Elementary statistics technology update 11th edition part 2

6-1

Review and Preview

6-2

The Standard Normal
Distribution

6-3

Applications of Normal
Distributions

6-4

Sampling Distributions
and Estimators

6-5

The Central Limit
Theorem

6-6

Normal as Approximation
to Binomial

6-7

Assessing Normality

Normal Probability
Distributions

248


CHAPTER PROBLEM

How do we design airplanes, boats, cars, and homes
for safety and comfort?
Ergonomics involves the study of people fitting into their environments. Ergonomics is
used in a wide variety of applications such as
these: Design a doorway so that most people
can walk through it without bending or hitting their head; design a car so that the dashboard is within easy reach of most drivers;
design a screw bottle top so that most people have sufficient grip strength to open it;
design a manhole cover so that most workers
can fit through it. Good ergonomic design results in an environment that is safe, functional, efficient, and comfortable. Bad ergonomic design can result in uncomfortable,
unsafe, or possibly fatal conditions. For example, the following real situations illustrate
the difficulty in determining safe loads in aircraft and boats.
• “We have an emergency for Air Midwest fiftyfour eighty,” said pilot Katie Leslie, just before her plane crashed in Charlotte, North
Carolina. The crash of the Beech plane killed
all of the 21 people on board. In the subsequent investigation, the weight of the
passengers was suspected as a factor that
contributed to the crash. This prompted
the Federal Aviation Administration to order airlines to collect weight information
from randomly selected flights, so that the
old assumptions about passenger weights
could be updated.

• Twenty passengers were killed when the
Ethan Allen tour boat capsized on New
York’s Lake George. Based on an assumed
mean weight of 140 lb, the boat was certified to carry 50 people. A subsequent investigation showed that most of the passengers weighed more than 200 lb, and
the boat should have been certified for a
much smaller number of passengers.
• A water taxi sank in Baltimore’s Inner Harbor. Among the 25 people on board, 5
died and 16 were injured. An investigation
revealed that the safe passenger load for
the water taxi was 3500 lb. Assuming a
mean passenger weight of 140 lb, the boat
was allowed to carry 25 passengers, but
the mean of 140 lb was determined 44
years ago when people were not as heavy
as they are today. (The mean weight of the
25 passengers aboard the boat that sank
was found to be 168 lb.) The National
Transportation and Safety Board suggested
that the old estimated mean of 140 lb be
updated to 174 lb, so the safe load of 3500
lb would now allow only 20 passengers instead of 25.
This chapter introduces the statistical
tools that are basic to good ergonomic design. After completing this chapter, we will
be able to solve problems in a wide variety of
different disciplines, including ergonomics.


250

Chapter 6

Normal Probability Distributions

6-1

Review and Preview

In Chapter 2 we considered the distribution of data, and in Chapter 3 we considered
some important measures of data sets, including measures of center and variation. In
Chapter 4 we discussed basic principles of probability, and in Chapter 5 we presented
the concept of a probability distribution. In Chapter 5 we considered only discrete
probability distributions, but in this chapter we present continuous probability distributions. To illustrate the correspondence between area and probability, we begin with
a uniform distribution, but most of this chapter focuses on normal distributions. Normal distributions occur often in real applications, and they play an important role in
methods of inferential statistics. In this chapter we present concepts of normal distributions that will be used often in the remaining chapters of this text. Several of the
statistical methods discussed in later chapters are based on concepts related to the
central limit theorem, discussed in Section 6-5. Many other sections require normally
distributed populations, and Section 6-7 presents methods for analyzing sample data
to determine whether or not the sample appears to be from such a normally distributed population.

If a continuous random variable has a distribution with a graph that is
symmetric and bell-shaped, as in Figure 6-1, and it can be described by the
equation given as Formula 6-1, we say that it has a normal distribution.

Figure 6-1

Curve is bell-shaped
and symmetric

The Normal Distribution

m

Value

Formula 6-1

y =

1 x-m 2
s )

e - 2(

s22p

Formula 6-1 is mathematically challenging, and we include it only to illustrate
that any particular normal distribution is determined by two parameters: the mean, m,
and standard deviation, s. Formula 6-1 is like many an equation with one variable y
on the left side and one variable x on the right side. The letters p and e represent the
constant values of 3.14159 Á and 2.71828 Á , respectively. The symbols m and s
represent fixed values for the mean and standard deviation, respectively. Once specific
values are selected for m and s, we can graph Formula 6-1 as we would graph any
equation relating x and y ; the result is a continuous probability distribution with the
same bell shape shown in Figure 6-1. From Formula 6-1 we see that a normal distribution is determined by the fixed values of the mean m and standard deviation s.
And that’s all we need to know about Formula 6-1!


6-2 The Standard Normal Distribution

The Placebo
Effect

The Standard Normal Distribution

6-2

Key Concept In this section we present the standard normal distribution, which has
these three properties:
1. Its graph is bell-shaped (as in Figure 6-1).
2.

Its mean is equal to 0 (that is, m = 0).

Its standard deviation is equal to 1 (that is, s = 1).
In this section we develop the skill to find areas (or probabilities or relative frequencies) corresponding to various regions under the graph of the standard normal distribution. In addition, we find z-scores that correspond to areas under the graph.
3.

Uniform Distributions
The focus of this chapter is the concept of a normal probability distribution, but we
begin with a uniform distribution. The uniform distribution allows us to see two very
important properties:
1. The area under the graph of a probability distribution is equal to 1.
There is a correspondence between area and probability (or relative frequency),
so some probabilities can be found by identifying the corresponding areas.
Chapter 5 considered only discrete probability distributions, but we now consider
continuous probability distributions, beginning with the uniform distribution.
2.

A continuous random variable has a uniform distribution if its values are

spread evenly over the range of possibilities. The graph of a uniform distribution results in a rectangular shape.

1

Home Power Supply The Newport Power and Light
Company provides electricity with voltage levels that are uniformly distributed between 123.0 volts and 125.0 volts. That is, any voltage amount between 123.0
volts and 125.0 volts is possible, and all of the possible values are equally likely. If
we randomly select one of the voltage levels and represent its value by the random
variable x, then x has a distribution that can be graphed as in Figure 6-2.

