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4.5 The Normal Distribution

One of the most commonly observed continuous random variables has a bell-shaped probability distribution (or bell curve), as shown in Figure 4.15. It is known as a normal random variable and its probability distribution is called a normal distribution.

Figure 4.15

A normal probability distribution

The normal distribution plays a very important role in the science of statistical inference. Moreover, many phenomena generate random variables with probability distributions that are very well approximated by a normal distribution. For example, the error made in measuring a person’s blood pressure may be a normal random variable, and the probability distribution for the yearly rainfall in a certain region might be approximated by a normal probability distribution. You can determine the adequacy of the normal approximation to an existing population of data by comparing the relative frequency distribution of a large sample of the data with the normal probability distribution. Methods for detecting disagreement between a set of data and the assumption of normality are presented in Section 4.6.

The normal distribution is perfectly symmetric about its mean $μ,$ $μ,$ as can be seen in the examples in Figure 4.16. Its spread is determined by the value of its standard deviation $σ .$ $σ .$ The formula for the normal probability distribution is shown in the next box. When plotted, this formula yields a curve like that shown in Figure 4.15.

Biography Carl F. Gauss (1777–1855)

The Gaussian Distribution

The normal distribution began in the 18th century as a theoretical distribution for errors in disciplines in which fluctuations in nature were believed to behave randomly. Although he may not have been the first to discover the formula, the normal distribution was named the Gaussian distribution after Carl Friedrich Gauss. A well-known and respected German mathematician, physicist, and astronomer, Gauss applied the normal distribution while studying the motion of planets and stars. Gauss’s prowess as a mathematician was exemplified by one of his most important discoveries: At the young age of 22, Gauss constructed a regular 17-gon by ruler and compasses—a feat that was the most major advance in mathematics since the time of the ancient Greeks. In addition to publishing close to 200 scientific papers, Gauss invented the heliograph as well as a primitive telegraph.

Probability Distribution for a Normal Random Variable `x`

Probability density function: $f (x) = \frac{1}{σ \sqrt{2 π}} e^{- (1 / 2) {[(x - μ) / σ]}^{2}}$ $f (x) = \frac{1}{σ \sqrt{2 π}} e^{- (1 / 2) {[(x - μ) / σ]}^{2}}$

where

[&*AS*|mu|*AP*|=|~rom~Mean of the normal random variable~normal~ x&][&*AS*|sig|*AP*|=|~rom~Standard deviation~normal~&][&*AS*|pi|*AP*|=|3.1416|elip|&][&*AS*e*AP*|=|2.71828|elip|&]

\begin{array}{l} μ & = & Mean of the normal random variable x \\ σ & = & Standard deviation \\ π & = & 3.1416 \dots \\ e & = & 2.71828 \dots \end{array}

$\begin{array}{l} μ & = & Mean of the normal random variable x \\ σ & = & Standard deviation \\ π & = & 3.1416 \dots \\ e & = & 2.71828 \dots \end{array}$

$P (x < a)$ $P (x < a)$ is obtained from a table of normal probabilities.

Note that the mean $μ$ $μ$ and standard deviation $σ$ $σ$ appear in this formula, so that no separate formulas for $μ$ $μ$ and $σ$ $σ$ are necessary. To graph the normal curve, we have to know the numerical values of $μ$ $μ$ and $σ$ $σ$ . Computing the area over intervals under the normal probability distribution is a difficult task.* Consequently, we will use either technology or the computed areas listed in Table II of Appendix B. Although there are an infinitely large number of normal curves—one for each pair of values of $μ$ $μ$ and $σ$ $σ$ —we have formed a single table that will apply to any normal curve.

Figure 4.16

Several normal distributions with different means and standard deviations

Table II is based on a normal distribution with mean $μ = 0$ $μ = 0$ and standard deviation $σ = 1,$ $σ = 1,$ called a standard normal distribution. A random variable with a standard normal distribution is typically denoted by the symbol z. The formula for the probability distribution of z is

[&f|pbo|z|pbc||=|*frac*{1}{*smrad*{2|pi|}}|thn|e^{|minus||pbo|1|sol|2|pbc|z^{2}} &]

f (z) = \frac{1}{\sqrt{2 π}} e^{- (1 / 2) z^{2}}

$f (z) = \frac{1}{\sqrt{2 π}} e^{- (1 / 2) z^{2}}$

Figure 4.17 shows the graph of a standard normal distribution.

Figure 4.17

Standard normal distribution: $μ = 0, σ = 1$ $μ = 0, σ = 1$

The standard normal distribution is a normal distribution with $μ = 0$ $μ = 0$ and $σ = 1.$ $σ = 1.$ A random variable with a standard normal distribution, denoted by the symbol z, is called a standard normal random variable.

Since we will ultimately convert all normal random variables to standard normal variables in order to use Table II to find probabilities, it is important that you learn to use Table II well. A partial reproduction of that table is shown in Table 4.5. Note that the values of the standard normal random variable z are listed in the left-hand column. The entries in the body of the table give the area (probability) between 0 and z. Examples 4.15–4.18 illustrate the use of the table.

Alternate View

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359

.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224

.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549

.7 .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852

.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133

.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389

**1.0** .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621

**1.1** .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .3830

**1.2** .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .4015

**1.3** .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177

**1.4** .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319

**1.5** .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441

Example 4.15 Using the Standard Normal Table to Find $P (- z_{0} < z < z_{0})$ $P (- z_{0} < z < z_{0})$

Problem

Find the probability that the standard normal random variable z falls between $- 1.33$ $- 1.33$ and $+ 1.33.$ $+ 1.33.$

Solution

The standard normal distribution is shown again in Figure 4.18. Since all probabilities associated with standard normal random variables can be depicted as areas under the standard normal curve, you should always draw the curve and then equate the desired probability to an area.

Figure 4.18

Areas under the standard normal curve for Example 4.15

In this example, we want to find the probability that z falls between $- 1.33$ $- 1.33$ and $+ 1.33,$ $+ 1.33,$ which is equivalent to the area between $- 1.33$ $- 1.33$ and $+ 1.33,$ $+ 1.33,$ shown highlighted in Figure 4.18. Table II gives the area between $z = 0$ $z = 0$ and any positive value of z, so that if we look up $z = 1.33$ $z = 1.33$ (the value in the 1.3 row and .03 column, as shown in Figure 4.19), we find that the area between $z = 0$ $z = 0$ and $z = 1.33$ $z = 1.33$ is .4082. This is the area labeled $A_{1}$ $A_{1}$ in Figure 4.18. To find the area $A_{2}$ $A_{2}$ located between $z = 0$ $z = 0$ and $z = - 1.33,$ $z = - 1.33,$ we note that the symmetry of the normal distribution implies that the area between $z = 0$ $z = 0$ and any point to the left is equal to the area between $z = 0$ $z = 0$ and the point equidistant to the right. Thus, in this example the area between $z = 0$ $z = 0$ and $z = - 1.33$ $z = - 1.33$ is equal to the area between $z = 0$ $z = 0$ and $z = + 1.33.$ $z = + 1.33.$ That is,

Figure 4.19

Finding $z = 1.33$ $z = 1.33$ in the standard normal table, Example 4.15

[&A_{1}|=|A_{2}|=|.4082 &]
$A_{1} = A_{2} = .4082$ $A_{1} = A_{2} = .4082$

The probability that z falls between $- 1.33$ $- 1.33$ and $+ 1.33$ $+ 1.33$ is the sum of the areas of $A_{1}$ $A_{1}$ and $A_{2}$ $A_{2}$ . We summarize in probabilistic notation:

[&*AS*P|pbo||minus|1.33|less|z|less||minus|1.33|pbc|*AP*|=|P|pbo||minus|1.33|less|z|less|0|pbc||+|P|pbo|0|less|z|leq|1.33|pbc| &][&*AS**AP*|=|A_{1}|+|A_{2}|=|.4082|+|.4082|=|.8164 &]
$\begin{array}{l} P (- 1.33) < z < - 1.33) & = & P (- 1.33 < z < 0) + P (0 < z \leq 1.33) \\ = & A_{1} + A_{2} = .4082 + .4082 = .8164 \end{array}$ $\begin{array}{l} P (- 1.33) < z < - 1.33) & = & P (- 1.33 < z < 0) + P (0 < z \leq 1.33) \\ = & A_{1} + A_{2} = .4082 + .4082 = .8164 \end{array}$

Look Back

Remember that “ $<$ $<$ ” and “ $\leq$ $\leq$ ” are equivalent in events involving z because the inclusion (or exclusion) of a single point does not alter the probability of an event involving a continuous random variable.

