Ch7_NormalDistribution

372

Outline7.1 Properties of the Normal

Distribution

7.2 The Standard NormalDistribution

7.3 Applications of the NormalDistribution

7.4 Assessing Normality

7.5 The Normal Approximationto the Binomial ProbabilityDistribution

You are interested in starting your ownMENSA-type club. To qualify for the club, the

potential member must have intelligence that isin the top 20% of all people. You must decide the

baseline score that allows an individual to qualify.See the Decisions project on page 421.

PUTTING IT TOGETHERIn Chapter 6, we introduced discrete probability distributions and, in particular, the binomial probability distri-bution.We computed probabilities for this discrete distribution using its probability distribution function.

However, we could also determine the probability of any discrete random variable from its probabilityhistogram. For example, the figure shows the probability histogram for the binomial random variable X with

and From the probability histogram, we can see

Notice that the width of each rectangle in the probability histogramis 1. Since the area of a rectangle equals height times width, we canthink of P(1) as the area of the rectangle corresponding to Thinking of probability in this fashion makes the transition fromcomputing discrete probabilities to continuous probabilities mucheasier.

In this chapter, we discuss two continuous distributions, theuniform distribution and the normal distribution. The greater part ofthe discussion will focus on the normal distribution, which has manyuses and applications.

x = 1.

P112 L 0.31.p = 0.35.n = 5

7 The NormalProbabilityDistribution

0 1 2 3 4 5

P(x)

0.4

0.3

0.2

0.1

0 x

Binomial ProbabilityHistogram;

n 5, p 0.35

M07_SULL8028_03_SE_C07.QXD 9/10/08 6:49 PM Page 372

Section 7.1 Properties of the Normal Distribution 373

Note to InstructorIf you like, you can print out and distrib-ute the Preparing for This Section quizlocated in the Instructors Resource Cen-ter. The purpose of the quiz is to verifythat the students have the prerequisiteknowledge for the section

7.1 PROPERTIES OF THE NORMAL DISTRIBUTIONPreparing for This Section Before getting started, review the following: Continuous variable (Section 1.1, p. 8)

Objectives 1 Utilize the uniform probability distribution2 Graph a normal curve3 State the properties of the normal curve4 Explain the role of area in the normal density function5 Describe the relation between a normal random variable and a

standard normal random variable

1 Utilize the Uniform Probability DistributionWe illustrate a uniform distribution using an example. Using the uniform distribu-tion makes it easy to see the relation between area and probability.

EXAMPLE 1 The Uniform Distribution

Imagine that a friend of yours is always late. Let the random variable X representthe time from when you are supposed to meet your friend until he shows up. Furthersuppose that your friend could be on time or up to 30 minutes late

with all intervals of equal time between and being equallylikely. For example, your friend is just as likely to be from 3 to 4 minutes late as he isto be 25 to 26 minutes late. The random variable X can be any value in the intervalfrom 0 to 30, that is, Because any two intervals of equal length between0 and 30, inclusive, are equally likely, the random variable X is said to follow auniform probability distribution.

When we compute probabilities for discrete random variables, we usually sub-stitute the value of the random variable into a formula.

Things are not as easy for continuous random variables. Since an infinite numberof outcomes are possible for continuous random variables, the probability of observinga particular value of a continuous random variable is zero. For example, the proba-bility that your friend is exactly 12.9438823 minutes late is zero. This result is basedon the fact that classical probability is found by dividing the number of ways an eventcan occur by the total number of possibilities. There is one way to observe12.9438823, and there are an infinite number of possible values between 0 and 30, sowe get a probability that is zero.To resolve this problem, we compute probabilities ofcontinuous random variables over an interval of values. For example, we mightcompute the probability that your friend is between 10 and 15 minutes late. To findprobabilities for continuous random variables, we use probability density functions.

0 x 30.

x = 30x = 01x = 302, 1x = 02

Note to InstructorTo help students understand the relationbetween area and probability, we beginwith a discussion of the uniform distribu-tion. The graph of the uniform distribu-tion is a rectangle. Tell students that wewill be using the concept of area in thissection and that the area of a rectangleis length times width.

Definition A probability density function (pdf) is an equation used to compute probabilitiesof continuous random variables. It must satisfy the following two properties:

1. The total area under the graph of the equation over all possible values of therandom variable must equal 1.2. The height of the graph of the equation must be greater than or equal to 0 for allpossible values of the random variable. That is, the graph of the equation must lieon or above the horizontal axis for all possible values of the random variable.

z-score (Section 3.4, pp. 167168)

The Empirical Rule (Section 3.2, pp. 150152) Rules for a discrete probability distribution (Section 6.1, p. 332)


The area under the graph of a density function over an interval represents theprobability of observing a value of the random variable in that interval.

374 Chapter 7 The Normal Probability Distribution

Property 1 is similar to the rule for discrete probability distributions that statedthe sum of the probabilities must add up to 1. Property 2 is similar to the rule thatstated all probabilities must be greater than or equal to 0.

Figure 1 illustrates the properties for the example about your friend who is al-ways late. Since all possible values of the random variable between 0 and 30 areequally likely, the graph of the probability density function for uniform randomvariables is a rectangle. Because the random variable is any number between 0 and30 inclusive, the width of the rectangle is 30. Since the area under the graph of theprobability density function must equal 1, and the area of a rectangle equals height

times width, the height of the rectangle must be 1

30.

In Other WordsTo find probabilities for continuousrandom variables, we do not useprobability distribution functions (as we did for discrete random variables).Instead, we use probability densityfunctions. The word density is usedbecause it refers to the number ofindividuals per unit of area.

0

Den

sity

Random Variable (time)

Area 1130

30 X

Figure 1

A pressing question remains: How do we use density functions to find probabil-ities of continuous random variables?

The following example illustrates this statement.

EXAMPLE 2 Area as a Probability

Problem: Refer to the situation presented in Example 1. What is the probabilitythat your friend will be between 10 and 20 minutes late the next time you meet him?

Approach: Figure 1 presented the graph of the density function. We need to findthe area under the graph between 10 and 20 minutes.

Solution: Figure 2 presents the graph of the density function with the area we wishto find shaded in green.

0 302010

Den

sity

130

Time (min)

X

Figure 2

The width of the shaded rectangle is 10 and its height is Therefore, the area

between 10 and 20 is The probability that your friend is between 10

and 20 minutes late is13

.

10a 130b = 1

3.

130

.

Now Work Problem 13



*The vertical scale on the graph, which indicates density, is purposely omitted. The vertical scale, whileimportant, will not play a role in any of the computations using this curve.

Historical NoteKarl Pearson coined the

phrase normal curve. He did not dothis to imply that a distribution that isnot normal is abnormal. Rather,Pearson wanted to avoid giving thename of the distribution a propername, such as Gaussian (as in CarlFriedrich Gauss).

We introduced the uniform density function so that we could associate proba-bility with area. We are now better prepared to discuss the most frequently usedcontinuous distribution, the normal distribution.

2 Graph a Normal CurveMany continuous random variables, such as IQ scores, birth weights of babies, orweights of M&Ms, have relative frequency histograms that have a shape similar toFigure 3. Relative frequency histograms that have a shape similar to Figure 3 aresaid to have the shape of a normal curve.

Definition A continuous random variable is normally distributed, or has a normal proba-bility distribution, if the relative frequency histogram of the random variablehas the shape of a normal curve.

Figure 4 shows a normal curve, demonstrating the roles that and play indrawing the curve. For any distribution, the mode represents the high point of thegraph of the distribution. The median represents the point where 50% of the areaunder the distribution is to the left and 50% of the area under the distribution is tothe right. The mean represents the balancing point of the graph of the distribution(see Figure 2 on page 131 in Section 3.1). For symmetric distributions with a singlepeak, such as the normal distribution, the Because of this,the mean, is the high point of the graph of the distribution.

The points at and are the inflection points on the normalcurve. The inflection points are the points on the curve where the curvature of thegraph changes. To the left of and to the right of the curve

is drawn upward or . In between and the curve is

drawn downward .*Figure 5 shows how changes in and change the position or shape of a nor-

mal curve. In Figure 5(a), two normal density curves are drawn with the location ofthe inflection points labeled. One density curve has and the other has

We can see that increasing the mean from 0 to 3 caused the graph toshift three units to the right but maintained its shape. In Figure 5(b), two normaldensity curves are drawn, again with the inflection points labeled. One density curvehas and the other has We can see that increasing thestandard deviation from 1 to 2 causes the graph to become flatter and more spreadout but maintained its location of center.

m = 0, s = 2.m = 0, s = 1,

m = 3, s = 1.m = 0, s = 1,

smBA

x = m + s,x = m - sBAx = m + s,x = m - s

x = m + sx = m - sm,

mean = median = mode.

sm

Inflectionpoint

Inflectionpoint

m s m m s X

Figure 4

m 3, s 1m 0, s 1

1 0 1 2(a)

3 4 X

Figure 5

m 0, s 1

m 0, s 2

6 24 0 2 4 6(b)

X

Figure 3

Now Work Problem 25



m 3s m sm 2s m m s m 2s

68% within1 standarddeviation

of the mean

m 3s

95% within2 standard deviations

of the mean

99.7% within3 standard deviations of the mean

Normal Distribution

34%

13.5%2.35%2.35%

13.5%

34%

Figure 6

3 State the Properties of the Normal CurveThe normal probability density function satisfies all the requirements that arenecessary to have a probability distribution. We list the properties of the normaldensity curve next.