P (x)
0. 5
Area ϭ 1
0
123 . 0

Voltage

125 . 0

Figure 6-2 Uniform Distribution of
Voltage Levels

251

x

It has long been believed
that placebos actually help
some patients. In fact, some

formal
studies
have
shown
that
when
given
a placebo
(a treatment with no medicinal value), many test subjects
show some improvement.
Estimates of improvement
rates have typically ranged
between one-third and twothirds of the patients. However, a more recent study
suggests that placebos have
no real effect. An article in
the New England Journal of
Medicine (Vol. 334, No. 21)
was based on research of 114
medical studies over 50 years.
The authors of the article concluded that placebos appear
to have some effect only for
relieving pain, but not for other
physical conditions. They concluded that apart from clinical trials, the use of placebos
“cannot be recommended.”


252

Chapter 6

Normal Probability Distributions

The graph of a continuous probability distribution, such as in Figure 6-2, is called a
density curve. A density curve must satisfy the following two requirements.
Requirements for a Density Curve
1.

The total area under the curve must equal 1.

2.

Every point on the curve must have a vertical height that is 0 or greater. (That
is, the curve cannot fall below the x-axis.)

By setting the height of the rectangle in Figure 6-2 to be 0.5, we force the enclosed area to be 2 * 0.5 = 1, as required. (In general, the area of the rectangle becomes 1 when we make its height equal to the value of 1>range.) The requirement
that the area must equal 1 makes solving probability problems simple, so the following statement is important:
Because the total area under the density curve is equal to 1, there is a correspondence between area and probability.

2

Voltage Level Given the uniform distribution illustrated in
Figure 6-2, find the probability that a randomly selected voltage level is greater
than 124.5 volts.

The shaded area in Figure 6-3 represents voltage levels that are
greater than 124.5 volts. Because the total area under the density curve is equal to
1, there is a correspondence between area and probability. We can find the desired probability by using areas as follows:
Area ϭ 0. 5 x 0. 5
ϭ 0. 25

P (x)
0. 5

0
123 . 0

123 . 5

Figure 6-3

124 . 0 124 . 5
Voltage

125 . 0

x

Using Area to Find Probability

P (voltage greater than 124.5 volts) = area of shaded region in Figure 6-3
= 0.5 * 0.5
= 0.25
The probability of randomly selecting a voltage level greater
than 124.5 volts is 0.25.


6-2 The Standard Normal Distribution

Standard Normal Distribution
The density curve of a uniform distribution is a horizontal line, so we can find the

area of any rectangular region by applying this formula: Area = width * height. Because the density curve of a normal distribution has a complicated bell shape as
shown in Figure 6-1, it is more difficult to find areas. However, the basic principle is
the same: There is a correspondence between area and probability. In Figure 6-4 we
show that for a standard normal distribution, the area under the density curve is
equal to 1.

The standard normal distribution is a normal probability distribution with
m = 0 and s = 1. The total area under its density curve is equal to 1. (See
Figure 6-4.)
It is not easy to find areas in Figure 6-4, so mathematicians have calculated
many different areas under the curve, and those areas are included in Table A-2 in
Appendix A.

Area ϭ 1
Ϫ3

Ϫ2

Ϫ1

0

1

2

3

z Score
Figure 6-4 Standard Normal Distribution:

Bell-Shaped Curve with M ‫ ؍‬0 and S ‫ ؍‬1

Finding Probabilities When Given z Scores
Using Table A-2 (in Appendix A and the Formulas and Tables insert card), we can find
areas (or probabilities) for many different regions. Such areas can also be found using
a TI-83>84 Plus calculator, or computer software such as STATDISK, Minitab, or
Excel. The key features of the different methods are summarized in Table 6-1 on the
next page. Because calculators or computer software generally give more accurate results
than Table A-2, we strongly recommend using technology. (When there are discrepancies, answers in Appendix D will generally include results based on Table A-2 as
well as answers based on technology.)
If using Table A-2, it is essential to understand these points:
1. Table A-2 is designed only for the standard normal distribution, which has a
mean of 0 and a standard deviation of 1.
2.

Table A-2 is on two pages, with one page for negative z scores and the other
page for positive z scores.

253


254

Chapter 6

Normal Probability Distributions

3.

Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z score.

4.

When working with a graph, avoid confusion between z scores and areas.
z score: Distance along the horizontal scale of the standard normal
distribution; refer to the leftmost column and top row of
Table A-2.
Area:

5.

Region under the curve; refer to the values in the body of
Table A-2.

The part of the z score denoting hundredths is found across the top row of
Table A-2.

CAUTION
When working with a normal distribution, avoid confusion between z scores and
areas.

Table 6-1

Methods for Finding Normal Distribution Areas

Table A-2, STATDISK,
Minitab, Excel

The procedure
Ta b l e A - 2

for using Table A-2 is described in
the text.

Gives the cumulative area
from the left up to a vertical
line above a specific value
of z.

z

Select Analysis,
S TAT D I S K
Probability Distributions, Normal
Distribution. Enter the z value,
then click on Evaluate.
Select Calc,
M I N I TA B
Probability Distributions, Normal.
In the dialog box, select Cumulative
Probability, Input Constant.
Select fx, StatistiE XC E L
cal, NORMDIST. In the dialog box,
enter the value and mean, the
standard deviation, and “true.”

T I - 8 3 / 8 4 Press F O
[2: normal cdf ( ], then enter the
two z scores separated by a comma,
as in (left z score, right z score).

TI-83/84 Plus Calculator
Gives area bounded on the
left and bounded on the
right by vertical lines above
any specific values.

Lower

Upper

The following example requires that we find the probability associated with a
z score less than 1.27. Begin with the z score of 1.27 by locating 1.2 in the left column;
next find the value in the adjoining row of probabilities that is directly below 0.07, as
shown in the following excerpt from Table A-2.


6-2

TABLE A-2

The Standard Normal Distribution

255

(continued) Cumulative Area from the LEFT

z

.00

.01

.02

.03

.04

.05

.06

.07

.08

.09

0.0
0.1
0.2

.5000
.5398
.5793

.5040
.5438
.5832

.5080
.5478
.5871

.5120
.5517
.5910

.5160
.5557
.5948

.5199
.5596
.5987

.5239
.5636
.6026

.5279
.5675
.6064

.5319
.5714
.6103

.5359
.5753

.6141

1.0
1.1
1.2
1.3
1.4

.8413
.8643
.8849
.9032
.9192

.8438
.8665
.8869
.9049
.9207

.8461
.8686
.8888
.9066
.9222

.8485
.8708
.8907
.9082

.9236

.8508
.8729
.8925
.9099
.9251

.8531
.8749
.8944
.9115
.9265

.8554
.8770
.8962
.9131
.9279

.8577
.8790
.8980
.9147
.9292

.8599
.8810
.8997
.9162

.9306

.8621
.8830
.9015
.9177
.9319

The area (or probability) value of 0.8980 indicates that there is a probability of
0.8980 of randomly selecting a z score less than 1.27. (The following sections will
consider cases in which the mean is not 0 or the standard deviation is not 1.)