Now Work Exercise 4.77e–f

Example 4.16 Using the Standard Normal Table to Find $P (z > z_{0})$ $P (z > z_{0})$

Problem

Find the probability that a standard normal random variable exceeds 1.64; that is, find $P (z > 1.64) .$ $P (z > 1.64) .$

Solution

The area under the standard normal distribution to the right of 1.64 is the highlighted area labeled $A_{1}$ $A_{1}$ in Figure 4.20. This area represents the probability that z exceeds 1.64. However, when we look up $z = 1.64$ $z = 1.64$ in Table II , we must remember that the probability given in the table corresponds to the area between $z = 0$ $z = 0$ and $z = 1.64$ $z = 1.64$ (the area labeled $A_{2}$ $A_{2}$ in Figure 4.20). From Table II , we find that $A_{2} = .4495 .$ $A_{2} = .4495 .$ To find the area $A_{1}$ $A_{1}$ to the right of 1.64, we make use of two facts:
1. The standard normal distribution is symmetric about its mean, $z = 0.$ $z = 0.$
2. The total area under the standard normal probability distribution equals 1.
Taken together, these two facts imply that the areas on either side of the mean, $z = 0,$ $z = 0,$ equal .5; thus, the area to the right of $z = 0$ $z = 0$ in Figure 4.20 is $A_{1} + A_{2} = .5 .$ $A_{1} + A_{2} = .5 .$ Then

[&P|pbo|z|gtr|1.64|pbc||=|A_{1}|=|.5|-|A_{2}|=|.5|-|.4495|=|.0505 &]
$P (z > 1.64) = A_{1} = .5 - A_{2} = .5 - .4495 = .0505$ $P (z > 1.64) = A_{1} = .5 - A_{2} = .5 - .4495 = .0505$

Look Back

To attach some practical significance to this probability, note that the implication is that the chance of a standard normal random variable exceeding 1.64 is only about .05.

Figure 4.20

Areas under the standard normal curve for Example 4.16

Now Work Exercise 4.78a

Example 4.17 Using the Standard Normal Table to Find $P (z < z_{0})$ $P (z < z_{0})$

Problem

Find the probability that a standard normal random variable lies to the left of .67.

Solution

The event sought is shown as the highlighted area in Figure 4.21. We want to find $P (z < .67) .$ $P (z < .67) .$ We divide the highlighted area into two parts: the area $A_{1}$ $A_{1}$ between $z = 0$ $z = 0$ and $z = .67,$ $z = .67,$ and the area $A_{2}$ $A_{2}$ to the left of $z = 0.$ $z = 0.$ We must always make such a division when the desired area lies on both sides of the mean $(z = 0)$ $(z = 0)$ because Table II contains areas between $z = 0$ $z = 0$ and the point you look up. We look up $z = .67$ $z = .67$ in Table II to find that $A_{1} = .2486 .$ $A_{1} = .2486 .$ The symmetry of the standard normal distribution also implies that half the distribution lies on each side of the mean, so the area $A_{2}$ $A_{2}$ to the left of $z = 0$ $z = 0$ is .5. Then

Figure 4.21

Areas under the standard normal curve for Example 4.17

[&P|pbo|z|less|.67|pbc||=|A_{1}|+|A_{2}|=|.2486|+|.5|=|.7486 &]
$P (z < .67) = A_{1} + A_{2} = .2486 + .5 = .7486$ $P (z < .67) = A_{1} + A_{2} = .2486 + .5 = .7486$

Look Back

Note that this probability is approximately .75. Thus, about 75% of the time, the standard normal random variable z will fall below .67. This statement implies that $z = .67$ $z = .67$ represents the approximate 75th percentile (or upper quartile) of the standard normal distribution.

Now Work Exercise 4.78h

Example 4.18 Using the Standard Normal Table to Find $P (| z | > z_{0})$ $P (| z | > z_{0})$

Problem

Find the probability that a standard normal random variable exceeds 1.96 in absolute value.

Solution

The event sought is shown highlighted in Figure 4.22. We want to find

Figure 4.22

Areas under the standard normal curve for Example 4.18

[&P|pbo||abso|z|absc||gtr|1.96|pbc||=|P|pbo|z|less||minus|1.96 ~rom~or~normal~ z|gtr|1.96|pbc| &]
$P (z > 1.96) = P (z < - 1.96 o r z > 1.96)$ $P (z > 1.96) = P (z < - 1.96 o r z > 1.96)$

Note that the total highlighted area is the sum of the two areas $A_{1}$ $A_{1}$ and $A_{2}$ $A_{2}$ —areas that are equal because of the symmetry of the normal distribution.

We look up $z = 1.96$ $z = 1.96$ and find the area between $z = 0$ $z = 0$ and $z = 1.96$ $z = 1.96$ to be .4750. Then $A_{2}$ $A_{2}$ , the area to the right of 1.96, is $.5 - .4750 = .0250$ $.5 - .4750 = .0250$ , so that

[&P|pbo||abso|z|absc||gtr|1.96|pbc||=|A_{1}|+|A_{2}|=|.0250|+|.0250|=|.05 &]
$P (| z | > 1.96) = A_{1} + A_{2} = .0250 + .0250 = .05$ $P (| z | > 1.96) = A_{1} + A_{2} = .0250 + .0250 = .05$

Look Back

We emphasize, again, the importance of sketching the standard normal curve in finding normal probabilities.

To apply Table II to a normal random variable x with any mean $μ$ $μ$ and any standard deviation $σ,$ $σ,$ we must first convert the value of x to a z-score. The population z-score for a measurement was defined in Section 2.6 asis the distance between the measurement and the population mean, divided by the population standard deviation. Thus, the z-score gives the distance between a measurement and the mean in units equal to the standard deviation. In symbolic form, the z-score for the measurement x is

[&z|=|*frac*{x|-||mu|}{|sig|} &]

z = \frac{x - μ}{σ}

$z = \frac{x - μ}{σ}$

Note that when $x = μ,$ $x = μ,$ we obtain $z = 0.$ $z = 0.$

An important property of the normal distribution is that if x is normally distributed with any mean and any standard deviation, z is always normally distributed with mean 0 and standard deviation 1. That is, z is a standard normal random variable.

Converting a Normal Distribution to a Standard Normal Distribution

If x is a normal random variable with mean $μ$ $μ$ and standard deviation $σ$ $σ$ , then the random variable z defined by the formula

[&z|=|*frac*{x|-||mu|}{|sig|} &]

z = \frac{x - μ}{σ}

$z = \frac{x - μ}{σ}$

has a standard normal distribution. The value z describes the number of standard deviations between x and $μ .$ $μ .$

Recall from Example 4.18 that $P (| z | > 1.96) = .05$ $P (| z | > 1.96) = .05$ . This probability, coupled with our interpretation of z, implies that any normal random variable lies more than 1.96 standard deviations from its mean only 5% of the time. Compare this statement to the empirical rule (Chapter 2), which tells us that about 5% of the measurements in mound-shaped distributions will lie beyond two standard deviations from the mean. The normal distribution actually provides the model on which the empirical rule is based, along with much “empirical” experience with real data that often approximately obey the rule, whether drawn from a normal distribution or not.

