The Normal Distribution

Then place the mean plus one standard deviation 26 to the right, the mean plus two standard deviations 28 to the right, the mean plus three standard deviations 30 to the right, the mean

The Normal Distribution

Introduction

A branch of mathematics that uses probability is called statistics Statistics is

the branch of mathematics that uses observations and measurements called

data to analyze, summarize, make inferences, and draw conclusions based on

the data gathered This chapter will explain some basic concepts of statistics

such as measures of average and measures of variation Finally, the

relationship between probability and normal distribution will be explained

in the last two sections

147

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Measures of Average

There are three statistical measures that are commonly used for average.They are the mean, median, and mode The mean is found by adding the datavalues and dividing by the number of values

EXAMPLE: Find the mean of 18, 24, 16, 15, and 12

SOLUTION:

Add the values: 18 þ 24 þ 16 þ 15 þ 12 ¼ 85Divide by the number of values, 5: 85  5 ¼ 17Hence the mean is 17

EXAMPLE: The ages of 6 executives are 48, 56, 42, 52, 53 and 52 Find themean

SOLUTION:

Add: 48 þ 56 þ 42 þ 52 þ 53 þ 52 ¼ 303Divide by 6: 303  6 ¼ 50.5

Hence the mean age is 50.5

The median is the middle data value if there is an odd number of datavalues or the number halfway between the two data values at the center, ifthere is an even number of data values, when the data values are arranged inorder

EXAMPLE: Find the median of 18, 24, 16, 15, and 12

SOLUTION:

Arrange the data in order: 12, 15, 16, 18, 24Find the middle value: 12, 15, 16, 18, 24The median is 16

EXAMPLE: Find the median of the number of minutes 10 people had to wait

in a checkout line at a local supermarket: 3, 0, 8, 2, 5, 6, 1, 4, 1, and 0

SOLUTION:

Arrange the data in order: 0, 0, 1, 1, 2, 3, 4, 5, 6, 8The middle falls between 2 and 3; hence, the median is (2 þ 3)  2 ¼ 2.5

The third measure of average is called the mode The mode is the data value

that occurs most frequently

EXAMPLE: Find the mode for 22, 27, 30, 42, 16, 30, and 18

In this example, 3 and 6 occur most often; hence, 3 and 6 are used as the

mode In this case, we say that the distribution is bimodal

EXAMPLE: Find the mode for 18, 24, 16, 15, and 12

SOLUTION:

Since no value occurs more than any other value, there is no mode

A distribution can have one mode, more than one mode, or no mode

Also, the mean, median, and mode for a set of values most often differ

somewhat

PRACTICE

1 Find the mean, median, and mode for the number of sick days nine

employees used last year The data are 3, 6, 8, 2, 0, 5, 7, 8, and 5

2 Find the mean, median, and mode for the number of rooms seven

hotels in a large city have The data are 332, 256, 300, 275, 216, 314,

and 192

3 Find the mean, median, and mode for the number of tornadoes that

occurred in a specific state over the last 5 years The data are

18, 6, 3, 9, and 10

4 Find the mean, median, and mode for the number of items 9 people

purchased at the express checkout register The data are

12, 8, 6, 1, 5, 4, 6, 2, and 6

5 Find the mean, median, and mode for the ages of 10 children who

participated in a field trip to the zoo The ages are 7, 12, 11, 11, 5,

8, 11, 7, 8, and 6

Mode ¼ 11

Measures of Variability

In addition to measures of average, statisticians are interested in measures ofvariation One measure of variability is called the range The range is thedifference between the largest data value and the smallest data value.EXAMPLE: Find the range for 27, 32, 18, 16, 19, and 40

Since the largest data value is 40 and the smallest data value is 16, the range is

40  16 ¼ 24

Another measure that is also used as a measure of variability for individual

data values is called the standard deviation This measure was also used in

Chapter 7

The steps for computing the standard deviation for individual data

values are

Step 1: Find the mean

Step 2: Subtract the mean from each value and square the differences

Step 3: Find the sum of the squares

Step 4: Divide the sum by the number of data values minus one

Step 5: Take the square root of the answer

EXAMPLE: Find the standard deviation for 32, 18, 15, 24, and 11

Recall from Chapter 7 that most data values fall within 2 standard

deviations of the mean In this case, 20  2(8.22) is 3.56 < most

values < 36.44 Looking at the data, you can see all the data values fallbetween 3.56 and 36.44.

