Section 5.6 The Central Limit Theorem 205

n = 2 n = 7

0.1

0.2

0.3

0.4

0.1

0.2

0.3

0.4

10−1−2−3 2 3 10−1−2−3 2 3

N(0, 1) pdf N(0, 1) pdf

Figure 5.6-3 pdfs of (X − μ)/(σ/√n ), underlying distribution triangular

N(0, 1) pdf

0.2

0.4

0.6

0.8

0.1

0.2

0.3

0.4

−1−2−3 3210 −1−2−3 3210

n = 2

n = 7

N(0, 1) pdf

Figure 5.6-4 pdfs of (X − μ)/(σ/√n ), underlying distribution U-shaped

So far, all the illustrations have concerned distributions of the continuous type.However, the hypotheses for the central limit theorem do not require the distribu-tion to be continuous. We shall consider applications of the central limit theorem fordiscrete-type distributions in the next section.

Exercises

5.6-1. Let X be the mean of a random sample of size12 from the uniform distribution on the interval (0, 1).Approximate P(1/2 ≤ X ≤ 2/3).

5.6-2. Let Y = X1 + X2 + · · · + X15 be the sum of a ran-dom sample of size 15 from the distribution whose pdf isf (x) = (3/2)x2, −1 < x < 1. Using the pdf of Y, we findthat P(−0.3 ≤ Y ≤ 1.5) = 0.22788. Use the central limittheorem to approximate this probability.

5.6-3. Let X be the mean of a random sample of size36 from an exponential distribution with mean 3.Approximate P(2.5 ≤ X ≤ 4).

5.6-4. Approximate P(39.75 ≤ X ≤ 41.25), where X isthe mean of a random sample of size 32 from a distribu-tion with mean μ = 40 and variance σ 2 = 8.

5.6-5. Let X1,X2, . . . ,X18 be a random sample of size 18from a chi-square distribution with r = 1. Recall thatμ = 1 and σ 2 = 2.

(a) How is Y =∑18i=1 Xi distributed?

(b) Using the result of part (a), we see from Table IV inAppendix B that

P(Y ≤ 9.390) = 0.05 and P(Y ≤ 34.80) = 0.99.

Compare these two probabilities with the approxima-tions found with the use of the central limit theorem.

5.6-6. A random sample of size n = 18 is taken from thedistribution with pdf f (x) = 1− x/2, 0 ≤ x ≤ 2.

(a) Find μ and σ 2. (b) Find P(2/3 ≤ X ≤ 5/6),approximately.

206 Chapter 5 Distributions of Functions of Random Variables

5.6-7. LetX equal the maximal oxygen intake of a humanon a treadmill, where the measurements are in millilitersof oxygen per minute per kilogram of weight. Assumethat, for a particular population, the mean of X is μ =54.030 and the standard deviation is σ = 5.8. Let X bethe sample mean of a random sample of size n = 47. FindP(52.761 ≤ X ≤ 54.453), approximately.

5.6-8. Let X equal the weight in grams of a minia-ture candy bar. Assume that μ = E(X) = 24.43 andσ 2 = Var(X) = 2.20. Let X be the sample mean of arandom sample of n = 30 candy bars. Find

(a) E(X). (b) Var(X). (c) P(24.17 ≤ X ≤ 24.82),approximately.

5.6-9. In Example 5.6-4, compute P(1.7 ≤ Y ≤ 3.2)with n = 4 and compare your answer with the normalapproximation of this probability.

5.6-10. Let X and Y equal the respective numbers ofhours a randomly selected child watches movies or car-toons on TV during a certain month. From experience,it is known that E(X) = 30, E(Y) = 50, Var(X) = 52,Var(Y) = 64, and Cov(X,Y) = 14. Twenty-five childrenare selected at random. Let Z equal the total number ofhours these 25 children watch TV movies or cartoons inthe next month. Approximate P(1970 < Z < 2090). Hint:Use the remark after Theorem 5.3-2.

5.6-11. A company has a one-year group life policy thatdivides its employees into two classes as follows:

Class Probability of Death Benefit Number in Class

A 0.01 $20,000 1000

B 0.03 $10,000 500

The insurance company wants to collect a premium thatequals the 90th percentile of the distribution of the totalclaims. What should that premium be?

5.6-12. At certain times during the year, a bus companyruns a special van holding 10 passengers from Iowa Cityto Chicago. After the opening of sales of the tickets, thetime (in minutes) between sales of tickets for the trip hasa gamma distribution with α = 3 and θ = 2.

(a) Assuming independence, record an integral that givesthe probability of being sold out within one hour.

(b) Approximate the answer in part (a).

5.6-13. The tensile strength X of paper, in pounds persquare inch, has μ = 30 and σ = 3. A random sampleof size n = 100 is taken from the distribution of tensilestrengths. Compute the probability that the sample meanX is greater than 29.5 pounds per square inch.

5.6-14. Suppose that the sick leave taken by the typicalworker per year has μ = 10, σ = 2, measured in days.A firm has n = 20 employees. Assuming independence,how many sick days should the firm budget if the finan-cial officer wants the probability of exceeding the numberof days budgeted to be less than 20%?

5.6-15. Let X1,X2,X3,X4 represent the random times indays needed to complete four steps of a project. Thesetimes are independent and have gamma distributions withcommon θ = 2 and α1 = 3,α2 = 2,α3 = 5,α4 = 3, respec-tively. One step must be completed before the next canbe started. Let Y equal the total time needed to completethe project.

(a) Find an integral that represents P(Y ≤ 25).

(b) Using a normal distribution, approximate the answerto part (a). Is this approach justified?

5.7 APPROXIMATIONS FOR DISCRETE DISTRIBUTIONSIn this section, we illustrate how the normal distribution can be used to approximateprobabilities for certain discrete-type distributions. One of the more important dis-crete distributions is the binomial distribution. To see how the central limit theoremcan be applied, recall that a binomial random variable can be described as the sumof Bernoulli random variables. That is, letX1,X2, . . . ,Xn be a random sample from aBernoulli distribution with meanμ = p and variance σ 2 = p(1−p), where 0 < p < 1.Then Y =∑n

i=1Xi is b(n,p). The central limit theorem states that the distribution of

W = Y − np√np(1− p) =

X − p√p(1− p)/n

is N(0, 1) in the limit as n→ ∞. Thus, if n is “sufficiently large,” the distribution ofY is approximately N[np,np(1− p)], and probabilities for the binomial distribution