Example 4.19 Finding a Normal Probability—Cell Phone Application

Problem

Assume that the length of time, x, between charges of a cellular phone is normally distributed with a mean of 10 hours and a standard deviation of 1.5 hours. Find the probability that the cell phone will last between 8 and 12 hours between charges.

Solution

The normal distribution with mean $μ = 10$ $μ = 10$ and $σ = 1.5$ $σ = 1.5$ is shown in Figure 4.23. The desired probability that the cell phone lasts between 8 and 12 hours is highlighted. In order to find that probability, we must first convert the distribution to a standard normal distribution, which we do by calculating the z-score:

Figure 4.23

Areas under the normal curve for Example 4.19

[&z|=|*frac*{x|-||mu|}{|sig|} &]
$z = \frac{x - μ}{σ}$ $z = \frac{x - μ}{σ}$

The z-scores corresponding to the important values of x are shown beneath the x values on the horizontal axis in Figure 4.23. Note that $z = 0$ $z = 0$ corresponds to the mean of $μ = 10$ $μ = 10$ hours, whereas the x values 8 and 12 yield z-scores of $- 1.33$ $- 1.33$ and $+ 1.33,$ $+ 1.33,$ respectively. Thus, the event that the cell phone lasts between 8 and 12 hours is equivalent to the event that a standard normal random variable lies between $- 1.33$ $- 1.33$ and $+ 1.33.$ $+ 1.33.$ We found this probability in Example 4.15 (see Figure 4.18) by doubling the area corresponding to $z = 1.33$ $z = 1.33$ in Table II . That is,

[&P|pbo|8|leq|x|leq|12|pbc||=|P|pbo||minus|1.33|leq|z|leq|1.33|pbc||=|2|pbo|.4082|pbc||=|.8164 &]
$P (8 \leq x \leq 12) = P (- 1.33 \leq z \leq 1.33) = 2 (.4082) = .8164$ $P (8 \leq x \leq 12) = P (- 1.33 \leq z \leq 1.33) = 2 (.4082) = .8164$

Now Work Exercise 4.82a

Table II in Appendix B provides good approximations to probabilities under the normal curve. However, if you do not have access to a normal table, you can always rely on statistical software or a graphing calculator to compute the desired probability. With most statistical software, you will need to specify the mean and standard deviation of the normal distribution, as well as the key values of the variable for which you desire probabilities. In Example 4.19, we desire $P (8 \leq x \leq 12)$ $P (8 \leq x \leq 12)$ , where $μ = 10$ $μ = 10$ and $σ = 1.5$ $σ = 1.5$ . To find this probability using MINITAB’s normal probability function, we enter 10 for the mean and 1.5 for the standard deviation, and then find two cumulative probabilities: $P (x \leq 12)$ $P (x \leq 12)$ and $P (x < 8)$ $P (x < 8)$ . These two probabilities are shown (shaded) on the MINITAB printout in Figure 4.24. The difference between the two probabilities yields the desired result:

Figure 4.24

MINITAB output with cumulative normal probabilities

[&P|pbo|8|leq|x|leq|12|pbc||=|P|pbo|x|leq|12|pbc||-|P|pbo|x|less|8|pbc||=|.908789|-|.0912112|=|.8175778 &]

P (8 \leq x \leq 12) = P (x \leq 12) - P (x < 8) = .908789 - .0912112 = .8175778

$P (8 \leq x \leq 12) = P (x \leq 12) - P (x < 8) = .908789 - .0912112 = .8175778$

Note that this probability agrees with the value computed using Table II to two decimal places. The difference is due to rounding of the probabilities given in Table II.

The steps to follow in calculating a probability corresponding to a normal random variable are shown in the following box:

Steps for Finding a Probability Corresponding to a Normal Random Variable

Sketch the normal distribution and indicate the mean of the random variable x. Then shade the area corresponding to the probability you want to find.
Convert the boundaries of the shaded area from x values to standard normal random variable z values by using the formula

[&z|=|*frac*{x|-||mu|}{|sig|} &]
$z = \frac{x - μ}{σ}$ $z = \frac{x - μ}{σ}$

Show the z values under the corresponding x values on your sketch.
Use technology or Table II in Appendix B to find the areas corresponding to the z values. If necessary, use the symmetry of the normal distribution to find areas corresponding to negative z values and the fact that the total area on each side of the mean equals .5 to convert the areas from Table II to the probabilities of the event you have shaded.

Example 4.20 Using Normal Probabilities to Make an Inference—Advertised Gas Mileage

Problem

Suppose an automobile manufacturer introduces a new model that has an advertised mean in-city mileage of 27 miles per gallon. Although such advertisements seldom report any measure of variability, suppose you write the manufacturer for the details of the tests and you find that the standard deviation is 3 miles per gallon. This information leads you to formulate a probability model for the random variable x, the in-city mileage for this car model. You believe that the probability distribution of x can be approximated by a normal distribution with a mean of 27 and a standard deviation of 3.
1. If you were to buy this model of automobile, what is the probability that you would purchase one that averages less than 20 miles per gallon for in-city driving? In other words, find $P (x < 20) .$ $P (x < 20) .$
2. Suppose you purchase one of these new models and it does get less than 20 miles per gallon for in-city driving. Should you conclude that your probability model is incorrect?

Solution

The probability model proposed for x, the in-city mileage, is shown in Figure 4.25. We are interested in finding the area A to the left of 20, since that area corresponds to the probability that a measurement chosen from this distribution falls below 20. In other words, if this model is correct, the area A represents the fraction of cars that can be expected to get less than 20 miles per gallon for in-city driving. To find A, we first calculate the z value corresponding to $x = 20.$ $x = 20.$ That is,

Figure 4.25

Area under the normal curve for Example 4.20

[&z|=|*frac*{x|-||mu|}{|sig|}|=|*frac*{20|-|27}{3}|=||minus|*frac*{7}{3}|=||minus|2.33 &]
$z = \frac{x - μ}{σ} = \frac{20 - 27}{3} = - \frac{7}{3} = - 2.33$ $z = \frac{x - μ}{σ} = \frac{20 - 27}{3} = - \frac{7}{3} = - 2.33$

Then

[&P|pbo|x|less|20|pbc||=|P|pbo|z|less||minus|2.33|pbc| &]
$P (x < 20) = P (z < - 2.33)$ $P (x < 20) = P (z < - 2.33)$

as indicated by the highlighted area in Figure 4.25. Since Table II gives only areas to the right of the mean (and because the normal distribution is symmetric about its mean), we look up 2.33 in Table II and find that the corresponding area is .4901. This is equal to the area between $z = 0$ $z = 0$ and $z = - 2.33,$ $z = - 2.33,$ so we find that

[&P|pbo|x|less|20|pbc||=|A|=|.5|-|.4901|=|.0099|app|.01 &]
$P (x < 20) = A = .5 - .4901 = .0099 \approx .01$ $P (x < 20) = A = .5 - .4901 = .0099 \approx .01$

According to this probability model, you should have only about a 1% chance of purchasing a car of this make with an in-city mileage under 20 miles per gallon.
Now you are asked to make an inference based on a sample: the car you purchased. You are getting less than 20 miles per gallon for in-city driving. What do you infer? We think you will agree that one of two possibilities exists:
1. The probability model is correct. You simply were unfortunate to have purchased one of the cars in the 1% that get less than 20 miles per gallon in the city.
2. The probability model is incorrect. Perhaps the assumption of a normal distribution is unwarranted, or the mean of 27 is an overestimate, or the standard deviation of 3 is an underestimate, or some combination of these errors occurred. At any rate, the form of the actual probability model certainly merits further investigation.
  
  You have no way of knowing with certainty which possibility is correct, but the evidence points to the second one. We are again relying on the rare-event approach to statistical inference that we introduced earlier. The sample (one measurement in this case) was so unlikely to have been drawn from the proposed probability model that it casts serious doubt on the model. We would be inclined to believe that the model is somehow in error.