Properties of the Normal Density Curve1. It is symmetric about its mean,2. Because there is a single peak and the highestpoint occurs at 3. It has inflection points at and 4. The area under the curve is 1.5. The area under the curve to the right of equals the area under the curveto the left of which equals 6. As x increases without bound (gets larger and larger), the graph approaches,but never reaches, the horizontal axis. As x decreases without bound (getslarger and larger in the negative direction), the graph approaches, but neverreaches, the horizontal axis.7. The Empirical Rule: Approximately 68% of the area under the normalcurve is between and Approximately 95% of the areaunder the normal curve is between and Approxi-mately 99.7% of the area under the normal curve is between and

See Figure 6.x = m + 3s.x = m - 3s

x = m + 2s.x = m - 2sx = m + s.x = m - s

12 .m,

m

m + s.m - sx = m.

mean = median = mode,m.

Historical Note

Abraham de Moivre wasborn in France on May 26, 1667. He isknown as a great contributor to theareas of probability and trigonometry.In 1685, he moved to England. DeMoivre was elected a fellow of theRoyal Society in 1697. He was partof the commission to settle the dis-pute between Newton and Leibnizregarding who was the discoverer ofcalculus. He published The Doctrineof Chance in 1718. In 1733, hedeveloped the equation that describesthe normal curve. Unfortunately, deMoivre had a difficult time beingaccepted in English society (perhapsdue to his accent) and was able tomake only a meager living tutoringmathematics. An interesting piece ofinformation regarding de Moivre; hecorrectly predicted the day of hisdeath, November 27, 1754.

4 Explain the Role of Area in the NormalDensity FunctionLets look at an example of a normally distributed random variable.

EXAMPLE 3 A Normal Random Variable

Problem: The relative frequency distribution given in Table 1 represents theheights of a pediatricians 200 three-year-old female patients. The raw data indicatethat the mean height of the patients is inches with standard deviation

inches.(a) Draw a relative frequency histogram of the data. Comment on the shape of thedistribution.

s = 3.17m = 38.72



(b) Draw a normal curve with inches and inches on the relativefrequency histogram. Compare the area of the rectangle for heights between 40 and40.9 inches to the area under the normal curve for heights between 40 and 40.9 inches.

Approach(a) Draw the relative frequency histogram. If the histogram is shaped like Figure 3,we say that height is approximately normal. We say approximately normal, ratherthan normal, because the normal curve is an idealized description of the data,and data rarely follow the curve exactly.(b) Draw the normal curve on the histogram with the high point at and the in-flection points at and Shade the rectangle corresponding to heightsbetween 40 and 40.9 inches, and compare the area of the shaded region to the areaunder the normal curve between 40 and 40.9.

Solution(a) Figure 7 shows the relative frequency distribution. The relative frequencyhistogram is symmetric and bell-shaped.

m + s.m - sm

s = 3.17m = 38.72

0.140.16

0

0.100.12

0.080.060.040.02

Rel

ativ

e Fr

eque

ncy

Height (inches)4829 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47

Height of 3-Year-Old Females

Figure 7

(b) In Figure 8, the normal curve with and is superimposed onthe relative frequency histogram. The figure demonstrates that the normal curvedescribes the heights of 3-year-old females fairly well. We conclude that the heightsof 3-year-old females are approximately normal with and

Figure 8 also shows the rectangle corresponding to heights between 40 and40.9 inches.The area of this rectangle represents the proportion of 3-year-old femalesbetween 40 and 40.9 inches. Notice that the area of this shaded region is very close tothe area under the normal curve for the same region, so we can use the area underthe normal curve to approximate the proportion of 3-year-old females with heightsbetween 40 and 40.9 inches!

s = 3.17.m = 38.72

s = 3.17m = 38.72

Table 1

RelativeFrequency

Height (Inches)

29.029.9 0.005

30.030.9 0.005

31.031.9 0.005

32.032.9 0.025

33.033.9 0.02

34.034.9 0.055

35.035.9 0.075

36.036.9 0.09

37.037.9 0.115

38.038.9 0.15

39.039.9 0.12

40.040.9 0.11

41.041.9 0.07

42.042.9 0.06

43.043.9 0.035

44.044.9 0.025

45.045.9 0.025

46.046.9 0.005

47.047.9 0.005

Height of 3-Year-Old Females

29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Height (in inches)X

Figure 8

The normal curve drawn in Figure 8 is a model. In mathematics, a model is anequation, table, or graph that is used to describe reality. The normal distribution, ornormal curve, is a model that is used to describe variables that are approximatelynormally distributed. For example, the normal curve drawn in Figure 8 does a goodjob of describing the observed distribution of heights of 3-year-old females.

In Other WordsModels are not always mathematical. Forexample, a map can be thought of as amodel of a highway system. The modeldoes not show every detail of thehighway system (such as traffic lights),but it does serve the purpose ofdescribing how to get from point A topoint B. Mathematical models do thesame thing: They make assumptions tosimplify the mathematics, while stilltrying to accomplish the goal ofaccurately describing reality.



The equation (model) that is used to determine the probability of a continu-ous random variable is called a probability density function (or pdf). The normalprobability density function is given by

where is the mean and is the standard deviation of the normal random variable.This equation represents the normal distribution. Dont feel threatened by this equa-tion, because we will not be using it in this text. Instead, we will use the normal distri-bution in graphical form by drawing the normal curve, as we did in Figure 5.

We now summarize the role area plays in the normal curve.

sm

=1

s22p e

- 12

ax-msb2

y =1

s22p e

-1x-m222s2

Area under a Normal CurveSuppose that a random variableX is normally distributed with mean and stan-dard deviation The area under the normal curve for any interval of values ofthe random variable X represents either

the proportion of the population with the characteristic described by theinterval of values or

the probability that a randomly selected individual from the population willhave the characteristic described by the interval of values.

s.m

EXAMPLE 4 Interpreting the Area under a Normal Curve

Problem: The serum total cholesterol for males 20 to 29 years old is approximatelynormally distributed with mean and based on data obtainedfrom the National Health and Nutrition Examination Survey.(a) Draw a normal curve with the parameters labeled.(b) An individual with total cholesterol greater than 200 is considered to have highcholesterol. Shade the region under the normal curve to the right of (c) Suppose that the area under the normal curve to the right of is 0.2903.(You will learn how to find this area in Section 7.3.) Provide two interpretations ofthis result.

Approach(a) Draw the normal curve with the mean labeled at the high point and theinflection points at and (b) Shade the region under the normal curve to the right of (c) The two interpretations of this shaded region are (1) the proportion of 20- to29-year-old males who have high cholesterol and (2) the probability that a randomlyselected 20- to 29-year-old male has high cholesterol.

Solution(a) Figure 9(a) shows the graph of the normal curve.

x = 200.m + s = 180 + 36.2 = 216.2.m - s = 180 - 36.2 = 143.8

m = 180

x = 200x = 200.

s = 36.2,m = 180

Historical Note

The normal probabilitydistribution is often referred to as theGaussian distribution in honor of CarlGauss, the individual thought to havediscovered the idea. However, it wasactually Abraham de Moivre who first wrote down the equation of thenormal distribution. Gauss was bornin Brunswick, Germany, on April 30,1777. Mathematical prowess wasevident early in Gausss life. At age 8he was able to instantly add the first100 integers. In 1799, Gauss earned his doctorate. The subject of hisdissertation was the FundamentalTheorem of Algebra. In 1809, Gausspublished a book on the mathematicsof planetary orbits. In this book, hefurther developed the theory of least-squares regression by analyzing theerrors. The analysis of these errorsled to the discovery that errors followa normal distribution. Gauss wasconsidered to be glacially cold asa person and had troubled relation-ships with his family. Gauss died onFebruary 23, 1855.

X X(a) (b)180143.8 216.2 180 200

Figure 9



(b) Figure 9(b) shows the region under the normal curve to the right of shaded.

(c) The two interpretations for the area of this shaded region are (1) the propor-tion of 20- to 29-year-old males that have high cholesterol is 0.2903 and (2) theprobability that a randomly selected 20- to 29-year-old male has high cholesterolis 0.2903.

x = 200

5 Describe the Relation between a NormalRandom Variable and a Standard NormalRandom VariableAt this point, we know that a random variable X is approximately normally distributed if its relative frequency histogram has the shape of a normal curve.We use a normal random variable with mean and standard deviation tomodel the distribution of X. The area below the normal curve (model of X) rep-resents the proportion of the population with a given characteristic or the prob-ability that a randomly selected individual from the population will have a givencharacteristic.

The question now becomes, How do I find the area under the normal curve?Finding the area under a curve requires techniques introduced in calculus, which arebeyond the scope of this text. An alternative would be to use a series of tables tofind areas. However, this would result in an infinite number of tables being createdfor each possible mean and standard deviation!