3

Scientific Thermometers The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0°C at the freezing point of water. Tests on a large sample of these instruments reveal that at the freezing point of water, some thermometers give readings
below 0° (denoted by negative numbers) and some give readings above 0° (denoted by positive numbers). Assume that the mean reading is 0°C and the standard deviation of the readings is 1.00°C. Also assume that the readings are normally distributed. If one thermometer is randomly selected, find the probability
that, at the freezing point of water, the reading is less than 1.27°.

The probability distribution of readings is a standard normal distribution, because the readings are normally distributed with m = 0 and s = 1. We
need to find the area in Figure 6-5 below z = 1.27. The area below z = 1.27 is
equal to the probability of randomly selecting a thermometer with a reading less than
1.27°. From Table A-2 we find that this area is 0.8980.
Figure 6-5
Finding the Area Below
z ‫ ؍‬1.27

Area ϭ 0.8980
(from Table A-2)
0

z ϭ 1. 27

The probability of randomly selecting a thermometer with a
reading less than 1.27° (at the freezing point of water) is equal to the area of 0.8980
shown as the shaded region in Figure 6-5. Another way to interpret this result is to
conclude that 89.80% of the thermometers will have readings below 1.27°.


256

Chapter 6

Normal Probability Distributions

4

Scientific Thermometers Using the thermometers from
Example 3, find the probability of randomly selecting one thermometer that reads
(at the freezing point of water) above -1.23°.

We again find the desired probability by finding a corresponding
area. We are looking for the area of the region that is shaded in Figure 6-6, but Table
A-2 is designed to apply only to cumulative areas from the left. Referring to Table A-2
for the page with negative z scores, we find that the cumulative area from the left up
to z = - 1.23 is 0.1093 as shown. Because the total area under the curve is 1, we
can find the shaded area by subtracting 0.1093 from 1. The result is 0.8907. Even
though Table A-2 is designed only for cumulative areas from the left, we can use it to
find cumulative areas from the right, as shown in Figure 6-6.

Area

ϭ 1Ϫ 0. 1093
ϭ 0. 8907

Area found
in Table A-2
0.1093

z ϭ Ϫ1. 23

0

Figure 6-6 Finding the Area Above z ‫ ؍‬؊ 1.23

Because of the correspondence between probability and area,
we conclude that the probability of randomly selecting a thermometer with a reading
above -1.23° at the freezing point of water is 0.8907 (which is the area to the right of
z = - 1.23). In other words, 89.07% of the thermometers have readings above - 1.23°.

Example 4 illustrates a way that Table A-2 can be used indirectly to find a cumulative area from the right. The following example illustrates another way that we can
find an area indirectly by using Table A-2.

5

Scientific Thermometers Make a random selection
from the same sample of thermometers from Example 3. Find the probability that the
chosen thermometer reads (at the freezing point of water) between - 2.00° and 1.50°.

We are again dealing with normally distributed values having a
mean of 0° and a standard deviation of 1°. The probability of selecting a thermometer that reads between -2.00° and 1.50° corresponds to the shaded area in Figure 6-7.
Table A-2 cannot be used to find that area directly, but we can use the table to find

that z = - 2.00 corresponds to the area of 0.0228, and z = 1.50 corresponds to
the area of 0.9332, as shown in the figure. From Figure 6-7 we see that the shaded
area is the difference between 0.9332 and 0.0228. The shaded area is therefore
0.9332 - 0.0228 = 0.9104.


6-2 The Standard Normal Distribution

257

Figure 6-7

(2) Total area from left up to
z ϭ 1. 50 is 0. 9332 (from Table A-2)

Finding the Area Between
Two Values

(1) Area is
0. 0228
(from Table A-2)

(3)
Area
ϭ 0.9332 Ϫ 0. 0228
ϭ 0.9104

z ϭϪ2 . 00

0

z ϭ1. 50

Using the correspondence between probability and area, we
conclude that there is a probability of 0.9104 of randomly selecting one of the thermometers with a reading between -2.00° and 1.50° at the freezing point of water.
Another way to interpret this result is to state that if many thermometers are selected
and tested at the freezing point of water, then 0.9104 (or 91.04%) of them will read
between -2.00° and 1.50°.
Example 5 can be generalized as the following rule: The area corresponding to
the region between two specific z scores can be found by finding the difference
between the two areas found in Table A-2. Figure 6-8 illustrates this general rule.
Note that the shaded region B can be found by calculating the difference between two
areas found from Table A-2: area A and B combined (found in Table A-2 as the area
corresponding to z Right) and area A (found in Table A-2 as the area corresponding to
z Left). Study hint: Don’t try to memorize a rule or formula for this case. Focus on understanding how Table A-2 works. If necessary, first draw a graph, shade the desired
area, then think of a way to find that area given the condition that Table A-2 provides
only cumulative areas from the left.
Figure 6-8 Finding the
Area Between Two z
Scores

B
A

z Left

0

z Right

Shaded area B ϭ (areas A and B combined) — (area A)
ϭ (area from Table A-2 using z Right ) — (area from Table A-2 using z Left )

Probabilities such as those in the preceding examples can also be expressed with
the following notation.
Notation

P (a 6 z 6 b)
denotes the probability that the z score is between a and b.
P(z 7 a)
denotes the probability that the z score is greater than a.
P(z 6 a)
denotes the probability that the z score is less than a.
Using this notation, we can express the result of Example 5 as: P(-2.00 6 z 6
1.50) = 0.9104, which states in symbols that the probability of a z score falling between


258

Chapter 6

Normal Probability Distributions

-2.00 and 1.50 is 0.9104. With a continuous probability distribution such as the
normal distribution, the probability of getting any single exact value is 0. That is,
P(z = a) = 0. For example, there is a 0 probability of randomly selecting someone
and getting a person whose height is exactly 68.12345678 in. In the normal distribution, any single point on the horizontal scale is represented not by a region under
the curve, but by a vertical line above the point. For P(z = 1.50) we have a vertical
line above z = 1.50, but that vertical line by itself contains no area, so
P(z = 1.50) = 0. With any continuous random variable, the probability of any

one exact value is 0, and it follows that P(a … z … b) = P(a 6 z 6 b). It also
follows that the probability of getting a z score of at most b is equal to the probability
of getting a z score less than b. It is important to correctly interpret key phrases such as
at most, at least, more than, no more than, and so on.