Look Back

In applying the rare-event approach, the calculated probability must be small (say, less than or equal to .05) in order to infer that the observed event is, indeed, unlikely.

Now Work Exercise 4.88

Occasionally you will be given a probability and will want to find the values of the normal random variable that correspond to that probability. For example, suppose the scores on a college entrance examination are known to be normally distributed and a certain prestigious university will consider for admission only those applicants whose scores exceed the 90th percentile of the test score distribution. To determine the minimum score for consideration for admission, you will need to be able to use Table II or statistical software in reverse, as demonstrated in the next example.

Example 4.21 Finding a `z`-Value Associated with a 1-Tail Normal Probability

Problem

Find the value of z—call it $z_{0}$ $z_{0}$ —in the standard normal distribution that will be exceeded only 10% of the time. That is, find $z_{0}$ $z_{0}$ such that $P (z \geq z_{0}) = .10 .$ $P (z \geq z_{0}) = .10 .$

Solution

In this case, we are given a probability, or an area, and are asked to find the value of the standard normal random variable that corresponds to the area. Specifically, we want to find the value $z_{0}$ $z_{0}$ such that only 10% of the standard normal distribution exceeds $z_{0} .$ $z_{0} .$ (See Figure 4.26.)

Figure 4.26

Areas under the standard normal curve for Example 4.21

We know that the total area to the right of the mean, $z = 0,$ $z = 0,$ is .5, which implies that $z_{0}$ $z_{0}$ must lie to the right of 0 $(z_{0} > 0) .$ $(z_{0} > 0) .$ To pinpoint the value, we use the fact that the area to the right of $z_{0}$ $z_{0}$ is .10, which implies that the area between $z = 0$ $z = 0$ and $z_{0}$ $z_{0}$ is $.5 - .1 = .4 .$ $.5 - .1 = .4 .$ But areas between $z = 0$ $z = 0$ and some other z value are exactly the types given in TableII . Therefore, we look up the area .4000 in the body of Table II and find that the corresponding z value is (to the closest approximation) $z_{0} = 1.28.$ $z_{0} = 1.28.$ The implication is that the point 1.28 standard deviations above the mean is the 90th percentile of a normal distribution.

We can also arrive at this answer using statistical software. In MINITAB, we use the inverse cumulative distribution function for a normal random variable and specify the cumulative probability,

[&P|pbo|z|leq|z_{0}|pbc||=|.9 &]
$P (z \leq z_{0}) = .9$ $P (z \leq z_{0}) = .9$

The value of z₀ is shown (shaded) on the MINITAB printout in Figure 4.27. You can see that this value agrees with our solution using the normal table.

Figure 4.27

MINITAB Output for Example 4.21

Look Back

As with earlier problems, it is critical to correctly draw the normal probability sought on the normal curve. The placement of $z_{0}$ $z_{0}$ to the left or right of 0 is the key. Be sure to shade the probability (area) involving $z_{0} .$ $z_{0} .$ If it does not agree with the probability sought (i.e., if the shaded area is greater than .5 and the probability sought is smaller than .5), then you need to place $z_{0}$ $z_{0}$ on the opposite side of 0.

Example 4.22 Finding a `z`-Value Associated with a 2-Tail Normal Probability

Problem

Find the value of $z_{0}$ $z_{0}$ such that 95% of the standard normal z values lie between $- z_{0}$ $- z_{0}$ and $+ z_{0};$ $+ z_{0};$ that is, find $P (- z_{0} \leq z \leq z_{0}) = .95 .$ $P (- z_{0} \leq z \leq z_{0}) = .95 .$

Solution

Here we wish to move an equal distance $z_{0}$ $z_{0}$ in the positive and negative directions from the mean $z = 0$ $z = 0$ until 95% of the standard normal distribution is enclosed. This means that the area on each side of the mean will be equal to $\frac{1}{2} (.95) = .475,$ $\frac{1}{2} (.95) = .475,$ as shown in Figure 4.28. Since the area between $z = 0$ $z = 0$ and $z_{0}$ $z_{0}$ is .475, we look up .475 in the body of Table II to find the value $z_{0} = 1.96.$ $z_{0} = 1.96.$ Thus, as we found in reverse order in Example 5.6, 95% of a normal distribution lies between $+ 1.96$ $+ 1.96$ and $- 1.96$ $- 1.96$ standard deviations of the mean.

Figure 4.28

Areas under the standard normal curve for Example 4.22

Now Work Exercise 4.80a

Now that you have learned to find a standard normal z value that corresponds to a specified probability, we demonstrate a practical application in Example 4.23.

Example 4.23 Finding the Value of a Normal Random Variable—College Entrance Exam Application

Problem

Suppose the scores x on a college entrance examination are normally distributed with a mean of 550 and a standard deviation of 100. A certain prestigious university will consider for admission only those applicants whose scores exceed the 90th percentile of the distribution. Find the minimum score an applicant must achieve in order to receive consideration for admission to the university.

Solution

In this example, we want to find a score $x_{0}$ $x_{0}$ such that 90% of the scores (x values) in the distribution fall below $x_{0}$ $x_{0}$ and only 10% fall above $x_{0} .$ $x_{0} .$ That is,

[&P|pbo|x|leq|x_{0}|pbc||=|.90 &]
$P (x \leq x_{0}) = .90$ $P (x \leq x_{0}) = .90$

Converting x to a standard normal random variable where $μ = 550$ $μ = 550$ and $σ = 100,$ $σ = 100,$ we have

[&*AS*P|pbo|x|leq|x_{0}|pbc|*AP*|=|P|pbo|z|leq|*frac*{x_{0}|-||mu|}{|sig|}|pbc| &][&*AS**AP*|=|P|pbo|z|leq|*frac*{x_{0}|-|550}{100}|pbc||=|.90 &]
$\begin{array}{l} P (x \leq x_{0}) & = & P (z \leq \frac{x_{0} - μ}{σ}) \\ = & P (z \leq \frac{x_{0} - 550}{100}) = .90 \end{array}$ $\begin{array}{l} P (x \leq x_{0}) & = & P (z \leq \frac{x_{0} - μ}{σ}) \\ = & P (z \leq \frac{x_{0} - 550}{100}) = .90 \end{array}$

In Example 4.21 (see Figure 4.26), we found the 90th percentile of the standard normal distribution to be $z_{0} = 1.28.$ $z_{0} = 1.28.$ That is, we found that $P (z \leq 1.28) = .90 .$ $P (z \leq 1.28) = .90 .$ Consequently, we know that the minimum test score $x_{0}$ $x_{0}$ corresponds to a z-score of 1.28; in other words,

[&*frac*{x_{0}|-|550}{100}|=|1.28 &]
$\frac{x_{0} - 550}{100} = 1.28$ $\frac{x_{0} - 550}{100} = 1.28$

If we solve this equation for $x_{0},$ $x_{0},$ we find that

[&x_{0}|=|550|+|1.28|pbo|100|pbc||=|550|+|128|=|678 &]
$x_{0} = 550 + 1.28 (100) = 550 + 128 = 678$ $x_{0} = 550 + 1.28 (100) = 550 + 128 = 678$

This x value is shown in Figure 4.29. Thus, the 90th percentile of the test score distribution is 678. That is to say, an applicant must score at least 678 on the entrance exam to receive consideration for admission by the university.

Figure 4.29

Area under the normal curve for Example 4.23

Look Back

As the example shows, in practical applications of the normal table in reverse, first find the value of $z_{0}$ $z_{0}$ and then use the z-score formula in reverse to convert the value to the units of x.