A solution to the problem lies in the z-score. Recall that the z-score allows us totransform a random variable X with mean and standard deviation into a ran-dom variable Z with mean 0 and standard deviation 1.

sm

sm

Note to InstructorThe material in Objective 5 can be skippedif you only plan on using technology tofind areas under normal curves.

Standardizing a Normal Random VariableSuppose that the random variable X is normally distributed with mean andstandard deviation Then the random variable

is normally distributed with mean and standard deviation Therandom variable Z is said to have the standard normal distribution.

s = 1.m = 0

Z =X - ms

s.m

This result is powerful! We need only one table of areas corresponding to thestandard normal distribution. If a normal random variable has mean different from0 or standard deviation different from 1, we transform the normal random variableinto a standard normal random variable Z, and then we use a table to find the areaand, therefore, the probability.

We demonstrate the idea behind standardizing a normal random variable in thenext example.

In Other WordsTo find the area under any normal curve,we first find the z-score of the normalrandom variable. Then we use a table tofind the area.

EXAMPLE 5 Relation between a Normal Random Variable and a StandardNormal Random Variable

Problem: The heights of a pediatricians 200 three-year-old female patients areapproximately normal with mean inches and inches. We wish todemonstrate that the area under the normal curve between 35 and 38 inches is equal

s = 3.17m = 38.72

Now Work Problems 29 and 33



to the area under the standard normal curve between the z-scores corresponding toheights of 35 and 38 inches.

ApproachStep 1: Draw a normal curve and shade the area representing the proportion of 3-year-old females between 35 and 38 inches tall.Step 2: Standardize the values and using

Step 3: Draw the standard normal curve with the standardized values of andlabeled. Shade the area that represents the proportion of 3-year-old

females between 35 and 38 inches tall. Comment on the relation between the twoshaded regions.

SolutionStep 1: Figure 10(a) shows the normal curve with mean and The region between and is shaded.

Step 2: With and the standardized version of is

The standardized version of is

Step 3: Figure 10(b) shows the standard normal curve with the region betweenand shaded.z = -0.23z = -1.17

z =x - ms

=38 - 38.72

3.17= -0.23

x = 38

z =x - ms

=35 - 38.72

3.17= -1.17

x = 35s = 3.17,m = 38.72x = 38x = 35

s = 3.17.m = 38.72

x = 38x = 35

z =x - ms

x = 38x = 35

Recall that we round z-scoresto two decimal places.

m 38.72

38 38.723.17

x 35 0.23 1.17X Z

z

35 38.723.17

z

x 38

(a) (b)

32 33 34 36 37 39 40 41 42 43 44 4 3 2 1 0 1 2 3 445

These twoareas are

equal

Figure 10

The area under the normal curve with inches and inchesbounded to the left by and bounded to the right by is equal to thearea under the standard normal curve bounded to the left by and boundedto the right by z = -0.23.

z = -1.17x = 38x = 35

s = 3.17m = 38.72

Now Work Problem 35



7.1 ASSESS YOUR UNDERSTANDING

Concepts and Vocabulary1. State the two characteristics of the graph of a probability

density function.

2. To find the probabilities for continuous random variables, wedo not use probability distribution functions, but instead weuse probability density functions.

3. Provide two interpretations of the area under the graph of aprobability density function.

4. Why do we standardize normal random variables to find thearea under any normal curve?

5. The points at and are the inflection points on the normal curve.

6. As increases, the normal density curve becomes morespread out. Knowing that the area under the density curvemust be 1, what effect does increasing have on the heightof the curve?

For Problems 712, determine whether the graph can represent anormal density function. If it cannot, explain why.

7. No

s

s

x =x =

12. No

X

8. Yes

X

9. No

X

10. No

X

11. Yes

X

X

Skill Building

Problems 1316 use the information presented in Examples 1 and 2.

13. Find the probability that your friend is between 5 and 10 minutes late. 1/6

14. Find the probability that your friend is between 15 and 25 minutes late. 1/3

15. Find the probability that your friend is at least 20 minuteslate. 1/3

16. Find the probability that your friend is no more than 5 min-utes late. 1/6

17. Uniform Distribution The random-number generator oncalculators randomly generates a number between 0 and 1.The random variable X, the number generated, follows auniform probability distribution.

(a) Draw the graph of the uniform density function.(b) What is the probability of generating a number between

0 and 0.2? 0.2(c) What is the probability of generating a number between

0.25 and 0.6? 0.35(d) What is the probability of generating a number greater

than 0.95? 0.05(e) Use your calculator or statistical software to randomly

generate 200 numbers between 0 and 1. What propor-tion of the numbers are between 0 and 0.2? Comparethe result with part (b).

NW

m - s m + s



500 1000 1500 2000 2500 3000 3500 4000 4500 5000

Rel

ativ

e Fr

eque

ncy

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

0

Birth Weight (grams)

Birth Weights of BabiesWhose Term Was 36 Weeks

18. Uniform Distribution The reaction time X (in minutes) of acertain chemical process follows a uniform probability distri-bution with

(a) Draw the graph of the density curve.

(b) What is the probability that the reaction time is between6 and 8 minutes? 2/5

(c) What is the probability that the reaction time is between5 and 8 minutes? 3/5

(d) What is the probability that the reaction time is less than6 minutes? 1/5

In Problems 1922, determine whether or not the histogram indi-cates that a normal distribution could be used as a model for thevariable.

19. Birth Weights The following relative frequency histogramrepresents the birth weights (in grams) of babies whose termwas 36 weeks. Normal

5 X 10.

0 642 8 10 12 14

Rel

ativ

e F

requ

ency

0.40

0.30

0.20

0.10

0

Length (in minutes)

Length ofPhone Calls

20. Waiting in Line The following relative frequency histogramrepresents the waiting times in line (in minutes) for theDemon Roller Coaster for 2000 randomly selected peopleon a Saturday afternoon in the summer. Not normal

0.20

0.15

0.10

0.05

0.00

Rel

ativ

e F

requ

ency

Rhode Island Red HenIncubation Times

Incubation Time (in hours)

432 456 480 504 528 552 576

0 50 100

Rel

ativ

e Fr

eque

ncy

0.20

0.10

0

Waiting Time

Waiting Timefor the Demon Roller Coaster

21. Length of Phone Calls The following relative frequency his-togram represents the length of phone calls on my wifes cellphone during the month of September. Not normal

22. Incubation Times The following relative frequency his-togram represents the incubation times of a random sampleof Rhode Island Red hens eggs. Normal

A

B

10 X

23. One graph in the following figure represents a normal distri-bution with mean and standard deviation The other graph represents a normal distribution with mean

and standard deviation Determine whichgraph is which and explain how you know.A: m 10, s 2; B: m 10, s 3

s = 2.m = 10

s = 3.m = 10

24. One graph in the following figure represents a normal distri-bution with mean and standard deviation Theother graph represents a normal distribution with mean

and standard deviation Determine whichgraph is which and explain how you know.A: m 8, s 2; B: m 14, s 2

s = 2.m = 14

s = 2.m = 8



26. m 5, s 2

Applying the Concepts29. You Explain It! Cell Phone Rates Monthly charges for cell

phone plans in the United States are normally distributedwith mean and standard deviation Source: Based on information from Consumer Reports

s = $18.m = $62

BA

8 14 X

1 1197531 X

7 4 1 11852 X

27. m 100, s 15

14513011555 1008570 X

28. m 530, s 100

830730630230 530430330 X

NW

(a) Draw a normal curve with the parameters labeled.(b) Shade the region that represents the proportion of plans

that charge less than $44.(c) Suppose that the area under the normal curve to the left

of is 0.1587. Provide two interpretations of thisresult.

30. You Explain It! Refrigerators The lives of refrigerators arenormally distributed with mean years and standarddeviation years.Source: Based on information from Consumer Reports(a) Draw a normal curve with the parameters labeled.(b) Shade the region that represents the proportion of re-

frigerators that last for more than 17 years.(c) Suppose that the area under the normal curve to the

right of is 0.1151. Provide two interpretations ofthis result.

31. You Explain It! Birth Weights The birth weights of full-termbabies are normally distributed with mean and Source: Based on data obtained from the National Vital Sta-tistics Report, Vol. 48, No. 3(a) Draw a normal curve with the parameters labeled.(b) Shade the region that represents the proportion of full-

term babies who weigh more than 4,410 grams.(c) Suppose that the area under the normal curve to the

right of is 0.0228. Provide two interpretationsof this result.

32. You Explain It! Height of 10-Year-Old Males The heights of 10-year-old males are normally distributed with mean

and (a) Draw a normal curve with the parameters labeled.(b) Shade the region that represents the proportion of

10-year-old males who are less than 46.5 inches tall.(c) Suppose that the area under the normal curve to the left

of is 0.0496. Provide two interpretations of thisresult.

33. You Explain It! Gestation Period The lengths of humanpregnancies are normally distributed with and

(a) The following figure represents the normal curve withand The area to the right of

is 0.1908. Provide two interpretations of thisarea.x = 280

s = 16 days.m = 266 days

s = 16 days.m = 266 days

x = 46.5

s = 5.7 inches.m = 55.9 inches

x = 4,410

s = 505 grams.m = 3,400 grams

x = 17

s = 2.5m = 14

x = $44

m 266 280

Area 0.1908

X

In Problems 2528, the graph of a normal curve is given. Use thegraph to identify the value of and

25. m 2, s 3

s.m

NW

NW

(b) The following figure represents the normal curvewith and The area betweens = 16 days.m = 266 days



(b) The following figure represents the normal curve withper gallon and per gallon.