EXAMPLE: Find the standard deviation for the number of minutes 10 peoplewaited in a checkout line at a local supermarket The times in minutes are

2 Eight students were asked how many hours it took them to write a

research paper Their times (in hours) are shown: 6, 10, 3, 5, 7, 8, 2, 7

Find the range and standard deviation

3 The high temperatures for 10 selected cities are shown: 32, 19,

57, 48, 44, 50, 42, 49, 53, 46 Find the range and standard deviation

4 The times in minutes it took a driver to get to work last week are

shown: 32, 35, 29, 31, 33 Find the range and standard deviation

5 The number of hours 8 part-time employees worked last week is

shown: 26, 28, 15, 25, 32, 36, 19, 11 Find the range and standard

11¼6:73

ffiffiffiffiffiffiffiffiffi6:73p

¼2:59 ðroundedÞ

7 ¼6:86

ffiffiffiffiffiffiffiffiffi6:86

9 ¼122:67

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi122:67p

¼11:08

4 ¼5

ffiffiffi5

7 ¼72

ffiffiffiffiffi72p

¼8:49 ðroundedÞ

The Normal Distribution

Recall from Chapter 7 that a continuous random variable can assume allvalues between any two given values For example, the heights of adultmales is a continuous random variable since a person’s height can beany number We are, however, limited by our measuring instruments.The variable temperature is a continuous variable since temperature canassume any numerical value between any two given numbers Manycontinuous variables can be represented by formulas and graphs or curves.These curves represent probability distributions In order to find probabilitiesfor values of a variable, the area under the curve between two given values

is used

One of the most often used continuous probability distributions iscalled the normal probability distribution Many variables are approxi-mately normally distributed and can be represented by the normal distribu-tion It is important to realize that the normal distribution is a perfecttheoretical mathematical curve but no real-life variable is perfectly normallydistributed

The real-life normally distributed variables can be described by thetheoretical normal distribution This is not so unusual when you think about

it Consider the wheel It can be represented by the mathematically perfectcircle, but no real-life wheel is perfectly round The mathematics of the circle,then, is used to describe the wheel

The normal distribution has the following properties:

1 It is bell-shaped

2 The mean, median, and mode are at the center of the distribution

3 It is symmetric about the mean (This means that it is a reflection ofitself if a mean was placed at the center.)

4 It is continuous; i.e., there are no gaps

5 It never touches the x axis

6 The total area under the curve is 1 or 100%

7 About 0.68 or 68% of the area under the curve falls within onestandard deviation on either side of the mean (Recall that
is thesymbol for the mean and  is the symbol for the standard deviation.)About 0.95 or 95% of the area under the curve falls within twostandard deviations of the mean

About 1.00 or 100% of the area falls within three standard deviations

of the mean (Note: It is somewhat less than 100%, but for simplicity,100% will be used here.) See Figure 9-1

EXAMPLE: The mean commuting time between a person’s home and office

is 24 minutes The standard deviation is 2 minutes Assume the variable is

normally distributed Find the probability that it takes a person between

24 and 28 minutes to get to work

SOLUTION:

Draw the normal distribution and place the mean, 24, at the center Then

place the mean plus one standard deviation (26) to the right, the mean plus

two standard deviations (28) to the right, the mean plus three standard

deviations (30) to the right, the mean minus one standard deviation (22) to

the left, the mean minus two standard deviations (20) to the left, and the

mean minus three standard deviations (18) to the left, as shown in Figure 9-2

Using the areas shown in Figure 9-1, the area under the curve between 24

and 28 minutes is 0.341 þ 0.136 ¼ 0.477 or 47.7% Hence the probability that

the commuter will take between 24 and 28 minutes is about 48%

Fig 9-1.

Fig 9-2.

EXAMPLE: According to a study by A.C Neilson, children between 2 and 5years of age watch an average of 25 hours of television per week Assume thevariable is approximately normally distributed with a standard deviation

of 2 If a child is selected at random, find the probability that the childwatched more than 27 hours of television per week

SOLUTION:

Draw the normal distribution curve and place 25 at the center; then place

27, 29, and 31 to the right corresponding to one, two, and three standarddeviations above the mean, and 23, 21, and 19 to the left corresponding toone, two, and three standard deviations below the mean Now place theareas (percents) on the graph See Figure 9-3

Since we are finding the probabilities for the number of hours greater than

27, add the areas of 0.136 þ 0.023 ¼ 0.159 or 15.9% Hence, the probability isabout 16%

EXAMPLE: The scores on a national achievement exam are normallydistributed with a mean of 500 and a standard deviation of 100 If a studentwho took the exam is randomly selected, find the probability that the studentscored below 600

SOLUTION:

Draw the normal distribution curve and place 500 at the center Place

600, 700, and 800 to the right and 400, 300, and 200 to the left, corresponding

to one, two, and three standard deviations above and below the mean tively Fill in the corresponding areas See Figure 9-4

respec-Fig 9-3.