Now Work Exercise 4.90b

Statistics in Action Revisited

Using the Normal Model to Maximize the Probability of a Hit with the Super Weapon

Recall that a defense contractor has developed a prototype gun for the U.S. Army that fires 1,100 flechettes with a single round. The specifications of the weapon are set so that when the gun is aimed at a target 500 meters away, the mean horizontal grid value of the flechettes is equal to the aim point. In the range test, the weapon was aimed at the center target in Figure SIA4.1; thus, $μ = 5$ $μ = 5$ feet. For three different tests, the standard deviation was set at $σ = 1$ $σ = 1$ foot, $σ = 2$ $σ = 2$ feet, and $σ = 4$ $σ = 4$ feet. From past experience, the defense contractor has found that the distribution of the horizontal flechette measurements is closely approximated by a normal distribution. Therefore, we can use the normal distribution to find the probability that a single flechette shot from the weapon will hit any one of the three targets. Recall from Figure SIA4.1 that the three targets range from $- 1$ $- 1$ to 1, 4 to 6, and 9 to 11 feet on the horizontal grid.

Consider first the middle target. Letting x represent the horizontal measurement for a flechette shot from the gun, we see that the flechette will hit the target if $4 \leq x \leq 6.$ $4 \leq x \leq 6.$ Using the normal probability table (Table II, Appendix B), we then find that the probability that this flechette will hit the target when $μ = 5$ $μ = 5$ and $σ = 1$ $σ = 1$ is

$\begin{array}{l} M i d d l e : P (4 \leq x \leq 6) & = & P (\frac{4 - 5}{1} < z < \frac{6 - 5}{1}) \\ σ = 1 \\ = & P (- 1 < z < 1) \\ = & 2 (.3413) = .6826 \end{array}$ $\begin{array}{l} M i d d l e : P (4 \leq x \leq 6) & = & P (\frac{4 - 5}{1} < z < \frac{6 - 5}{1}) \\ σ = 1 \\ = & P (- 1 < z < 1) \\ = & 2 (.3413) = .6826 \end{array}$

Similarly, we find the probabilities that the flechette hits the left and right targets shown in Figure SIA4.1:

$\begin{array}{l} L e f t : P (- 1 \leq x \leq 1) & = & P (\frac{- 1 - 5}{1} < z < \frac{1 - 5}{1}) \\ σ = 1 \\ = & P (- 6 < z < - 4) \approx 0 \\ R i g h t : P (9 \leq x \leq 11) & = & P (\frac{9 - 5}{1} < z < \frac{11 - 5}{1}) \\ σ = 1 \\ = & P (4 < z < 6) \approx 0 \end{array}$ $\begin{array}{l} L e f t : P (- 1 \leq x \leq 1) & = & P (\frac{- 1 - 5}{1} < z < \frac{1 - 5}{1}) \\ σ = 1 \\ = & P (- 6 < z < - 4) \approx 0 \\ R i g h t : P (9 \leq x \leq 11) & = & P (\frac{9 - 5}{1} < z < \frac{11 - 5}{1}) \\ σ = 1 \\ = & P (4 < z < 6) \approx 0 \end{array}$

You can see that there is about a 68% chance that a flechette will hit the middle target, but virtually no chance that one will hit the left or right target when the standard deviation is set at 1 foot.

To find these three probabilities for $σ = 2$ $σ = 2$ and $σ = 4,$ $σ = 4,$ we use the normal probability function in MINITAB. Figure SIA4.2 is a MINITAB worksheet giving the cumulative probabilities of a normal random variable falling below the x values in the first column. The cumulative probabilities for $σ = 2$ $σ = 2$ and $σ = 4$ $σ = 4$ are given in the columns named “sigma2” and “sigma4,” respectively.

Figure SIA4.2

MINITAB worksheet with cumulative normal probabilities

Using the cumulative probabilities in the figure to find the three probabilities when $σ = 2,$ $σ = 2,$ we have

$\begin{array}{l} M i d d l e : P (4 \leq x \leq 6) & = & P (x \leq 6) - P (x \leq 4) \\ σ = 2 \\ = & .6915 - .3085 = .3830 \\ L e f t : P (- 1 \leq x \leq 1) & = & P (x \leq 1) - P (x \leq - 1) \\ σ = 2 \\ = & .0227 - .0013 = .0214 \\ R i g h t : P (9 \leq x \leq 11) & = & P (x \leq 11) - P (x \leq 9) \\ σ = 2 \\ = & .9987 - .9773 = .0214 \end{array}$ $\begin{array}{l} M i d d l e : P (4 \leq x \leq 6) & = & P (x \leq 6) - P (x \leq 4) \\ σ = 2 \\ = & .6915 - .3085 = .3830 \\ L e f t : P (- 1 \leq x \leq 1) & = & P (x \leq 1) - P (x \leq - 1) \\ σ = 2 \\ = & .0227 - .0013 = .0214 \\ R i g h t : P (9 \leq x \leq 11) & = & P (x \leq 11) - P (x \leq 9) \\ σ = 2 \\ = & .9987 - .9773 = .0214 \end{array}$

Thus, when $σ = 2,$ $σ = 2,$ there is about a 38% chance that a flechette will hit the middle target, a 2% chance that one will hit the left target, and a 2% chance that one will hit the right target. The probability that a flechette will hit either the middle or the left or the right target is simply the sum of these three probabilities (an application of the additive rule of probability). This sum is $.3830 + .0214 + .0214 = .4258;$ $.3830 + .0214 + .0214 = .4258;$ consequently, there is about a 42% chance of hitting any one of the three targets when specifications are set so that $σ = 2.$ $σ = 2.$

Now we use the cumulative probabilities in Figure SIA5.2 to find the three hit probabilities when $σ = 4 :$ $σ = 4 :$

$\begin{array}{l} M i d d l e : P (4 \leq x \leq 6) & = & P (x \leq 6) - P (x \leq 4) \\ σ = 4 \\ = & .5987 - .4013 = .1974 \\ L e f t : P (- 1 \leq x \leq 1) & = & P (x \leq 1) - P (x \leq - 1) \\ σ = 4 \\ = & .1587 - .0668 = .0919 \\ R i g h t : P (9 \leq x \leq 11) & = & P (x \leq 11) - P (x \leq 9) \\ σ = 4 \\ = & .9332 - .8413 = .0919 \end{array}$ $\begin{array}{l} M i d d l e : P (4 \leq x \leq 6) & = & P (x \leq 6) - P (x \leq 4) \\ σ = 4 \\ = & .5987 - .4013 = .1974 \\ L e f t : P (- 1 \leq x \leq 1) & = & P (x \leq 1) - P (x \leq - 1) \\ σ = 4 \\ = & .1587 - .0668 = .0919 \\ R i g h t : P (9 \leq x \leq 11) & = & P (x \leq 11) - P (x \leq 9) \\ σ = 4 \\ = & .9332 - .8413 = .0919 \end{array}$

Thus, when $σ = 4,$ $σ = 4,$ there is about a 20% chance that a flechette will hit the middle target, a 9% chance that one will hit the left target, and a 9% chance that one will hit the right target. The probability that a flechette will hit any one of the three targets is $.1974 + .0919 + .0919 = .3812 .$ $.1974 + .0919 + .0919 = .3812 .$

These probability calculations reveal a few patterns. First, the probability of hitting the middle target (the target at which the gun is aimed) is reduced as the standard deviation is increased. Obviously, then, if the U.S. Army wants to maximize the chance of hitting the target that the prototype gun is aimed at, it will want specifications set with a small value of $σ .$ $σ .$ But if the Army wants to hit multiple targets with a single shot of the weapon, $σ$ $σ$ should be increased. With a larger $σ,$ $σ,$ not as many of the flechettes will hit the target aimed at, but more will hit peripheral targets. Whether $σ$ $σ$ should be set at 4 or 6 (or some other value) depends on how high of a hit rate is required for the peripheral targets.