The area under the curve between and is0.1107. Provide two interpretations of this area.

x = 21x = 18s = 3.2 milesm = 24.6 miles

35. A random variable X is normally distributed with and

(a) Compute for

(b) Compute for 0.67

(c) The area under the normal curve between andis 0.495. What is the area between and

36. A random variable X is normally distributed with and

(a) Compute for 1.17x1 = 18.z1 =x1 - ms

s = 6.m = 25z2?z1x2 = 12

x1 = 8

x2 = 12.z2 =x2 - ms

0.67x1 = 8.z1 =x1 - ms

s = 3.m = 10

m 24.6 X2118

Area 0.1107

NW

100 97 101 101 103 100 99 100 100

104 100 101 98 100 99 99 97 101

104 99 101 101 101 100 96 99 99

98 94 98 107 98 100 98 103 100

98 94 104 104 98 101 99 97 103

102 101 101 100 95 104 99 102 95

99 102 103 97 101 102 96 102 99

96 108 103 100 95 101 103 105 100

94 99 95

44.5 42.4 42.2 46.2 45.7 44.8 43.3 39.5

45.4 43.0 43.4 44.7 38.6 41.6 50.2 46.9

39.6 44.7 36.5 42.7 40.6 47.5 48.4 37.5

45.5 43.3 41.2 40.5 44.4 42.6 42.0 40.3

42.0 42.2 38.5 43.6 40.6 45.0 40.7 36.3

44.5 37.6 42.2 40.3 48.5 41.6 41.7 38.9

39.5 43.6 41.3 38.8 41.9 40.3 42.1 41.9

42.3 44.6 40.5 37.4 44.5 40.7 38.2 42.6

44.0 35.9 43.7 48.1 38.7 46.0 43.4 44.6

37.7 34.6 42.4 42.7 47.0 42.8 39.9 42.3

35. (c) 0.495 36. (c) 0.6760

(a) Use MINITAB or some other statistical software to con-struct a relative frequency histogram. Comment on theshape of the distribution.

(b) Use MINITAB or some other statistical software todraw a normal density function on the relative frequencyhistogram.

(c) Do you think the normal density function accurately de-scribes the distance Michael hits with a pitching wedge?Why?

38. Heights of 5-Year-Old Females The following frequencydistribution represents the heights (in inches) of eighty ran-domly selected 5-year-old females.

(a) Use MINITAB or some other statistical software to con-struct a relative frequency histogram. Comment on theshape of the distribution.

(b) Use MINITAB or some other statistical software todraw a normal density curve on the relative frequencyhistogram.

(c) Do you think the normal density function accurately describes the heights of 5-year-old females? Why?

m 24.6 X26

Area 0.3309

34. You Explain It! Miles per Gallon Elena conducts an experi-ment in which she fills up the gas tank on her Toyota Camry40 times and records the miles per gallon for each fill-up. Ahistogram of the miles per gallon indicates that a normal dis-tribution with mean of 24.6 miles per gallon and a standarddeviation of 3.2 miles per gallon could be used to model thegas mileage for her car.(a) The following figure represents the normal curve with

per gallon and per gallon.The area under the curve to the right of is 0.3309.Provide two interpretations of this area.

x = 26s = 3.2 milesm = 24.6 miles

(b) Compute for 0.83

(c) The area under the normal curve between andis 0.6760. What is the area between and

37. Hitting a Pitching Wedge In the game of golf, distance con-trol is just as important as how far a player hits the ball.Michael went to the driving range with his range finder andhit 75 golf balls with his pitching wedge and measured thedistance each ball traveled (in yards). He obtained the following data:

z2?z1x2 = 30x1 = 18

x2 = 30.z2 =x2 - ms

m 266 X260230

Area 0.3416

and is 0.3416. Provide two interpreta-tions of this area.

x = 260x = 230


In Section 7.1, we introduced the normal distribution. We learned that, if X is a nor-mally distributed random variable, we can use the area under the normal densityfunction to obtain the proportion of a population with a certain characteristic, or theprobability that a randomly selected individual from the population has the charac-teristic. To find the area under the normal curve by hand, we first convert the ran-dom variable X to a standard normal random variable Z and find the area under thestandard normal curve. In this section we discuss methods for finding area under thestandard normal curve.

Properties of the Standard Normal DistributionThe standard normal distribution has a mean of 0 and a standard deviation of 1.Thestandard normal curve, therefore, has its high point located at 0 and inflection pointslocated at and We use Z to represent a standard normal random variable.The graph of the standard normal curve is presented in Figure 11.

Although we stated the properties of normal curves in Section 7.1, it is worth-while to restate them here in terms of the standard normal curve.

+1.-1

1 0 1 Z

Figure 11

Properties of the Standard Normal Curve1. It is symmetric about its mean, and has standard deviation 2. The Its highest point occurs at 3. It has inflection points at and 4. The area under the curve is 1.5. The area under the curve to the right of equals the area under thecurve to the left of which equals 6. As the value of Z increases, the graph approaches, but never equals, zero.As the value of Z decreases, the graph approaches, but never equals, zero.7. The Empirical Rule: Approximately of the area under the standardnormal curve is between and . Approximately of the areaunder the standard normal curve is between and . Approxi-mately of the area under the standard normal curve is betweenand . See Figure 12.z = 3

z = -399.7%z = 2z = -295%z = 1z = -1

68%

12 .m = 0,

m = 0

m + s = 0 + 1 = 1.m - s = 0 - 1 = -1z = 0.mean = median = mode = 0.

s = 1.m = 0,

Z123 0 1 2 3

0.0215

0.1359

0.3413

0.0215

0.1359

0.3413

Figure 12

We now discuss the procedure for finding area under the standard normal curve.

1 Find the Area under the Standard Normal CurveWe discuss two methods for finding area under the standard normal curve. The firstmethod uses a table of areas that has been constructed for various values of Z. Thesecond method involves the use of statistical software or a graphing calculator withadvanced statistical features.

Table V, which can be found on the inside back cover of the text and inAppendix A, gives areas under the standard normal curve for values to the left ofa specified z-score, as shown in Figure 13.z,0z Z

Figure 13

Objectives 1 Find the area under the standard normal curve2 Find z-scores for a given area3 Interpret the area under the standard normal curve as a probability

7.2 THE STANDARD NORMAL DISTRIBUTIONPreparing for This Section Before getting started, review the following: The Complement Rule (Section 5.2, pp. 279281)

Note to InstructorIf you like, you can print out and distributethe Preparing for This Section quiz locatedin the Instructors Resource Center. Thepurpose of the quiz is to verify that thestudents have the prerequisite knowledgefor the section.

Section 7.2 The Standard Normal Distribution 385


386 Chapter 7 The Normal Probability Distribution386 Chapter 7 The Normal Probability Distribution

EXAMPLE 1 Finding Area under the Standard Normal Curve to the Left of a z-Score

Problem: Find the area under the standard normal curve that lies to the left of

ApproachStep 1: Draw a standard normal curve with labeled, and shade the areaunder the curve to the left of

Step 2: The rows in Table V represent the ones and tenths digits of z, while thecolumns represent the hundredths digit. To find the area under the curve to the leftof we split 1.68 into 1.6 and 0.08. Find the row that represents 1.6 and thecolumn that represents 0.08 in Table V. Identify where the row and column intersect.This value is the area.

SolutionStep 1: Figure 14 shows the graph of the standard normal curve with labeled. The area to the left of is shaded.

Step 2: A portion of Table V is presented in Figure 15.We have enclosed the row thatrepresents 1.6 and the column that represents 0.08.The value located where the row andcolumn intersect is the area we are seeking.The area to the left of is 0.9535.z = 1.68

z = 1.68z = 1.68

z = 1.68,

z = 1.68.z = 1.68

z = 1.68.

The shaded region represents the area under the standard normal curve to the leftof When finding area under a normal curve, you should always sketch the normalcurve and shade the area in question.This practice will help to minimize errors.z.

Z1.680

0.9535

Figure 14

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706

z .01.00 .02 .03 .04 .05 .06 .07 .08 .09

Figure 15

The area under a standard normal curve may also be determined using statisti-cal software or a graphing calculator with advanced statistical features.

EXAMPLE 2 Finding the Area under a Standard Normal Curve Using Technology

Problem: Find the area under the standard normal curve to the left of usingstatistical software or a graphing calculator with advanced statistical features.

Approach: We will use MINITAB and a TI-84 Plus graphing calculator to find the area. The steps for determining the area under the standard normal curve using MINITAB, Excel, and the TI-83/84 Plus graphing calculators are given in theTechnology Step-by-Step on page 396.