Finding z Scores from Known Areas
So far in this section, all of the examples involving the standard normal distribution
have followed the same format: Given z scores, find areas under the curve. These
areas correspond to probabilities. In many cases, we have the reverse: Given the area
(or probability), find the corresponding z score. In such cases, we must avoid confusion between z scores and areas. Remember, z scores are distances along the horizontal
scale, whereas areas (or probabilities) are regions under the curve. (Table A-2 lists
z-scores in the left column and across the top row, but areas are found in the body of
the table.) Also, z scores positioned in the left half of the curve are always negative. If
we already know a probability and want to determine the corresponding z score, we
find it as follows.
Procedure for Finding a z Score from a Known Area
1.

Draw a bell-shaped curve and identify the region under the curve that corresponds to the given probability. If that region is not a cumulative region from the
left, work instead with a known region that is a cumulative region from the left.

Using the cumulative area from the left, locate the closest probability in the
body of Table A-2 and identify the corresponding z score.
When referring to Table A-2, remember that the body of the table gives cumulative
areas from the left.
2.

6

Scientific Thermometers Use the same thermometers

from Example 3, with temperature readings at the freezing point of water that are
normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find
the temperature corresponding to P95, the 95th percentile. That is, find the temperature separating the bottom 95% from the top 5%. See Figure 6-9.

Area ϭ 0.95
0

zϭ?

Figure 6-9 Finding the 95th Percentile


6-2 The Standard Normal Distribution

259

Figure 6-9 shows the z score that is the 95th percentile, with 95%
of the area (or 0.95) below it. Referring to Table A-2, we search for the area of 0.95 in
the body of the table and then find the corresponding z score. In Table A-2 we find
the areas of 0.9495 and 0.9505, but there’s an asterisk with a special note indicating
that 0.9500 corresponds to a z score of 1.645. We can now conclude that the z score
in Figure 6-9 is 1.645, so the 95th percentile is the temperature reading of 1.645°C.

When tested at freezing, 95% of the readings will be less
than or equal to 1.645°C, and 5% of them will be greater than or equal to 1.645°C.
Note that in the preceding solution, Table A-2 led to a z score of 1.645, which is
midway between 1.64 and 1.65. When using Table A-2, we can usually avoid interpolation by simply selecting the closest value. Special cases are listed in the accompanying table because they are often used in a wide variety of applications. (For one of
those special cases, the value of z = 2.576 gives an area slightly closer to the area of
0.9950, but z = 2.575 has the advantage of being the value midway between
z = 2.57 and z = 2.58.) Except in these special cases, we can select the closest value

in the table. (If a desired value is midway between two table values, select the larger
value.) For z scores above 3.49, we can use 0.9999 as an approximation of the cumulative area from the left; for z scores below -3.49, we can use 0.0001 as an approximation of the cumulative area from the left.

Table A-2 Special Cases

z Score

Cumulative
Area from
the Left

1.645

0.9500

- 1.645

0.0500

2.575

0.9950

- 2.575

0.0050

Above 3.49

0.9999

Below - 3.49

0.0001

7

Scientific Thermometers Using the same thermometers
from Example 3, find the temperatures separating the bottom 2.5% and the top 2.5%.

The required z scores are shown in Figure 6-10. To find the z
score located to the left, we search the body of Table A-2 for an area of 0.025. The
result is z = -1.96. To find the z score located to the right, we search the body of
Table A-2 for an area of 0.975. (Remember that Table A-2 always gives cumulative
areas from the left.) The result is z = 1.96. The values of z = -1.96 and z = 1.96
separate the bottom 2.5% and the top 2.5%, as shown in Figure 6-10.
Figure 6-10
Finding z Scores

Area ϭ 0. 025

Area ϭ 0. 025
z ϭ Ϫ1. 96

0

z ϭ 1. 96

To find this z score,
locate the cumulative

area to the left in
Table A–2 . Locate 0. 975
in the body of Table A–2 .

When tested at freezing, 2.5% of the thermometer readings
will be equal to or less than -1.96°, and 2.5% of the readings will be equal to or
greater than 1.96°. Another interpretation is that at the freezing point of water, 95%
of all thermometer readings will fall between -1.96° and 1.96°.


260

Chapter 6

Normal Probability Distributions

Critical Values For a normal distribution, a critical value is a z score on the borderline separating the z scores that are likely to occur from those that are unlikely.
Common critical values are z = -1.96 and z = 1.96, and they are obtained as
shown in Example 7. In Example 7, the values below z = -1.96 are not likely to
occur, because they occur in only 2.5% of the readings, and the values above
z = 1.96 are not likely to occur because they also occur in only 2.5% of the readings. The reference to critical values is not so important in this chapter, but will
become extremely important in the following chapters. The following notation is
used for critical z values found by using the standard normal distribution.
Notation
The expression z a denotes the z score with an area of a to its right. (a is the Greek letter
alpha.)

8

Finding z A In the expression z a, let a = 0.025 and find the

value of z 0.025.

The notation of z 0.025 is used to represent the z score with an area
of 0.025 to its right. Refer to Figure 6-10 and note that the value of z = 1.96 has an
area of 0.025 to its right, so z 0.025 = 1.96.

USING
T E C H N O LO GY

Caution: When using Table A-2 for finding a value of z a for a particular value of
a, note that a is the area to the right of z a, but Table A-2 lists cumulative areas to the
left of a given z score. To find the value of z a by using Table A-2, resolve that conflict
by using the value of 1 - a. In Example 8, the value of z 0.025 can be found by locating the area of 0.9750 in the body of the table.
The examples in this section were created so that the mean of 0 and the standard
deviation of 1 coincided exactly with the properties of the standard normal distribution. In reality, it is unusual to find such convenient parameters, because typical normal distributions involve means different from 0 and standard deviations different
from 1. In the next section we introduce methods for working with such normal distributions, which are much more realistic and practical.