Exercises 4.73–4.102

Understanding the Principles

4.73 Describe the shape of a normal probability distribution.
4.74 If x has a normal distribution with mean $μ$ $μ$ and standard deviation $σ,$ $σ,$ describe the distribution of $z = (x - μ) / σ$ $z = (x - μ) / σ$ .
4.75 What is the name given to a normal distribution when $μ = 0$ $μ = 0$ and $σ = 1 ?$ $σ = 1 ?$ standard normal

Learning the Mechanics

4.76 Find the area under the standard normal probability distribution between the following pairs of z-scores:
1. $z = 0 and z = 2.00$ $z = 0 and z = 2.00$
2. $z = 0 and z = 1.00$ $z = 0 and z = 1.00$
3. $z = 0 and z = 3$ $z = 0 and z = 3$
4. $z = 0 and z = .58$ $z = 0 and z = .58$
5. $z = - 2.00 and z = 0$ $z = - 2.00 and z = 0$
6. $z = - 1.00 and z = 0$ $z = - 1.00 and z = 0$
7. $z = - 1.69 and z = 0$ $z = - 1.69 and z = 0$
8. $z = - .58 and z = 0$ $z = - .58 and z = 0$
4.77 Find each of the following probabilities for a standard normal random variable z:
1. a. $P (z = 1)$ $P (z = 1)$
2. b. $P (z \leq 1)$ $P (z \leq 1)$
3. c. $P (z < 1)$ $P (z < 1)$
4. d. $P (z > 1)$ $P (z > 1)$
5. e. $P (- 1 \leq z \leq 1)$ $P (- 1 \leq z \leq 1)$
6. f. $P (- 2 \leq z \leq 2)$ $P (- 2 \leq z \leq 2)$
7. g. $P (- 2.16 \leq z \leq .55)$ $P (- 2.16 \leq z \leq .55)$
8. h. $P (- .42 < z < 1.96)$ $P (- .42 < z < 1.96)$
4.78 Find the following probabilities for the standard normal random variable z:
1. a. $P (z > 1.46)$ $P (z > 1.46)$
2. b. $P (z < - 1.56)$ $P (z < - 1.56)$
3. c. $P (.67 \leq z \leq 2.41)$ $P (.67 \leq z \leq 2.41)$
4. d. $P (- 1.96 \leq z < - .33)$ $P (- 1.96 \leq z < - .33)$
5. e. $P (z \geq 0)$ $P (z \geq 0)$
6. f. $P (- 2.33 < z < 1.50)$ $P (- 2.33 < z < 1.50)$
7. g. $P (z \geq - 2.33)$ $P (z \geq - 2.33)$
8. h. $P (z < 2.33)$ $P (z < 2.33)$
4.79 Give the z-score for a measurement from a normal distribution for the following:
1. 1 standard deviation above the mean
2. 1 standard deviation below the mean
3. Equal to the mean
4. 2.5 standard deviations below the mean
5. 3 standard deviations above the mean
4.80 Find a value $z_{0}$ $z_{0}$ of the standard normal random variable z such that
1. a. $P (z \leq z_{0}) = .0401$ $P (z \leq z_{0}) = .0401$
2. b. $P (- z_{0} \leq z \leq z_{0}) = .95$ $P (- z_{0} \leq z \leq z_{0}) = .95$
3. c. $P (- z_{0} \leq z \leq z_{0}) = .90$ $P (- z_{0} \leq z \leq z_{0}) = .90$
4. d. $P (- z_{0} \leq z \leq z_{0}) = .8740$ $P (- z_{0} \leq z \leq z_{0}) = .8740$
5. e. $P (- z_{0} \leq z \leq 0) = .2967$ $P (- z_{0} \leq z \leq 0) = .2967$
6. f. $P (- 2 < z < z_{0}) = .9710$ $P (- 2 < z < z_{0}) = .9710$
7. g. $P (z \geq z_{0}) = .5$ $P (z \geq z_{0}) = .5$
8. h. $P (z \geq z_{0}) = .0057$ $P (z \geq z_{0}) = .0057$
4.81 Find a value $z_{0}$ $z_{0}$ of the standard normal random variable z such that
1. $P (z \geq z_{0}) = .05$ $P (z \geq z_{0}) = .05$
2. $P (z \geq z_{0}) = .025$ $P (z \geq z_{0}) = .025$
3. $P (z \leq z_{0}) = .025$ $P (z \leq z_{0}) = .025$
4. $P (z \geq z_{0}) = .10$ $P (z \geq z_{0}) = .10$
5. $P (z > z_{0}) = .10$ $P (z > z_{0}) = .10$
4.82 Suppose x is a normally distributed random variable with $μ = 11$ $μ = 11$ and $σ = 2.$ $σ = 2.$ Find each of the following:
1. a. $P (10 \leq x \leq 12)$ $P (10 \leq x \leq 12)$
2. b. $P (6 \leq x \leq 10)$ $P (6 \leq x \leq 10)$
3. c. $P (13 \leq x \leq 16)$ $P (13 \leq x \leq 16)$
4. d. $P (7.8 \leq x \leq 12.6)$ $P (7.8 \leq x \leq 12.6)$
5. e. $P (x \geq 13.24)$ $P (x \geq 13.24)$
6. f. $P (x \geq 7.62)$ $P (x \geq 7.62)$
4.83 Suppose x is a normally distributed random variable with $μ = 30$ $μ = 30$ and $σ = 8.$ $σ = 8.$ Find a value $x_{0}$ $x_{0}$ of the random variable x such that
1. $P (x \geq x_{0}) = .5$ $P (x \geq x_{0}) = .5$
2. $P (x < x_{0}) = .025$ $P (x < x_{0}) = .025$
3. $P (x > x_{0}) = .10$ $P (x > x_{0}) = .10$
4. $P (x > x_{0}) = .95$ $P (x > x_{0}) = .95$
5. 10% of the values of x are less than $x_{0} .$ $x_{0} .$
6. 80% of the values of x are less than $x_{0} .$ $x_{0} .$
7. 1% of the values of x are greater than $x_{0} .$ $x_{0} .$
4.84 The random variable x has a normal distribution with standard deviation 25. It is known that the probability that x exceeds 150 is .90. Find the mean $μ$ $μ$ of the probability distribution.

Applet Exercise 4.6

Open the applet entitled Sample from a Population. On the pull-down menu to the right of the top graph, select Bell shaped. The box to the left of the top graph displays the population mean, median, and standard deviation.

1. Run the applet for each available value of n on the pull-down menu for the sample size. Go from the smallest to the largest value of n. For each value of n, observe the shape of the graph of the sample data and record the mean, median, and standard deviation of the sample.
2. Describe what happens to the shape of the graph and the mean, median, and standard deviation of the sample as the sample size increases.