Solution: Figure 16(a) shows the results from MINITAB, and Figure 16(b) showsthe results from a TI-84 Plus graphing calculator.

z = 1.68


Notice that the output for MINITAB is titled Cumulative Distribution Function.Remember, the word cumulative means less than or equal to, so MINITAB is givingthe area under the standard normal curve for values of Z less than or equal to 1.68.Thecommand required by the TI-84 Plus is normalcdf(.The cdf stands for cumulativedistribution function. The TI-graphing calculators require a left and right bound. Weuse 1E99 for the left bound when obtaining areas less than or equal to some value.-

Cumulative Distribution Function

(a)

Normal with mean 0 and standard deviation 1

x

1.68P( X x)

0.953521

Figure 16

(b)

Now Work Problem 5

Often, rather than being interested in the area under the standard normal curve tothe left of we are interested in obtaining the area to the right of The solution to thistype of problem uses the Complement Rule and the fact that the area under the entirestandard normal curve is 1.Therefore,

P1Z 7 z2 = 1 - P1Z z2 aArea under the normal curveto the right of z b = 1 - a

area to the leftof z b

z.z,

In Other WordsArea right = 1 - area left

EXAMPLE 3 Finding Area under the Standard Normal Curve to the Right of a z-Score

Problem: Find the area under the standard normal curve to the right of ApproachStep 1: Draw a standard normal curve with labeled, and shade the areaunder the curve to the right of Step 2: Find the row that represents and the column that represents 0.06 inTable V. Identify where the row and column intersect. This value is the area to theleft of Step 3: The area under the standard normal curve to the right of is1 minus the area to the left of

SolutionStep 1: Figure 17 shows the graph of the standard normal curve with labeled. The area to the right of is shaded.Step 2: A portion of Table V is presented in Figure 18. We have enclosed the rowthat represents and the column that represents 0.06. The value where the row and column intersect is the area to the left of The area to the left of

is 0.3228.z = -0.46z = -0.46.

-0.4

z = -0.46z = -0.46

z = -0.46.z = -0.46

z = -0.46.

-0.4z = -0.46.

z = -0.46

z = -0.46.

0.3228

0.6772

0.46 0 Z

Figure 17

3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.00023.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.00033.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.00053.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.00073.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010

0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776

0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.31210.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.34830.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.38590.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.42470.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09Figure 18




Step 3: The area under the standard normal curve to the right of is1 minus the area to the left of

The area to the right of is 0.6772.z = -0.46

= 0.6772 = 1 - 0.3228

Area right of -0.46 = 1 - 1area left of -0.462z = -0.46.

z = -0.46

Using TechnologyA TI-84 Plus can directly find thearea between two z-values. Forexample, the command normalcdfwill give the results of Example 4.MINITAB and Excel require theuser to first find the area to the left of2.01 and then subtract the area to theleft of , as we did in Example 4.-1.35

(-1.35, 2.01, 0, 1)

Now Work Problem 7

Using TechnologyStatistical software and graphingcalculators can also find areas underthe standard normal curve to theright of a z-score. A TI-84 Plus candirectly find the area to the right.MINITAB and Excel require theuser to find the area to the left anduse this result to find the area to theright (as we did in Example 3).

The next example presents a situation in which we are interested in the areabetween two z-scores.

EXAMPLE 4 Find the Area under the Standard Normal Curve between Two z-Scores

Problem: Find the area under the standard normal curve between and

ApproachStep 1: Draw a standard normal curve with and labeled. Shadethe area under the curve between and

Step 2: Find the area to the left of Find the area to the left of

Step 3: The area under the standard normal curve between andis the area to the left of minus the area to the left of

SolutionStep 1: Figure 19 shows the standard normal curve with the area between and shaded.

Step 2: Based upon Table V, the area to the left of is 0.0885. The area tothe left of is 0.9778.

Step 3: The area between and is

1Area between z = -1.35 and z = 2.012 = 1area left of z = 2.01) - 1area left of z = -1.352z = 2.01z = -1.35

z = 2.01z = -1.35

z = 2.01z = -1.35

z = -1.35.z = 2.01z = 2.01z = -1.35

z = 2.01.z = -1.35.

z = 2.01.z = -1.35z = 2.01z = -1.35

z = 2.01.z = -1.35

Z1.35 2.01

Figure 19

1.35 2.01 Z

2.01 Z 1.35 Z

0.8893 0.9978 0.0885

The area between and is 0.8893.z = 2.01z = -1.35

= 0.8893

= 0.9778 - 0.0885

Now Work Problem 9

We summarize the methods for obtaining area under the standard normal curvein Table 2.

Because the normal curve extends indefinitely in both directions on the Z-axis,there is no z-value for which the area under the curve to the left of the z-value is 1.For example, the area to the left of is less than 1, even though graphing cal-culators and statistical software state that the area is 1, because they can only com-pute a limited number of decimal places. We will follow the practice of stating thearea to the left of or to the right of as The area underthe standard normal curve to the left of or to the right of will bestated as 70.9999.

z = -3.90z = 3.9060.0001.z = 3.90z = -3.90

z = 10State the area under thestandard normal curve to the left of

as (not 0). Statethe area under the standard normalcurve to the left of as70.9999 1not 12.

z = 3.90

60.0001z = -3.90



Table 2 Finding Areas under the Standard Normal Curve

Problem Approach Solution

Find the area to the left of Shade the area to the left of Use Table V to find the row andcolumn that correspond to Thearea is the value where the rowand column intersect. Or, usetechnology to find the area.

Find the area to the right of Shade the area to the right of Use Table V to find the area leftof The area to the right of is1 minus the area to the left of Or,use technology to find the area.

Find the area between and Shade the area between and Use Table V to find the area tothe left of and to the left of

The area between and is

Or, usetechnology to find the area.1area to the left of z12.1area to the left of z22 -

z2z1z2 .z1

z2 .z1z2 .z1

z .zz .

z .z .

z .z .z .

z

z1 z2

2 Find z-Scores for a Given AreaUp to this point, we have found areas given the value of a z-score. Often, we areinterested in finding a z-score that corresponds to a given area. The procedure tofollow is the reverse of the procedure for finding areas given z-scores.

EXAMPLE 5 Finding a z-Score from a Specified Area to Its Left

Problem: Find the z-score so that the area to the left of the z-score is 0.32.

ApproachStep 1: Draw a standard normal curve with the area and corresponding unknownz-score labeled.

Step 2: Look for the area in the table closest to 0.32.

Step 3: Find the z-score that corresponds to the area closest to 0.32.

SolutionStep 1: Figure 20 shows the graph of the standard normal curve with the area of0.32 labeled. We know that must be less than 0. Do you know why?

Step 2: We refer to Table V and look in the body of the table for an area closestto 0.32. The area closest to 0.32 is 0.3192. Figure 21 on the following page shows aportion of Table V with 0.3192 labeled.

z

Note to InstructorIf you are using the table to find z-scores,be sure to emphasize that the exact areawe are looking for may not be in the table.Therefore, the closest area should beused. If the area is equidistant from twoareas, the mean z-score can be used.

z Z

Area 0.32

Figure 20

z



The z-score that corresponds to an area of 0.32 to its left is so z = -0.47.-0.47,

Step 3: From reading the table in Figure 21, we see that the approximate z-score that corresponds to an area of 0.32 to its left is So, SeeFigure 22.

Statistical software or graphing calculators with advanced statistical featurescan also be used to determine a z-score corresponding to a specified area.

z = -0.47.-0.47.Area 0.32

0.47 Z

Figure 22

3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.00023.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.00033.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.00053.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.00073.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010

0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.21480.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.24510.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776

0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.31210.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.34830.2 0.4027 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09Figure 21

EXAMPLE 6 Finding a z-Score from a Specified Area to Its Left Using Technology

Problem: Find the z-score such that the area to the left of the z-score is 0.32 usingstatistical software or a graphing calculator with advanced statistical features.

Approach: We will use Excel and a TI-84 Plus graphing calculator to find the z-score. The steps for determining the z-score for MINITAB, Excel, and the TI-83/84Plus graphing calculators are given in the Technology Step-by-Step on page 396.

Solution: Figure 23(a) shows the results from Excel, and Figure 23(b) shows theresults from a TI-84 Plus graphing calculator.

Here is thez-score

(a)

Figure 23

(b)

Now Work Problem 15

It is useful to remember that if the area to the left of the z-score is less than 0.5the z-score must be less than 0. If the area to the left of the z-score is greater than0.5, the z-score must be greater than 0.



The next example deals with situations in which the area to the right of someunknown z-score is given. The solution uses the fact that the total area under thenormal curve is 1.

EXAMPLE 7 Finding a z-Score from a Specified Area to Its Right

Problem: Find the z-score so that the area to the right of the z-score is 0.4332.ApproachStep 1: Draw a standard normal curve with the area and corresponding unknownz-score labeled.Step 2: Determine the area to the left of the unknown z-score.Step 3: Look for the area in the table closest to the area determined in Step 2 andrecord the z-score that corresponds to the closest area.

SolutionStep 1: Figure 24 shows the standard normal curve with the area and unknown z-score labeled.Step 2: Since the area under the entire normal curve is 1, the area to the left of theunknown z-score is 1 minus the area right of the unknown z-score. Therefore,

Step 3: We look in the body of Table V for an area closest to 0.5668. See Figure 25.The area closest to 0.5668 is 0.5675.

