x s


Compute the coefficient of variation

Objective 7


Coefficient of Variation

The coefficient of variation (CV for short) tells how large the standard deviation is relative to the mean. It can be used to compare the spreads of data sets whose values have different units.

The coefficient of variation is found by dividing the standard deviation by the mean.

CV


Example 3.21

National Weather service records show that over a thirty-year period, the annual precipitation in Atlanta, Georgia had a mean of 49.8 inches with a standard deviation of 7.6 inches, and the annual temperature had a mean of 62.2 degrees Fahrenheit with a standard deviation of 1.3 degrees. Compute the coefficient of variation for precipitation and for temperature. Which has greater spread relative to its mean?


Solution

We compute the following:

The CV for precipitation is larger than the CV for temperature. Therefore precipitation has a greater spread relative to its mean.

standard deviation for precipitation 7.6CV for precipitation = 0.15

mean precipitation 49.8

standard deviation for temperature 1.3CV for temperature = 0.02

mean temperature 62.2


Do You Know…

• How to compute the range of a data set?• How to compute the variance of a population and

a sample and the appropriate notation?• How to compute the standard deviation of a

population and a sample and the appropriate notation?

• How to approximate the standard deviation using grouped data?

• How to use the Empirical Rule to summarize data?• How to use Chebyshev’s Inequality to describe a

data set?• How to compute the coefficient of variation?Copyright © The McGraw-Hill Companies, Inc. Permission

required for reproduction or display.

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