Applying the Concepts—Basic

4.85 Dental anxiety study. To gauge their fear of going to a dentist, a random sample of adults completed the Modified Dental Anxiety Scale questionnaire (BMC Oral Health, Vol. 9, 2009). Scores on the scale range from zero (no anxiety) to 25 (extreme anxiety). The mean score was 11 and the standard deviation was 4. Assume that the distribution of all scores on the Modified Dental Anxiety Scale is approximately normal with $μ = 11 and σ = 4$ $μ = 11 and σ = 4$ .
1. Suppose you score a 10 on the Modified Dental Anxiety Scale. Find the z-value for your score.
2. Find the probability that someone scores between 10 and 15 on the Modified Dental Anxiety Scale.
3. Find the probability that someone scores above 20 on the Modified Dental Anxiety Scale.
4.86 Tomato as a taste modifier. Miraculin—a protein naturally produced in a rare tropical fruit—can convert a sour taste into a sweet taste; thus, it has the potential to be an alternative low-calorie sweetener. In Plant Science (May 2010), a group of Japanese environmental scientists investigated the ability of a hybrid tomato plant to produce miraculin. For a particular generation of the tomato plant, the amount x of miraculin produced (measured in micrograms per gram of fresh weight) had a mean of 105.3 and a standard deviation of 8.0. Assume that x is normally distributed.
1. Find $P (x > 120) .$ $P (x > 120) .$
2. Find $P (100 < x < 110) .$ $P (100 < x < 110) .$
3. Find the value a for which $P (x < a) = .25$ $P (x < a) = .25$ .
4.87 Deep mixing of soil. Deep mixing is a ground improvement method developed for soft soils like clay, silt, and peat. Swedish civil engineers investigated the properties of soil improved by deep mixing with lime-cement columns in the journal Giorisk (Vol. 7, 2013). The mixed soil was tested by advancing a cylindrical rod with a cone tip down into the soil. During penetration, the cone penetrometer measures the cone tip resistance (megapascals, MPa). The researchers established that tip resistance for the deep mixed soil followed a normal distribution with $μ = 2.2$ $μ = 2.2$ MPa and $σ = .9$ $σ = .9$ MPa.
1. Find the probability that the tip resistance will fall between 1.3 and 4.0 MPa.
2. Find the probability that the tip resistance will exceed 1.0 MPa.
3. Find a value of tip resistance, T, such that 35% of all soil samples have tip resistance values that exceed T.
4.88 Casino gaming. In Chance (Spring 2005), University of Denver statistician R. C. Hannum discussed casino gaming and the laws of probability. Casino games of pure chance (e.g., craps, roulette, baccarat, and keno) always yield a “house advantage.” For example, in the game of double-zero roulette, the expected casino win percentage is 5.26% on bets made on whether the outcome will be either black or red. (This percentage implies that for every $5 bet on black or red, the casino will earn a net of about 25 cents.) It can be shown that in 100 roulette plays on black/red, the average casino win percentage is normally distributed with mean 5.26% and standard deviation 10%. Let x represent the average casino win percentage after 100 bets on black/red in double-zero roulette.
1. Find $P (x > 0) .$ $P (x > 0) .$ (This is the probability that the casino wins money.)
2. Find $P (5 < x < 15) .$ $P (5 < x < 15) .$
3. Find $P (x < 1) .$ $P (x < 1) .$
4. If you observed an average casino win percentage of $- 25 %$ $- 25 %$ after 100 roulette bets on black/red, what would you conclude?
4.89 Shopping vehicle and judgment. Refer toConsider the Journal of Marketing Research (Dec. 2011) study of whether you are more likely to choose a vice product (e.g., a candy bar) when your arm is flexed (as when carrying a shopping basket) than when your arm is extended (as when pushing a shopping cart), Exercise 2.111 (p. 78). The study measured choice scores (on a scale of 0 to 100, where higher scores indicate a greater preference for vice options) for consumers shopping under each of the two conditions. Recall that the average choice score for consumers with a flexed arm was 59, while the average for consumers with an extended arm was 43. For both conditions, assume that the standard deviation of the choice scores is 5. Also assume that both distributions are approximately normal.
1. In the flexed arm condition, what is the probability that a consumer has a choice score of 60 or greater?
2. In the extended arm condition, what is the probability that a consumer has a choice score of 60 or greater?
4.90 Shell lengths of sea turtles. Refer to the Aquatic Biology (Vol. 9, 2010) study of green sea turtles inhabiting the Grand Cayman South Sound lagoon, Exercise 2.85 (p. 69). Researchers discovered that the curved carapace (shell) length of these turtles is approximately normally distributed with mean 55.7 centimeters and standard deviation 11.5 centimeters.
1. a. The minimum and maximum size limits for captured turtles in the legal marine turtle fishery are 40 cm and 60 cm, respectively. How likely are you to capture a green sea turtle that is considered illegal?
2. b. What maximum limit, L, should be set so that only 10% of the turtles captured have shell lengths greater than L?
4.91 Flicker in an electrical power system. An assessment of the quality of the electrical power system in Turkey was the topic of an article published in Electrical Engineering (Mar. 2013). One measure of quality is the degree to which voltage fluctuations cause light flicker in the system. The perception of light flicker x when the system is set at 380 kV was measured periodically (over 10-minute intervals). For transformers supplying heavy industry plants, the light flicker distribution was found to follow (approximately) a normal distribution with $μ = 2.2 %$ $μ = 2.2 %$ and $σ = .5 %$ $σ = .5 %$ . If the perception of light flicker exceeds 3%, the transformer is shut down and the system is reset. How likely is it for a transformer supplying a heavy industry plant to be shut down due to light flicker?

Applying the Concepts—Intermediate

4.92 Optimal goal target in soccer. When attempting to score a goal in soccer, where should you aim your shot? Should you aim for a goalpost (as some soccer coaches teach), the middle of the goal, or some other target? To answer these questions, Chance (Fall 2009) utilized the normal probability distribution. Suppose the accuracy x of a professional soccer player’s shots follows a normal distribution with a mean of zero feet and a standard deviation of 3 feet. (For example, if the player hits his target, $x = 0$ $x = 0$ ; if he misses his target by 2 feet to the right, $x = 2$ $x = 2$ ; and if he misses 1 foot to the left, $x = - 1$ $x = - 1$ .) Now, a regulation soccer goal is 24 feet wide. Assume that a goalkeeper will stop (save) all shots within 9 feet of where he is standing; all other shots on goal will score. Consider a goalkeeper who stands in the middle of the goal.
1. If the player aims for the right goalpost, what is the probability that he will score?
2. If the player aims for the center of the goal, what is the probability that he will score?
3. If the player aims for halfway between the right goalpost and the outer limit of the goalkeeper’s reach, what is the probability that he will score?
4.93 Voltage sags and swells. Refer to the Electrical Engineering (Vol. 95, 2013) study of the power quality of a transformer, Exercise 2.127 (p. 82). Recall that two causes of poor power quality are “sags” and “swells.” (A sag is an unusual dip and a swell is an unusual increase in the voltage level of a transformer.) For Turkish transformers built for heavy industry, the mean number of sags per week was 353 and the mean number of swells per week was 184. As in Exercise 2.127 , assume the standard deviation of the sag distribution is 30 sags per week and the standard deviation of the swell distribution is 25 swells per week. Also, assume that the number of sags and number of swells are both normally distributed. Suppose one of the transformers is randomly selected and found to have 400 sags and 100 swells in a week.
1. What is the probability that the number of sags per week is less than 400?
2. What is the probability that the number of swells per week is greater than 100?

4.94 Safety of underground tunnels. Research published in the journal Tunnelling and Underground Space Technology (July 2014) evaluated the safety of underground tunnels built in rigid soils. A Factor of Safety (FS), measured as the ratio of capacity over demand, was determined for three different areas of tunnels made from shotcrete: tunnel face, tunnel walls, and tunnel crown. FS was determined to be normally distributed in each area, with means and standard deviations shown in the table. Tunnel failure is considered to occur when FS is lower than or equal to 1. Which tunnel area is more likely to result in failure? Why?