= 0.5668 = 1 - 0.4332

Area to the left = 1 - area to the right

To find a z-score, given the area to the right, you must firstdetermine the area to the left if youare using Table V.

z Z

Area 0.4332Area 0.5668

Figure 24

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

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

Figure 25

Z0.17

Area 0.4332Area 0.5668

Figure 26

The approximate z-score that corresponds to a right-tail area of 0.4332 is 0.17.Therefore, See Figure 26.z = 0.17.

Now Work Problem 19

In upcoming chapters, we will be interested in finding z-scores that separate themiddle area of the standard normal curve from the area in its tails.

EXAMPLE 8 Finding z-Scores from an Area in the Middle

Problem: Find the z-scores that separates the middle 90% of the area under thestandard normal curve from the area in the tails.

ApproachStep 1: Draw a standard normal curve with the middle of the area sep-arated from the area of in each of the two tails. Label the unknown z-scores and z2 .z1

5% = 0.0590% = 0.9



Step 2: Look in the body of Table V to find the area closest to 0.05.Step 3: Determine the z-score for the left tail.

Step 4: The area to the right of is 0.05. Therefore, the area to the left of is 0.95.Look in Table V for an area of 0.95 and find the corresponding z-score.

SolutionStep 1: Figure 27 shows the standard normal curve with the middle 90% of the areaseparated from the area in the two tails.

Step 2: We look in the body of Table V for an area closest to 0.05. See Figure 28.Notice that 0.0495 and 0.0505 are equally close to 0.05.

z2z2

Note to InstructorRemind students that the standard nor-mal curve is symmetric about its mean,0. Therefore, once the left z-score isknown, the right is found by symmetry.

Area 0.05

Area 0.9

Area 0.05

z1 z2 Z

Figure 27

3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.00023.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.00033.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005

1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.02941.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.03671.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.04551.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09Figure 28

Step 3: We agree to take the mean of the two z-scores corresponding to the areas.The z-score corresponding to an area of 0.0495 is The z-score corresponding toan area of 0.0505 is Therefore, the approximate z-score corresponding to anarea of 0.05 to its left is

Step 4: The area to the right of is 0.05. Therefore, the area to the left ofIn Table V, we find an area of 0.9495 corresponding to

and an area of 0.9505 corresponding to . Consequently, theapproximate z-score corresponding to an area of 0.05 to the right is

See Figure 29.

z2 =1.65 + 1.64

2= 1.645

z = 1.65z = 1.64z2 = 1 - 0.05 = 0.95.

z2

z =-1.65 + 1-1.642

2= -1.645

-1.64.-1.65.

Z1.645 1.645

Area 0.05 Area 0.05

Area 0.9

Figure 29

We could also obtain the solution to Example 8 using symmetry. Because thestandard normal curve is symmetric about its mean, 0, the z-score that correspondsto an area of 0.05 to the left will be the additive inverse (i.e., opposite) of the z-scorethat corresponds to an area of 0.05 to the right. Since the area to the left of

is 0.05, the area to the right of is also 0.05.z = 1.645z = -1.645

Now Work Problem 23

The notation (pronounced z sub alpha) is the z-score such that the areaunder the standard normal curve to the right of is Figure 30 illustrates thenotation.

a.zaza

za

Area a

Z

Figure 30

We are often interested in finding the z-score that has a specified area to theright. For this reason, we have special notation to represent this situation.



By Hand Solution: The area tothe right of the unknown z-value is0.10, so the area to the left of the z-value is .We lookin Table V for the area closest to0.90. The closest area is 0.8997,which corresponds to a z-value of1.28. Therefore, .z0.10 = 1.28

1 - 0.10 = 0.90

EXAMPLE 9 Finding the Value of

Problem: Find the value of Approach: We wish to find the z-value such that the area under the standardnormal curve to the right of the z-value is 0.10.

z0.10 .

zzA

Figure 31

0 z0.10 1.28

Area 0.10

Z

Figure 32

Figure 32 shows the z-value on the normal curve.

Now Work Problem 27

Technology Solution: The area to the right of the unknown z-value is 0.10, so thearea to the left is . A TI-84 Plus is used to find that the z-value suchthat the area to the left is 0.90 is 1.28. See Figure 31. Therefore, .z0.10 = 1.28

1 - 0.10 = 0.90

3 Interpret the Area under the StandardNormal Curve as a ProbabilityRecall that the area under a normal curve can be interpreted either as a probabilityor as the proportion of the population with a given characteristic (as represented byan interval of numbers). When interpreting the area under the standard normalcurve as a probability, we use the notation introduced in Chapter 6. For example, inExample 8, we found that the area under the standard normal curve to the left of

is 0.05; therefore, the probability of randomly selecting a standardnormal random variable, Z, that is less than is 0.05. We write this statementwith the notation

We will use the following notation to denote probabilities of a standard normalrandom variable, Z.

P1Z 6 -1.6452 = 0.05. -1.645z = -1.645

Notation for the Probability of a Standard Normal Random Variable

represents the probability that a standard normal randomvariable is between a and b.

represents the probability that a standard normal randomvariable is greater than a.

represents the probability that a standard normal randomvariable is less than a.

P1Z 6 a2P1Z 7 a2P1a Z b2

EXAMPLE 10 Finding Probabilities of Standard Normal Random Variables

Problem: Evaluate:

ApproachStep 1: Draw a standard normal curve with the area that we desire shaded.

P1Z 6 1.262




Note to InstructorBe sure to emphasize this concept withstudents.

Concepts and Vocabulary1. State the properties of the standard normal curve.

2. If the area under the standard normal curve to the left ofis 0.8849, what is the area under the standard

normal curve to the right of 0.1151

3. True or False: The area under the standard normal curve tothe left of is 1. Support your answer. False

4. Explain why

Skill BuildingIn Problems 512, find the indicated areas. For each problem, besure to draw a standard normal curve and shade the area that is tobe found.

5. Determine the area under the standard normal curve thatlies to the left of(a) 0.0071(b) 0.3336(c) 0.9115(d) 0.9998

6. Determine the area under the standard normal curve thatlies to the left of(a) 0.0002(b) 0.0233(c) 0.8212(d) 0.9981

7. Determine the area under the standard normal curve thatlies to the right of(a) 0.9987(b) 0.9441z = -1.59

z = -3.01

z = 2.90z = 0.92z = -1.99z = -3.49

z = 3.49z = 1.35z = -0.43z = -2.45

P1Z 6 -1.302 = P1Z -1.302.z = 5.30

z = 1.20?z = 1.20

(c) 0.0375(d) 0.0009

8. Determine the area under the standard normal curve thatlies to the right of(a) 0.9998(b) 0.7088(c) 0.0129(d) 0.0003

9. Determine the area under the standard normal curve thatlies between(a) 0.9586(b) 0.2088(c) 0.8479

10. Determine the area under the standard normal curve thatlies between(a) 0.9892(b) 0.4525(c) 0.9749

11. Determine the total area under the standard normal curve(a) to the left of or to the right of 0.0456(b) to the left of or to the right of (c) to the left of or to the right of

12. Determine the total area under the standard normal curve(a) to the left of or to the right of (b) to the left of or to the right of (c) to the left of or to the right of z = 1.23z = -0.88

z = 3.05z = -1.68z = 2.94z = -2.94

z = 1.20z = -0.24z = 2.56z = -1.56

z = 2z = -2

z = -3.03 and z = 1.98z = -1.67 and z = 0z = -2.55 and z = 2.55

z = -1.04 and z = 2.76z = -0.55 and z = 0z = -2.04 and z = 2.04

z = 3.45z = 2.23z = -0.55z = -3.49

z = 3.11z = 1.78

Step 2: Use Table V to find the area of the shaded region. This area represents theprobability.

SolutionStep 1: Figure 33 shows the standard normal curve with the area to the left of

shaded.

Step 2: Using Table V, we find that the area under the standard normal curve to theleft of is 0.8962. Therefore,P1Z 6 1.262 = 0.8962.z = 1.26

z = 1.26

0 1.26 Z

Figure 33

Now Work Problem 33

For any continuous random variable, the probability of observing a specificvalue of the random variable is 0. For example, for a standard normal random vari-able, for any value of a.This is because there is no area under the standardnormal curve associated with a single value, so the probability must be 0. Therefore,the following probabilities are equivalent:

For example,P1Z 6 1.262 = P1Z 1.262 = 0.8962.P1a 6 Z 6 b2 = P1a Z 6 b2 = P1a 6 Z b2 = P1a Z b2

P1a2 = 0

11. (b) 0.0646 11. (c) 0.5203 12. (a) 0.003212. (b) 0.0476 12. (c) 0.2987

NW

NW

NW


In Problems 13 and 14, find the area of the shaded region for eachstandard normal curve.

13. (a) 0.8877


(c) 0.9500

0 2.011.34 Z

(b) 0.0500

1.961.96 Z0

(c) 0.9802

0 2.332.33 Z

14. (a) 0.0198

(b) 0.3686

0 2.332.33 Z

01.12 Z

1.961.96 Z0

In Problems 1526, find the indicated z-score. Be sure to draw astandard normal curve that depicts the solution.