When working with the standard normal distribution, a technology can be used to find z scores or areas, so the technology can be
used instead of Table A-2. The following instructions describe
how to find such z scores or areas.
Select Analysis, Probability DistribuS TAT D I S K
tions, Normal Distribution. Either enter the z score to find
corresponding areas, or enter the cumulative area from the left to
find the z score. After entering a value, click on the Evaluate
button. See the accompanying STATDISK display for an entry
of z = 2.00.

STATDISK

6-2

The Standard Normal Distribution

M I N I TA B
• To find the cumulative area to the left of a z score (as in Table
A-2), select Calc, Probability Distributions, Normal, Cumulative probabilities. Then enter the mean of 0 and standard deviation of 1. Click on the Input Constant button and
enter the z score.

261

z score). Example 5 could be solved with the command of
normalcdf( ؊2.00, 1.50), which yields a probability of 0.9104
(rounded) as shown in the accompanying screen.

TI-83/84 PLUS

• To find a z score corresponding to a known probability, select
Calc, Probability Distributions, Normal. Then select
Inverse cumulative probabilities and the option Input constant. For the input constant, enter the total area to the left of
the given value.
E XC E L
• To find the cumulative area to the left of a z score (as in Table A-2),
click on f x, then select Statistical, NORMSDIST, and enter
the z score. (In Excel 2010, select NORM.S.DIST.)
• To find a z score corresponding to a known probability,
select f x, Statistical, NORMSINV, and enter the total
area to the left of the given value. (In Excel 2010, select
NORM.S.INV.)

To find a z score corresponding to a known probability, press F
O and select invNorm. Proceed to enter the total area to the left
of the z score. For example, the command of invNorm(0.975)
yields a z score of 1.959963986, which is rounded to 1.96, as in
Example 6.

To find the area between two
TI-83/84 PLUS
z scores, press F O and select normalcdf. Proceed to enter
the two z scores separated by a comma, as in (left z score, right

6-2

Basic Skills and Concepts

Statistical Literacy and Critical Thinking
1. Normal Distribution When we refer to a “normal” distribution, does the word “normal”

have the same meaning as in ordinary language, or does it have a special meaning in statistics?
What exactly is a normal distribution?
2. Normal Distribution A normal distribution is informally described as a probability distribution that is “bell-shaped” when graphed. Describe the “bell shape.”
3. Standard Normal Distribution What requirements are necessary for a normal probabil-

ity distribution to be a standard normal probability distribution?
4. Notation What does the notation z a indicate?

Continuous Uniform Distribution. In Exercises 5–8, refer to the continuous uniform distribution depicted in Figure 6-2. Assume that a voltage level between
123.0 volts and 125.0 volts is randomly selected, and find the probability that the
given voltage level is selected.

5. Greater than 124.0 volts
6. Less than 123.5 volts
7. Between 123.2 volts and 124.7 volts
8. Between 124.1 volts and 124.5 volts


262

Chapter 6

Normal Probability Distributions

Standard Normal Distribution. In Exercises 9–12, find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.
9.

10.

z ϭ Ϫ0 . 75

z ϭ 0 . 75
11.

12.

z ϭ 1 . 20

z ϭ Ϫ0 . 60

z ϭ 1 . 60

z ϭ Ϫ0 . 90

Standard Normal Distribution. In Exercises 13–16, find the indicated z score.
The graph depicts the standard normal distribution with mean 0 and standard
deviation 1.
13.

14.

0.2546
0.9798
z

z

15.

16.

0.1075

z

0.9418
z

Standard Normal Distribution. In Exercises 17–36, assume that thermometer
readings are normally distributed with a mean of 0°C and a standard deviation
of 1.00°C. A thermometer is randomly selected and tested. In each case, draw a
sketch, and find the probability of each reading. (The given values are in Celsius

degrees.) If using technology instead of Table A-2, round answers to four decimal
places.
17. Less than -1.50

18. Less than -2.75

19. Less than 1.23

20. Less than 2.34

21. Greater than 2.22

22. Greater than 2.33

23. Greater than -1.75

24. Greater than -1.96

25. Between 0.50 and 1.00

26. Between 1.00 and 3.00

27. Between -3.00 and -1.00

28. Between -1.00 and -0.50

29. Between -1.20 and 1.95

30. Between -2.87 and 1.34

31. Between -2.50 and 5.00

32. Between -4.50 and 1.00

33. Less than 3.55

34. Greater than 3.68

35. Greater than 0

36. Less than 0

Basis for the Range Rule of Thumb and the Empirical Rule. In Exercises 37–40,
find the indicated area under the curve of the standard normal distribution, then


6-2

The Standard Normal Distribution

convert it to a percentage and fill in the blank. The results form the basis for the
range rule of thumb and the empirical rule introduced in Section 3-3.
37. About _____% of the area is between z = - 1 and z = 1 (or within 1 standard deviation

of the mean).
38. About _____% of the area is between z = - 2 and z = 2 (or within 2 standard devia-

tions of the mean).
39. About _____% of the area is between z = - 3 and z = 3 (or within 3 standard devia-

tions of the mean).
40. About _____% of the area is between z = - 3.5 and z = 3.5 (or within 3.5 standard de-

viations of the mean).

Finding Critical Values. In Exercises 41–44, find the indicated value.
41. z 0.05

42. z 0.01

43. z 0.10

44. z 0.02

Finding Probability. In Exercises 45–48, assume that the readings on the thermometers are normally distributed with a mean of 0°C and a standard deviation
of 1.00°. Find the indicated probability, where z is the reading in degrees.
45. P ( -1.96 6 z 6 1.96)

46. P(z 6 1.645)

47. P (z 6 - 2.575 or z 7 2.575)

48. P(z 6 - 1.96 or z 7 1.96)

Finding Temperature Values. In Exercises 49–52, assume that thermometer readings are normally distributed with a mean of 0°C and a standard deviation of
1.00°C. A thermometer is randomly selected and tested. In each case, draw a
sketch, and find the temperature reading corresponding to the given information.
49. Find P95, the 95th percentile. This is the temperature reading separating the bottom 95%

from the top 5%.

50. Find P1, the 1st percentile. This is the temperature reading separating the bottom 1%

from the top 99%.
51. If 2.5% of the thermometers are rejected because they have readings that are too high and

another 2.5% are rejected because they have readings that are too low, find the two readings
that are cutoff values separating the rejected thermometers from the others.
52. If 0.5% of the thermometers are rejected because they have readings that are too low and

another 0.5% are rejected because they have readings that are too high, find the two readings
that are cutoff values separating the rejected thermometers from the others.