Since we are interested in the probability of a student scoring less than

600, add 0.341 þ 0.341 þ 0.136 þ 0.023 ¼ 0.841 ¼ 84.1% Hence, the

prob-ability of a randomly selected student scoring below 600 is 84%

PRACTICE

1 To qualify to attend a fire academy, an applicant must take a written

exam If the mean of all test scores is 80 and the standard deviation is

5, find the probability that a randomly selected applicant scores

between 75 and 95 Assume the test scores are normally distributed

2 The average time it takes an emergency service to respond to calls

in a certain municipality is 13 minutes If the standard deviation is

3 minutes, find the probability that for a randomly selected call, the

service takes less than 10 minutes Assume the times are normally

distributed

3 If the measure of systolic blood pressure is normally distributed with

a mean of 120 and a standard deviation of 10, find the probability

that a randomly selected person will have a systolic blood pressure

below 140 Assume systolic blood pressure is normally distributed

4 If an automobile gets an average of 25 miles per gallon on a trip and

the standard deviation is 2 miles per gallon, find the probability that

on a randomly selected trip, the automobile will get between 21 and

29 miles per gallon Assume the variable is normally distributed

5 If adult Americans spent on average $60 per year for books and the

standard deviation of the variable is $5, find the probability that a

randomly selected adult spent between $50 and $65 last year on

books Assume the variable is normally distributed

Fig 9-4.

4 The required area is shown in Figure 9-8.

Probability ¼ 0.136 þ 0.341 þ 0.341 þ 0.136 ¼ 0.954 or 95.4%

5 The required area is shown in Figure 9-9

Probability ¼ 0.136 þ 0.341 þ 0.341 ¼ 0.818 or 81.8%

The Standard Normal Distribution

The normal distribution can be used as a model to solve many problems

about variables that are approximately normally distributed Since each

variable has its own mean and standard deviation, statisticians use what is

called the standard normal distribution to solve the problems

The standard normal distribution has all the properties of a normal

dis-tribution, but the mean is zero and the standard deviation is one See

Figure 9-10

Fig 9-9.

Fig 9-8.

A value for any variable that is approximately normallydistributed can be transformed into a standard normal value by using thefollowing formula:

z ¼ value  mean

standard deviationThe standard normal values are called z values or z scores

EXAMPLE: Find the corresponding z value for a value of 18 if the mean

of a variable is 12 and the standard deviation is 4

z values are negative for values of variables that are below the mean.EXAMPLE: Find the corresponding z value for a value of 9 if the mean of avariable is 12 and the standard deviation is 4

In addition to finding probabilities for values that are between zero, one,two, and three standard deviations of the mean, probabilities for other values

Fig 9-10.

can be found by converting them to z values and using the standard normal

distribution

Areas between any two given z values under the standard normal

distribution curve can be found by using calculus instead; however, tables

for specific z values can be found in any statistics textbook An abbreviated

table of areas is shown in Table 9-1

Table 9-1 Approximate Cumulative Areas for the Standard Normal Distribution

z Area z Area z Area z Area

This table gives the approximate cumulative areas for z values between

3 and þ3 The next three examples will show how to find the area (andcorresponding probability in decimal form)

EXAMPLE: Find the area under the standard normal distribution curve tothe left of z ¼ 1.3

SOLUTION:

The area is shown in Figure 9-11

In order to find the area under the standard normal distribution curve tothe left of any given z value, just look it up directly in Table 9-1 The area is0.903 or 90.3%

EXAMPLE: Find the area under the standard normal distribution curvebetween z ¼ 1.6 and z ¼ 0.8

SOLUTION:

The area is shown in Figure 9-12

Fig 9-11.

Fig 9-12.

To find the area under the standard normal distribution curve between any

given two z values, look up the areas in Table 9-1 and subtract the smaller

area from the larger In this case the area corresponding to z ¼ 1.6 is 0.055,

and the area corresponding to z ¼ 0.8 is 0.788, so the area between z ¼ 1.6

and z ¼ 0.8 is 0.788  0.055 ¼ 0.733 ¼ 73.3% In other words, 73.3% of the

area under the standard normal distribution curve is between z ¼ 1.6 and

z ¼0.8

EXAMPLE: Find the area under the standard normal distribution curve to

the right of z ¼ 0.5

SOLUTION:

The area is shown in Figure 9-13

To find the area under the standard normal distribution curve to the right

of any given z value, look up the area in the table and subtract that from

1 The area corresponding to z ¼ 0.5 is 0.309 Hence 1  0.309 ¼ 0.691

The area to the right of z ¼ 0.5 is 0.691 In other words, 69.1% of

the area under the standard normal distribution curve lies to the right of

z ¼ 0.5

Using Table 9-1 and the formula for transforming values for variables that

are approximately normally distributed, you can find the probabilities of

various events

EXAMPLE: The scores on a national achievement exam are normally

dis-tributed with a mean of 500 and a standard deviation of 100 If a

student is selected at random, find the probability that the student scored

below 680

Fig 9-13.