15. Find the z-score such that the area under the standard normalcurve to its left is 0.1.

16. Find the z-score such that the area under the standard normalcurve to its left is 0.2.

17. Find the z-score such that the area under the standard normalcurve to its left is 0.98. 2.05

18. Find the z-score such that the area under the standard normalcurve to its left is 0.85. 1.04

19. Find the z-score such that the area under the standard normalcurve to its right is 0.25. 0.67

20. Find the z-score such that the area under the standard normalcurve to its right is 0.35. 0.39

21. Find the z-score such that the area under the standard normalcurve to its right is 0.89.

22. Find the z-score such that the area under the standard normalcurve to its right is 0.75.

23. Find the z-scores that separate the middle 80% of the dis-tribution from the area in the tails of the standard normaldistribution.




In Problems 2732, find the value of za .

1.88; 1.88

2.575; 2.575

1.04; 1.04

1.28; 1.28

0.67

1.23

0.84

1.28

27. 1.645 28. 0.39z0.35z0.05

29. 2.33 30. 2.05z0.02z0.01

31. 0.84 32. 1.04z0.15z0.20

In Problems 3344, find the indicated probability of the standardnormal random variable Z.

33. 0.9732

34. 0.2709

35. 0.9986

36. 0.1788

37. 0.8753

38. 0.0499

39. 0.0329

40. 0.8212

41. 0.7642

42. 0.0036P1Z -2.692P1Z 0.722P1Z -0.922P1Z 1.842P11.23 6 Z 1.562P1-1.20 Z 6 2.342P1Z 7 0.922P1Z 7 -2.982P1Z 6 -0.612P1Z 6 1.932

NW

NW

NW

NW

NW



47. According to Table V, the area under the standard normalcurve to the left of is 0.0054. Without consultingTable V, determine the area under the standard normal curveto the right of 0.0054

48. According to Table V, the area under the standard normalcurve between and is 0.4332. Without con-sulting Table V, determine the area under the standard nor-mal curve between and 0.4332

49. According to Table V, the area under the standard normalcurve between and is 0.1906. Withoutconsulting Table V, determine the area under the standardnormal curve between and 0.1906

50. (a) If find a. 2.50(b) If find a. 0.15(c) If find b. 1.53(d) If , find a. 2.14P(0 Z a) = 0.4838

P1-b 6 Z 6 b2 = 0.8740,P1Z a2 = 0.4404,P1Z 6 a2 = 0.9938,

z = 1.24.z = 0.53

z = -0.53z = -1.24

z = 1.50.z = 0

z = 0z = -1.50

z = 2.55.

z = -2.5543. 0.0875

44. 0.3788

Applying the Concepts45. The Empirical Rule The Empirical Rule states that about

68% of the data in a bell-shaped distribution lies within 1 standard deviation of the mean. For the standard normaldistribution, this means about 68% of the data lies between

and Verify this result. Verify that about 95%of the data lies within 2 standard deviations of the mean. Fi-nally, verify that about 99.7% of the data lies within 3 stan-dard deviations of the mean.

46. According to Table V, the area under the standard normalcurve to the left of is 0.0901. Without consultingTable V, determine the area under the standard normalcurve to the right of 0.0901z = 1.34.

z = -1.34

z = 1.z = -1

P1Z 6 -0.38 or Z 7 1.932P1Z 6 -2.56 or Z 7 1.392

TI-83/84 PlusFinding Areas under the Standard Normal Curve1. From the HOME screen, press 2nd VARS toaccess the DISTRibution menu.2. Select 2:normalcdf(3. With normalcdf( on the HOME screen,type lowerbound, upperbound, 0, 1). Forexample, to find the area left of underthe standard normal curve, type

normalcdf(-1E99, 1.26, 0, 1)and hit ENTER.

Note: When there is no lowerbound, enter1E99. When there is no upperbound, enter

1E99. The E shown is scientific notation; it isselected by pressing 2nd then ,.

Finding Z-Scores Corresponding to an Area1. From the HOME screen, press 2nd VARS toaccess the DISTRibution menu.2. Select 3:invNorm(.3. With invNorm( on the HOME screen,type area left, 0, 1). For example, to find the z-score such that the area under the normalcurve to the left of the z-score is 0.79, type

invNorm(0.79, 0 , 1)and hit ENTER.

MINITABFinding Areas under the Standard Normal Curve1. MINITAB will find an area to the left ofa specified z-score. Select the Calc menu,highlight Probability Distributions, andhighlight Normal. . . .

-

z = 1.26

TECHNOLOGY STEP-BY-STEP The Standard Normal Distribution

2. Select Cumulative Probability. Set the meanto 0 and the standard deviation to 1. SelectInput Constant, and enter the specified z-score.Click OK.

Finding z-Scores Corresponding to an Area1. MINITAB will find the z-score for an areato the left of an unknown z-score. Select theCalc menu, highlight Probability Distributions,and highlight Normal. . . .2. Select Inverse Cumulative Probability. Setthe mean to 0 and the standard deviation to 1.Select Input Constant, and enter the specifiedarea. Click OK.

ExcelFinding Areas under the Standard Normal Curve1. Excel will find the area to the left of aspecified z-score. Select the fx button from thetool bar. In Function Category:, selectStatistical. In Function Name:, selectNORMSDIST. Click OK.2. Enter the specified z-score. Click OK.

Finding z-Scores Corresponding to an Area1. Excel will find the z-score for an area to theleft of an unknown z-score. Select the fx buttonfrom the tool bar. In Function Category:, selectStatistical. In Function Name:, selectNORMSINV. Click OK.2. Enter the specified area. Click OK.


Section 7.3 Applications of the Normal Distribution 397

Objectives 1 Find and interpret the area under a normal curve2 Find the value of a normal random variable

1 Find and Interpret the Area under a Normal CurveFrom the discussions in Section 7.1, we know that finding the area under a normalcurve by hand requires that we transform values of a normal random variable Xwith mean and standard deviation into values of the standard normal randomvariable Z with mean 0 and standard deviation 1. This is accomplished by letting

and using Table V to find the area under the standard normal curve.This

idea is illustrated in Figure 34.

Z =X - ms

sm

Now that we have the ability to find the area under a standard normal curve, wecan find the area under any normal curve. We summarize the procedure next.

7.3 APPLICATIONS OF THE NORMAL DISTRIBUTIONPreparing for This Section Before getting started, review the following: Percentiles (Section 3.4, p. 168)

Note to InstructorIf you like, you can print out and distrib-ute the Preparing for This Section quizlocated in the Instructors ResourceCenter. The purpose of the quiz is to verify that the students have the prereq-uisite knowledge for the section.

0Xa bm

These areasare equal

Za ms

za b ms

zb

Figure 34

Note to InstructorBe sure to emphasize to students thatthey should draw the nonstandard normalcurve and the corresponding standardnormal curve to reinforce their conceptualunderstanding for finding area under anynormal curve.

Finding the Area under Any Normal CurveStep 1: Draw a normal curve and shade the desired area.

Step 2: Convert the values of X to z-scores using

Step 3: Draw a standard normal curve and shade the desired area.Step 4: Find the area under the standard normal curve. This area is equal to thearea under the normal curve drawn in Step 1.

z =x - ms

.

EXAMPLE 1 Finding Area under a Normal Curve

Problem: A pediatrician obtains the heights of her 200 three-year-old femalepatients.The heights are approximately normally distributed, with mean 38.72 inchesand standard deviation 3.17 inches. Use the normal model to determine the propor-tion of the 3-year-old females that have a height less than 35 inches.

Approach: Follow Steps 1 through 4.



SolutionStep 1: Figure 35 shows the normal curve with the area to the left of 35 shaded.

Step 2: We convert to a z-score.

Step 3: Figure 36 shows the standard normal curve with the area to the left ofshaded.The area to the left of is equal to the area to the left of

x = 35.z = -1.17z = -1.17

z =x - ms

=35 - 38.72

3.17= -1.17

x = 35

Xm 38.72

X323130313029282726 33 34 35 36 37 3938 40 41 42 43 44 45 46 47 48 49 50 51

Figure 35

Z1.17

4 3 2 1 0 1 2 3 4

Figure 36

Step 4: Using Table V, we find that the area to the left of is 0.1210. Thenormal model indicates that the proportion of the pediatricians 3-year-old femalesthat is less than 35 inches tall is 0.1210.

z = -1.17

According to the results of Example 1, the proportion of 3-year-old femaleswho are shorter than 35 inches is 0.1210. If the normal curve is a good model for de-termining proportions (or probabilities), then about 12.1% of the 200 three-year-olds in Table 1 should be shorter than 35 inches. For convenience, the informationprovided in Table 1 is repeated in Table 3.

From the relative frequency distribution in Table 3, we determine thatof the 3-year-old

females are less than 35 inches tall. The results based on the normal curve are inclose agreement with the actual results. The normal curve accurately models theheights of 3-year-old females.

Because the area under the normal curve represents a proportion, we can alsouse the area to find percentile ranks of scores. Recall that the kth percentile dividesthe lower k% of a data set from the upper In Example 1, 12% ofthe females have a height less than 35 inches, and 88% of the females have aheight greater than 35 inches. Therefore, a child whose height is 35 inches is at the12th percentile.