	Mean $(μ)$ $(μ)$	Standard Deviation $(σ)$ $(σ)$
Tunnel Face	1.2	.16
Tunnel Walls	1.4	.20
Tunnel Crown	2.1	.70

4.95 Ambulance response time. Ambulance response time is measured as the time (in minutes) between the initial call to emergency medical services (EMS) and when the patient is reached by ambulance. Geographical Analysis (Vol. 41, 2009) investigated the characteristics of ambulance response time for EMS calls in Edmonton, Alberta. For a particular EMS station (call it Station A), ambulance response time is known to be normally distributed with $μ = 7.5$ $μ = 7.5$ minutes and $σ = 2.5$ $σ = 2.5$ minutes.
1. Regulations require that 90% of all emergency calls should be reached in 9 minutes or less. Are the regulations met at EMS Station A? Explain.
2. A randomly selected EMS call in Edmonton has an ambulance response time of 2 minutes. Is it likely that this call was serviced by Station A? Explain.
4.96 Rating employee performance. Almost all companies utilize some type of year-end performance review for their employees. Human Resources (HR) at the University of Texas Health Science Center provides guidelines for supervisors rating their subordinates. For example, raters are advised to examine their ratings for a tendency to be either too lenient or too harsh. According to HR, “if you have this tendency, consider using a normal distribution—10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable.” Suppose you are rating an employee’s performance on a scale of 1 (lowest) to 100 (highest). Also, assume the ratings follow a normal distribution with a mean of 50 and a standard deviation of 15.
1. What is the lowest rating you should give to an “exemplary” employee if you follow the University of Texas HR guidelines?
2. What is the lowest rating you should give to a “competent” employee if you follow the University of Texas HR guidelines?
4.97 California’s electoral college votes. During a presidential election, each state is allotted a different number of votes to the electoral college depending on population. For example, California is allotted 55 votes (the most) while several states (including the District of Columbia) are allotted 3 votes each (the least). When a presidential candidate wins the popular vote in a state, the candidate wins all the electoral college votes in that state. To become president, a candidate must win 270 of the total of 538 votes in the electoral college. Chance (Winter 2010) demonstrated the impact of winning California on the presidential election. Assuming a candidate wins California’s 55 votes, the number of additional electoral college votes the candidate will win can be approximated by a normal distribution with $μ = 241.5$ $μ = 241.5$ votes and $σ = 49.8$ $σ = 49.8$ votes. If a presidential candidate wins the popular vote in California, what are the chances that he or she becomes the next U.S. president?
4.98 Alcohol, threats, and electric shocks. A group of Florida State University psychologists examined the effects of alcohol on the reactions of people to a threat (Journal of Abnormal Psychology, Vol. 107, 1998). After obtaining a specified blood alcohol level, the psychologists placed experimental subjects in a room and threatened them with electric shocks. Using sophisticated equipment to monitor the subjects’ eye movements, the psychologists recorded the startle response (measured in milliseconds) of each subject. The mean and standard deviation of the startle responses were 37.9 and 12.4, respectively. Assume that the startle response x for a person with the specified blood alcohol level is approximately normally distributed.
1. Find the probability that x is between 40 and 50 milliseconds.
2. Find the probability that x is less than 30 milliseconds.
3. Give an interval for x centered around 37.9 milliseconds so that the probability that x falls in the interval is .95.
4. Ten percent of the experimental subjects have startle responses above what value?

Applying the Concepts—Advanced

4.99 Box plots and the standard normal distribution. What relationship exists between the standard normal distribution and the box-plot methodology (Section 2.7) for describing distributions of data by means of quartiles? The answer depends on the true underlying probability distribution of the data. Assume for the remainder of this exercise that the distribution is normal.
1. Calculate the values $z_{L}$ $z_{L}$ and $z_{U}$ $z_{U}$ of the standard normal random variable z that correspond, respectively, to the hinges of the box plot (i.e., the lower and upper quartiles $Q_{L}$ $Q_{L}$ and $Q_{U}$ $Q_{U}$ ) of the probability distribution.
2. Calculate the z values that correspond to the inner fences of the box plot for a normal probability distribution.
3. Calculate the z values that correspond to the outer fences of the box plot for a normal probability distribution.
4. What is the probability that an observation lies beyond the inner fences of a normal probability distribution? The outer fences?
5. Can you now better understand why the inner and outer fences of a box plot are used to detect outliers in a distribution? Explain.
4.100 Load on frame structures. In the Journal of the International Association for Shell and Spatial Structures (Apr. 2004), Japanese environmental researchers studied the performance of truss-and-frame structures subjected to uncertain loads. The load was assumed to have a normal distribution with a mean of 20 thousand pounds. Also, the probability that the load is between 10 and 30 thousand pounds is .95. On the basis of this information, find the standard deviation of the load distribution.
4.101 Range of women’s heights. In Chance (Winter 2007), Yale Law School professor Ian Ayres published the results of a study he conducted with his son and daughter on whether college students could estimate a range for women’s heights. The students were shown a graph of a normal distribution of heights and were asked, “The average height of women over 20 years old in the United States is 64 inches. Using your intuition, please give your best estimate of the range of heights that would include C% of women over 20 years old. Please make sure that the center of the range is the average height of 64 inches.” The value of C was randomly selected as 50%, 75%, 90%, 95%, or 99% for each student surveyed.
1. Give your estimate of the range for $C = 50 %$ $C = 50 %$ of women’s heights.
2. Give your estimate of the range for $C = 75 %$ $C = 75 %$ of women’s heights.
3. Give your estimate of the range for $C = 90 %$ $C = 90 %$ of women’s heights.
4. Give your estimate of the range for $C = 95 %$ $C = 95 %$ of women’s heights.
5. Give your estimate of the range for $C = 99 %$ $C = 99 %$ of women’s heights.
6. The standard deviation of heights for women over 20 years old is known to be 2.6 inches. Use this information to revise your answers to parts a–e
7. Which value of C has the most accurate estimated range? (Note: The researchers found that college students were most accurate for $C = 90 %$ $C = 90 %$ and $C = 95 % .$ $C = 95 % .$ )
4.102 Executive coaching and meeting effectiveness. Poor executive leadership during business meetings can result in counterproductive outcomes (e.g., negative employee attitudes, ambiguous objectives). Can executive coaching help improve business meeting effectiveness? This was the question of interest in an article published in Consulting Psychology Journal: Practice and Research (Vol. 61, 2009). Certain behaviors by leaders during meetings were categorized as content behaviors (e.g., seeking information, disagreeing/attacking) or process behaviors (e.g., asking clarifying questions, summarizing). The goal of executive coaching was to reduce content behaviors in favor of process behaviors. The study reported that prior to receiving executive coaching, the percentage of observed content behaviors of leaders has a mean of 75% with a standard deviation of 8.5%. In contrast, after receiving executive coaching, the percentage of observed content behaviors of leaders has a mean of 52% with a standard deviation of 7.5%. Assume that the percentage of observed content behaviors is approximately normally distributed for both leaders with and without executive coaching. Suppose you observe 70% content behaviors by the leader of a business meeting. Give your opinion on whether the leader has received executive coaching.

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z	.00	.01	.02	.03	.04	.05	.06	.07	.08	.09
.0	.0000	.0040	.0080	.0120	.0160	.0199	.0239	.0279	.0319	.0359
.1	.0398	.0438	.0478	.0517	.0557	.0596	.0636	.0675	.0714	.0753
.2	.0793	.0832	.0871	.0910	.0948	.0987	.1026	.1064	.1103	.1141
.3	.1179	.1217	.1255	.1293	.1331	.1368	.1406	.1443	.1480	.1517
.4	.1554	.1591	.1628	.1664	.1700	.1736	.1772	.1808	.1844	.1879
.5	.1915	.1950	.1985	.2019	.2054	.2088	.2123	.2157	.2190	.2224
.6	.2257	.2291	.2324	.2357	.2389	.2422	.2454	.2486	.2517	.2549
.7	.2580	.2611	.2642	.2673	.2704	.2734	.2764	.2794	.2823	.2852
.8	.2881	.2910	.2939	.2967	.2995	.3023	.3051	.3078	.3106	.3133
.9	.3159	.3186	.3212	.3238	.3264	.3289	.3315	.3340	.3365	.3389
1.0	.3413	.3438	.3461	.3485	.3508	.3531	.3554	.3577	.3599	.3621
1.1	.3643	.3665	.3686	.3708	.3729	.3749	.3770	.3790	.3810	.3830
1.2	.3849	.3869	.3888	.3907	.3925	.3944	.3962	.3980	.3997	.4015
1.3	.4032	.4049	.4066	.4082	.4099	.4115	.4131	.4147	.4162	.4177
1.4	.4192	.4207	.4222	.4236	.4251	.4265	.4279	.4292	.4306	.4319
1.5	.4332	.4345	.4357	.4370	.4382	.4394	.4406	.4418	.4429	.4441

Table of Contents for 4.5 The Normal Distribution

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Table of Contents for
4.5 The Normal Distribution