6-2

Beyond the Basics

53. For a standard normal distribution, find the percentage of data that are
a. within 2 standard deviations of the mean.
b. more than 1 standard deviation away from the mean.
c. more than 1.96 standard deviations away from the mean.
d. between m- 3s and m+3s.
e. more than 3 standard deviations away from the mean.
54. If a continuous uniform distribution has parameters of m = 0 and s = 1, then the min-

imum is - 23 and the maximum is 23.

a. For this distribution, find P ( -1 6 x 6 1).
b. Find P (-1 6 x 6 1) if you incorrectly assume that the distribution is normal instead of

uniform.

c. Compare the results from parts (a) and (b). Does the distribution affect the results very much?

263


264

Chapter 6

Normal Probability Distributions

55. Assume that z scores are normally distributed with a mean of 0 and a standard devia-

tion of 1.
a. If P(z 6 a) = 0.9599, find a.
b. If P (z 7 b) = 0.9772, find b.
c. If P (z 7 c) = 0.0668, find c.
d. If P (-d 6 z 6 d ) = 0.5878, find d.
e. If P (-e 6 z 6 e) = 0.0956, find e.
56. In a continuous uniform distribution,

m =

range
minimum + maximum
and s =
2
212

Find the mean and standard deviation for the uniform distribution represented in Figure 6-2.

6-3

Applications of Normal Distributions

Key Concept In this section we introduce real and important applications involving nonstandard normal distributions by extending the procedures presented in
Section 6-2. We use a simple conversion (Formula 6-2) that allows us to standardize any normal distribution so that the methods of the preceding section can be
used with normal distributions having a mean that is not 0 or a standard deviation
that is not 1. Specifically, given some nonstandard normal distribution, we should
be able to find probabilities corresponding to values of the variable x, and given
some probability value, we should be able to find the corresponding value of the
variable x.
To work with a nonstandard normal distribution, we simply standardize values to
use the procedures from Section 6-2.
If we convert values to standard z-scores using Formula 6-2, then procedures for working with all normal distributions are the same as those for
the standard normal distribution.

Formula 6-2

z =

x - m
s

(round z scores to 2 decimal places)

Some calculators and computer software programs do not require the above
conversion to z scores because probabilities can be found directly. However, if you
use Table A-2 to find probabilities, you must first convert values to standard z
scores. Regardless of the method you use, you need to clearly understand the above

principle, because it is an important foundation for concepts introduced in the following chapters.
Figure 6-11 illustrates the conversion from a nonstandard to a standard normal
distribution. The area in any normal distribution bounded by some score x (as in
Figure 6-11(a)) is the same as the area bounded by the equivalent z score in the standard normal distribution (as in Figure 6-11(b)). This means that when working with
a nonstandard normal distribution, you can use Table A-2 the same way it was used
in Section 6-2, as long as you first convert the values to z scores.


6-3

zϭ

Applications of Normal Distributions

x Ϫm
s

P
(a)

265

P

m
x
Nonstandard
Normal Distribution

(b)

0
z
Standard
Normal Distribution

Figure 6-11 Converting from a Nonstandard
to a Standard Normal Distribution

When finding areas with a nonstandard normal distribution, use this procedure:
1. Sketch a normal curve, label the mean and the specific x values, then shade the
region representing the desired probability.
2.

For each relevant value x that is a boundary for the shaded region, use Formula 6-2
to convert that value to the equivalent z score.

Refer to Table A-2 or use a calculator or computer software to find the area of
the shaded region. This area is the desired probability.
The following example applies these three steps to illustrate the relationship between a typical nonstandard normal distribution and the standard normal distribution.
3.

1

Why Do Doorways Have a Height of 6 ft 8 in.? The
typical home doorway has a height of 6 ft 8 in., or 80 in. Because men tend to be
taller than women, we will consider only men as we investigate the limitations of
that standard doorway height. Given that heights of men are normally distributed
with a mean of 69.0 in. and a standard deviation of 2.8 in., find the percentage of
men who can fit through the standard doorway without bending or bumping

their head. Is that percentage high enough to continue using 80 in. as the standard
height? Will a doorway height of 80 in. be sufficient in future years?

Step 1: See Figure 6-12, which incorporates this information: Men have heights
that are normally distributed with a mean of 69.0 in. and a standard deviation of
2.8 in. The shaded region represents the men who can fit through a doorway that
has a height of 80 in.
Figure 6-12
Heights (in inches) of Men

Area ϭ 0. 9999
m ϭ 69. 0 in.
zϭ0

x (height)

x ϭ 80 in.

z scale

z ϭ 3. 93

Step 2: To use Table A-2, we first must use Formula 6-2 to convert from the nonstandard normal distribution to the standard normal distribution. The height of
80 in. is converted to a z score as follows:
x - m
80 - 69.0
= 3.93
=
z =
s

2.8

continued


266

Chapter 6

Multiple Lottery
Winners
Evelyn Marie Adams won
the New Jersey Lottery
twice in four months. This
happy event was reported
in the media as an
incredible coincidence with a
likelihood of
only 1
chance in
17 trillion.
But Harvard mathematicians Persi Diaconis
and Frederick Mosteller
note that there is 1 chance
in 17 trillion that a particular
person with one ticket in
each of two New Jersey
lotteries will win both
times. However, there is
about 1 chance in 30 that

someone in the United
States will win a lottery
twice in a four-month period. Diaconis and Mosteller
analyzed coincidences and
conclude that “with a large
enough sample, any outrageous thing is apt to happen.” More recently, according to the Detroit
News, Joe and Dolly Hornick won the Pennsylvania
lottery four times in 12
years for prizes of $2.5 million, $68,000, $206,217,
and $71,037.

Normal Probability Distributions

Step 3: Referring to Table A-2 and using z = 3.93, we find that this z score is in
the category of “3.50 and up,” so the cumulative area to the left of 80 in. is
0.9999 as shown in Figure 6-12.
If we use technology instead of Table A-2, we get the more accurate cumulative area
of 0.999957 (instead of 0.9999).

The proportion of men who can fit through the standard
doorway height of 80 in. is 0.9999, or 99.99%. Very few men will not be able to fit
through the doorway without bending or bumping their head. This percentage is
high enough to justify the use of 80 in. as the standard doorway height. However,
heights of men and women have been increasing gradually but steadily over the past
decades, so the time may come when the standard doorway height of 80 in. may no
longer be adequate.