Statistical software and graphing calculators with advanced statistical featurescan also be used to find areas under any normal curve.

1100 - k2%.

0.005 + 0.005 + 0.005 + 0.025 + 0.02 + 0.055 = 0.115 = 11.5%

Table 3

Height Relative(inches) Frequency

29.029.9 0.005

30.030.9 0.005

31.031.9 0.005

32.032.9 0.025

33.033.9 0.02

34.034.9 0.055

35.035.9 0.075

36.036.9 0.09

37.037.9 0.115

38.038.9 0.15

39.039.9 0.12

40.040.9 0.11

41.041.9 0.07

42.042.9 0.06

43.043.9 0.035

44.044.9 0.025

45.045.9 0.025

46.046.9 0.005

47.047.9 0.005

EXAMPLE 2 Finding Area under a Normal Curve Using Technology

Problem: Find the percentile rank of a 3-year-old female whose height is 43 inchesusing statistical software or a graphing calculator with advanced statistical features.From Example 1, we know that the heights are approximately normally distributed,with a mean of 38.72 inches and standard deviation of 3.17 inches.

Approach: We will use a TI-84 Plus graphing calculator to find the area. The stepsfor determining the area under any normal curve for MINITAB, Excel, and the



TI-83/84 Plus graphing calculators are given in the Technology Step-by-Step onpages 404405.

Solution: Figure 37 shows the results from a TI-84 Plus graphing calculator. Thearea under the normal curve to the left of 43 is approximately 0.91. Therefore, 91%of the heights are less than 43 inches, and 9% of the heights are more than 43 inches.A 3-year-old female whose height is 43 inches is at the 91st percentile.

Using TechnologyDue to rounding in the by-handapproach, technology answers willoften differ. For example, findingthe probability in Example 3 usingtechnology, we obtain

When answers between by-handcomputation and technology differ,we will report both results.

P(35 X 40) = 0.5365

In Other WordsThe normal probability density function is used to model random variables that appear to be normal (such as girlsheights). A good model is one that yieldsresults that are close to reality.

Figure 37

EXAMPLE 3 Finding the Probability of a Normal Random Variable

Problem: For the pediatrician presented in Example 1, use the normal distributionto compute the probability that a randomly selected 3-year-old female is between 35and 40 inches tall, inclusive. That is, find

Approach: We follow the Steps 1 through 4 on page 397.

Solution

Step 1: Figure 38 shows the normal curve with the area between andshaded.

Step 2: Convert the values and to z-scores.

Step 3: Figure 39 shows the graph of the standard normal curve with the areabetween and shaded.z2 = 0.40z1 = -1.17

z2 =x2 - ms

=40 - 38.72

3.17= 0.40

z1 =x1 - ms

=35 - 38.72

3.17= -1.17

x2 = 40x1 = 35

x2 = 40x1 = 35

P135 X 402.

m 38.72X32 33 34 35 36 37 3938 40 41 42 43 44 45

Figure 38

Z1.17

3 2 1 0 1 2 30.40

Figure 39

Step 4: Using Table V, we find that the area to the left of is 0.6554 and thearea to the left of is 0.1210. Therefore, the area between and

is We conclude that the probability that a ran-domly selected 3-year-old female is between 35 and 40 inches tall is 0.5344. That is,

If we randomly selected one3-year-old female 100 times, we would expect to select a child who is between 35 and40 inches tall about 53 times.

P135 X 402 = P1-1.17 Z 0.402 = 0.5344.0.6554 - 0.1210 = 0.5344.z2 = 0.40

z1 = -1.17z1 = -1.17z2 = 0.40

Now Work Problem 19

According to the relative frequency distribution in Table 3, the proportion ofthe 200 three-year-old females with heights between 35 and 40 inches is

This is very close to the proba-bility obtained in Example 3!0.075 + 0.09 + 0.115 + 0.15 + 0.12 = 0.55 = 55%.



2 Find the Value of a Normal Random VariableOften, rather than being interested in the proportion, probability or percentile fora value of a normal random variable, we are interested in calculating the value of anormal random variable required for the value to correspond to a certain propor-tion, probability, or percentile. For example, we might want to know the height of a3-year-old girl at the 20th percentile. This means that we want to know the height ofa 3-year-old girl who is taller than 20% of all 3-year-old girls.

Procedure for Finding the Value of a Normal Random VariableCorresponding to a Specified Proportion, Probability, or PercentileStep 1: Draw a normal curve and shade the area corresponding to the propor-tion, probability, or percentile.Step 2: Use Table V to find the z-score that corresponds to the shaded area.Step 3: Obtain the normal value from the formula *x = m + zs.

EXAMPLE 4 Finding the Value of a Normal Random Variable

Problem: The heights of a pediatricians 200 three-year-old females are approxi-mately normally distributed, with mean 38.72 inches and standard deviation3.17 inches. Find the height of a 3-year-old female at the 20th percentile. That is, findthe height of a 3-year-old female that separates the bottom 20% from the top 80%.

Approach: We follow Steps 1 through 3.

Solution

Step 1: Figure 40 shows the normal curve with the unknown value of X separatingthe bottom 20% of the distribution from the top 80% of the distribution.

Step 2: From Table V, the area closest to 0.20 is 0.2005. The corresponding z-scoreis

Step 3: The height of a 3-year-old female that separates the bottom 20% of the datafrom the top 80% is computed as follows:

The height of a 3-year-old female at the 20th percentile is 36.1 inches.

= 36.1 inches

= 38.72 + 1-0.84213.172 x = m + zs

-0.84.

Xx ?

Area 0.20 Area 0.80

Figure 40

EXAMPLE 5 Finding the Value of a Normal Random VariableUsing Technology

Problem: Use statistical software or a graphing calculator with advanced statis-tical features to verify the results of Example 4. That is, find the height for a 3-year-old female that is at the 20th percentile, assuming their heights are approximatelynormally distributed, with a mean of 38.72 inches and a standard deviation of3.17 inches.

*The formula provided in Step 3 is simply the formula for computing a z-score, solved for x.

Formula for standardizing a value, x, for a random variable XMultiply both sides by Add to both sides.mx = m + zs

s.zs = x - m

z =x - ms



Approach: We will use MINITAB to find the height at the 20th percentile.The stepsfor determining the value of a normal random variable given an area for MINITAB,Excel, and the TI-83/84 Plus graphing calculators are given in the Technology Step-by-Step on pages 404405.

Solution: Figure 41 shows the results obtained from MINITAB. The height of a3-year-old female at the 20th percentile is 36.1 inches.

Now Work Problem 25(a)

Inverse Cumulative Distribution FunctionNormal with mean 38.7200 and standard deviation 3.17000

P ( x x) 0.2000

x

36.0521

Figure 41

EXAMPLE 6 Finding the Value of a Normal Random Variable

Problem: The heights of a pediatricians 200 three-year-old females are approxi-mately normally distributed, with mean 38.72 inches and standard deviation3.17 inches. The pediatrician wishes to determine the heights that separate themiddle 98% of the distribution from the bottom 1% and top 1%. In other words,find the 1st and 99th percentiles.

Approach: We follow Steps 1 through 3 given on page 400.SolutionStep 1: Figure 42 shows the normal curve with the unknown values of x separatingthe bottom and top 1% of the distribution from the middle 98% of the distribution.

Step 2: First, we will find the z-score that corresponds to an area of 0.01 to the left.From Table V, the area closest to 0.01 is 0.0099. The corresponding z-score is The z-score that corresponds to an area of 0.01 to the right is the z-score that hasarea 0.99 to the left. The area closest to 0.99 is 0.9901. The corresponding z-scoreis 2.33.

Step 3: The height of a 3-year-old female that separates the bottom 1% of thedistribution from the top 99% is

The height of a 3-year-old female that separates the top 1% of the distributionfrom the bottom 99% is

A 3-year-old female whose height is less than 31.3 inches is in the bottom 1% ofall 3-year-old females, and a 3-year-old female whose height is more than 46.1 inchesis in the top 1% of all 3-year-old females. The pediatrician might use this informa-tion to identify those patients who have unusual heights.

= 46.1 inches

= 38.72 + 12.33213.172 x2 = m + zs

= 31.3 inches

= 38.72 + 1-2.33213.172 x1 = m + zs

-2.33.

If you are given a value of therandom variable and asked to find theprobability, proportion, or percentilecorresponding to the value, convert the

value to a z-score using

and find the area from the table. Ifyou are using a TI-83/84 Plus graphingcalculator, use normalcdf(.

If you are asked to find thevalue of a random variablecorresponding to a probability,proportion, or percentile, find the areathat represents the given probability,proportion, or percentile and use

to find the value of therandom variable. If you are using a TI-83/84 Plus graphing calculator,use invNorm(.

x = m + zs

z =x - ms

Now Work Problem 25(b)

Area 0.01 Area 0.01

x1 ? x2 ? X

Figure 42




Concepts and Vocabulary1. Describe the procedure for finding the area under any

normal curve.

2. Describe the procedure for finding the value of a normalrandom variable corresponding to a probability, proportion,or percentile.

Skill

Ch7_NormalDistribution

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