2

Birth Weights Birth weights in the United States are normally distributed with a mean of 3420 g and a standard deviation of 495 g. The

Newport General Hospital requires special treatment for babies that are less than
2450 g (unusually light) or more than 4390 g (unusually heavy). What is the percentage of babies who do not require special treatment because they have birth
weights between 2450 g and 4390 g? Under these conditions, do many babies require special treatment?

Figure 6-13 shows the shaded region representing birth weights between 2450 g and 4390 g. We can’t find that shaded area directly from Table A-2, but
we can find it indirectly by using the same basic procedures presented in Section 6-2,
as follows: (1) Find the cumulative area from the left up to 2450; (2) find the cumulative area from the left up to 4390; (3) find the difference between those two areas.
Total cumulative area from the
left is 0.9750.

0.0250

2450

␮ϭ3420

4390

z ϭ Ϫ1.96

zϭ0

z ϭ 1.96

x (birth weight)
z scale

Figure 6-13 Birth Weights

Find the cumulative area up to 2450:

x - m
2450 - 3420
=
= -1.96
s
495
Using Table A-2, we find that z = -1.96 corresponds to an area of 0.0250, as shown
in Figure 6-13.
z =


6-3 Applications of Normal Distributions

Find the cumulative area up to 4390:
x - m
4390 - 3420
=
= 1.96
s
495
Using Table A-2, we find that z = 1.96 corresponds to an area of 0.9750, as shown in
Figure 6-13.
Find the shaded area between 2450 and 4390:
z =

Shaded area = 0.9750 - 0.0250 = 0.9500

Expressing the result as a percentage, we conclude that
95.00% of the babies do not require special treatment because they have birth
weights between 2450 g and 4390 g. It follows that 5.00% of the babies do require

special treatment because they are unusually light or heavy. The 5.00% rate is probably not too high for typical hospitals.

Finding Values from Known Areas
Here are helpful hints for those cases in which the area (or probability or percentage)
is known and we must find the relevant value(s):
1. Don’t confuse z scores and areas. Remember, z scores are distances along the horizontal scale, but areas are regions under the normal curve. Table A-2 lists z scores in the
left columns and across the top row, but areas are found in the body of the table.
2.

Choose the correct (right>left) side of the graph. A value separating the top 10%
from the others will be located on the right side of the graph, but a value separating the bottom 10% will be located on the left side of the graph.

3.

A z score must be negative whenever it is located in the left half of the normal
distribution.

Areas (or probabilities) are positive or zero values, but they are never negative.
Graphs are extremely helpful in visualizing, understanding, and successfully working with normal probability distributions, so they should be used whenever possible.
4.

Procedure for Finding Values Using Table A-2 and Formula 6-2
1.

Sketch a normal distribution curve, enter the given probability or percentage in
the appropriate region of the graph, and identify the x value(s) being sought.

2.

Use Table A-2 to find the z score corresponding to the cumulative left area

bounded by x. Refer to the body of Table A-2 to find the closest area, then
identify the corresponding z score.

3.

Using Formula 6-2, enter the values for m, s, and the z score found in Step 2,
then solve for x. Based on Formula 6-2, we can solve for x as follows:
x = m + (z # s)
c

(another form of Formula 6-2)

(If z is located to the left of the mean, be sure that it is a negative number.)
4.

Refer to the sketch of the curve to verify that the solution makes sense in the
context of the graph and in the context of the problem.

267


268

Chapter 6

Normal Probability Distributions

The following example uses the procedure just outlined.
3

Designing Doorway Heights When designing an environment, one common criterion is to use a design that accommodates 95% of
the population. How high should doorways be if 95% of men will fit through
without bending or bumping their head? That is, find the 95th percentile of
heights of men. Heights of men are normally distributed with a mean of 69.0 in.
and a standard deviation of 2.8 in.

Step 1: Figure 6-14 shows the normal distribution with the height x that we want
to identify. The shaded area represents the 95% of men who can fit through the
doorway that we are designing.

Area ϭ 0. 9500

␮ϭ69. 0

xϭ?

zϭ0

z ϭ 1. 645

x (height)
z scale

Figure 6-14 Finding Height

Step 2: In Table A-2 we search for an area of 0.9500 in the body of the table. (The
area of 0.9500 shown in Figure 6-14 is a cumulative area from the left, and that is
exactly the type of area listed in Table A-2.) The area of 0.9500 is between the
Table A-2 areas of 0.9495 and 0.9505, but there is an asterisk and footnote indicating that an area of 0.9500 corresponds to z = 1.645.
Step 3: With z = 1.645, m = 69.0, and s = 2.8, we can solve for x by using

Formula 6-2:
x - m
x - 69.0
z =
becomes 1 .645 =
s
2.8
The result of x = 73.606 in. can be found directly or by using the following version of Formula 6-2:
x = m + (z

#

s) = 69.0 + (1.645

#

2.8) = 73.606

Step 4: The solution of x = 73.6 in. (rounded) in Figure 6-14 is reasonable because it is greater than the mean of 69.0 in.

A doorway height of 73.6 in. (or 6 ft 1.6 in.) would allow
95% of men to fit without bending or bumping their head. It follows that 5% of
men would not fit through a doorway with a height of 73.6 in. Because so many
men walk through doorways so often, this 5% rate is probably not practical.


6-3 Applications of Normal Distributions

269

4

Birth Weights The Newport General Hospital wants to
redefine the minimum and maximum birth weights that require special treatment because they are unusually low or unusually high. After considering relevant factors, a committee recommends special treatment for birth weights in the
lowest 3% and the highest 1%. The committee members soon realize that specific birth weights need to be identified. Help this committee by finding the
birth weights that separate the lowest 3% and the highest 1%. Birth weights in
the United States are normally distributed with a mean of 3420 g and a standard
deviation of 495 g.

Step 1: We begin with the graph shown in Figure 6-15. We have entered the mean
of 3420 g, and we have identified the x values separating the lowest 3% and the
highest 1%.
Figure 6-15

0. 03

Finding the Values Separating the Lowest 3% and
the Highest 1%

0. 01

xϭ?
z ϭ Ϫ1. 88

␮ϭ3420

xϭ?
z ϭ 2 . 33

x (birth weight)

z scale

Step 2: If using Table A-2, we must use cumulative areas from the left. For the
leftmost value of x, the cumulative area from the left is 0.03, so search for an area
of 0.03 in the body of the table to get z = -1.88 (which corresponds to the closest area of 0.0301). For the rightmost value of x, the cumulative area from the left
is 0.99, so search for an area of 0.99 in the body of the table to get z = 2.33 (which
corresponds to the closest area of 0.9901).
Step 3: We now solve for the two values of x by using Formula 6-2 directly or by
using the following version of Formula 6-2:
Leftmost value of x :
Rightmost value of x :

x = m + (z # s) = 3420 + (-1.88 # 495) = 2489.4
x = m + (z # s) = 3420 + (2.33

# 495) = 4573.35

Step 4: Referring to Figure 6-15, we see that the leftmost value of x = 2489.4 g
is reasonable because it is less than the mean of 3420 g. Also, the rightmost value
of 4573.35 is reasonable because it is above the mean of 3420 g. (Technology yields
the values of 2489.0 g and 4571.5 g.)

The birth weight of 2489 g (rounded) separates the lowest
3% of birth weights, and 4573 g (rounded) separates the highest 1% of birth weights.
The hospital now has well-defined criteria for determining whether a newborn baby
should be given special treatment for a birth weight that is unusually low or high.


270

Chapter 6

Normal Probability Distributions

When using the methods of this section with applications involving a normal
distribution, it is important to first determine whether you are finding a probability
(or area) from a known value of x or finding a value of x from a known probability (or
area). Figure 6-16 is a flowchart summarizing the main procedures of this section.

Applications with Normal Distributions
Start

Find a probability
(from a known value of x)

Table A-2

Convert to the
standard normal
distribution by
finding z:
z‫ ؍‬x ؊ μ
␴

Are
you using
technology or
Table A-2
?

What
do you want
to find
?

Technology

Find a value of x
(from known probability or area)

Identify the
cumulative
area to the
left of x.
x

Find the
probability
by using the
technology.

Table A-2

Look up z in Table
A-2 and find the
cumulative area
to the left of z.

Are
you using

technology or
Table A-2
?

Look up the cumulative
left area in Table A-2 and
find the corresponding
z score.

Solve for x:
x ‫ ؍‬μ ؉ z·␴

Figure 6-16 Procedures for Applications with Normal Distributions

Technology

Find x directly
from the
technology.


U S I N G T E C H N O LO GY

6-3

Applications of Normal Distributions

When working with a nonstandard normal distribution, a technology can be used to find areas or values of the relevant variable, so the
technology can be used instead of Table A-2. The following instructions describe how to use technology for such cases.
Select Analysis, Probability Distributions,

S TAT D I S K
Normal Distribution. Either enter the z score to find corresponding
areas, or enter the cumulative area from the left to find the z score.
After entering a value, click on the Evaluate button.
M I N I TA B
• To find the cumulative area to the left of a z score (as in Table A-2),
select Calc, Probability Distributions, Normal, Cumulative
probabilities. Enter the mean and standard deviation, then click
on the Input Constant button and enter the value.

271

TI-83/84 PLUS
• To find the area between two values, press 2nd, VARS, 2 (for
normalcdf ), then proceed to enter the two values, the mean, and
the standard deviation, all separated by commas, as in (left value,
right value, mean, standard deviation). Hint: If there is no left
value, enter the left value as - 999999, and if there is no right
value, enter the right value as 999999. In Example 1 we want
the area to the left of x = 80 in., so use the command
normalcdf ( ؊999999, 80, 69.0, 2.8) as shown in the accompanying screen display.

TI-83/84 PLUS

• To find a value corresponding to a known area, select Calc,
Probability Distributions, Normal, then select Inverse cumulative probabilities. Enter the mean and standard deviation. Select the option Input constant and enter the total area to the left
of the given value.
E XC E L
• To find the cumulative area to the left of a value (as in Table A-2),
click on f x, then select Statistical, NORMDIST. (In Excel 2010,

select NORM.DIST.) In the dialog box, enter the value for x,
enter the mean and standard deviation, and enter 1 in the “cumulative” space.

• To find a value corresponding to a known area, press 2nd, VARS,
the select invNorm, and proceed to enter the total area to the left
of the value, the mean, and the standard deviation in the format
of (total area to the left, mean, standard deviation) with the commas included.

• To find a value corresponding to a known area, select f x, Statistical,
NORMINV, (or NORM.INV in Excel 2010), and proceed to
make the entries in the dialog box. When entering the probability value, enter the total area to the left of the given value. See the
accompanying Excel display for Example 3.

EXCEL

6-3

Basic Skills and Concepts

Statistical Literacy and Critical Thinking
1. Normal Distributions What is the difference between a standard normal distribution and

a nonstandard normal distribution?
2. IQ Scores The distribution of IQ scores is a nonstandard normal distribution with a

mean of 100 and a standard deviation of 15, and a bell-shaped graph is drawn to represent
this distribution.
a. What is the area under the curve?
b. What is the value of the median?
c. What is the value of the mode?

272

Chapter 6

Normal Probability Distributions

3. Normal Distributions The distribution of IQ scores is a nonstandard normal distribu-

tion with a mean of 100 and a standard deviation of 15. What are the values of the mean and
standard deviation after all IQ scores have been standardized by converting them to z scores
using z = (x - m)>s?
4. Random Digits Computers are often used to randomly generate digits of telephone

numbers to be called when conducting a survey. Can the methods of this section be used to
find the probability that when one digit is randomly generated, it is less than 5? Why or why
not? What is the probability of getting a digit less than 5?

IQ Scores. In Exercises 5–8, find the area of the shaded region. The graphs depict
IQ scores of adults, and those scores are normally distributed with a mean of 100
and a standard deviation of 15 (as on the Wechsler test).
5.

6.

80

120
7.

8.

90

115

75

110

IQ Scores. In Exercises 9–12, find the indicated IQ score. The graphs depict IQ
scores of adults, and those scores are normally distributed with a mean of 100
and a standard deviation of 15 (as on the Wechsler test).
9.

10.

0.6

0.8
x

11.

x
12.

0.95

0.99

x

x

IQ Scores. In Exercises 13–20, assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the
Wechsler test). (Hint: Draw a graph in each case.)
13. Find the probability that a randomly selected adult has an IQ that is less than 115.
14. Find the probability that a randomly selected adult has an IQ greater than 131.5 (the

requirement for membership in the Mensa organization).
15. Find the probability that a randomly selected adult has an IQ between 90 and 110
(referred to as the normal range).
16. Find the probability that a randomly selected adult has an IQ between 110 and 120

(referred to as bright normal ).