Probability and statistics for engineers and scientists for engineers 9th edition johnson solutions

Probability and Statistics for Engineers and Scientists for Engineers 9th Edition Johnson Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/probability-and-statistics-for-engineers-and-scientists-for-engineers-9th-edition-j ohnson-solutions-manual/

Chapter 6

SAMPLING DISTRIBUTIONS 6.1 Only if the pieces are put on the assembly line such that every 20th piece is random with respect to the characteristic being measured. If twenty molds are dumping their contents, in sequential order, onto the assembly line then the sample would consist of output from a single mold. It would not be random.

6.2 The end is the last wrapping material made. If there is a drift in clarity long the whole 100 feet tending towards less clear, or towards more clear, the clarity of the end material would not be representative.

6.3 (a) A typical member of the population does not vacation on a luxury cruise. The sample taken would be biased. (b) This sample will very likely be biased. Those will high incomes will tend to respond while those with low incomes will tend to not respond. (c) Everyone feels that unfair things should be stopped. The way the question is phrased biases the responses.

6.4 Number the states from 1 to 50 according to their alphabetical order. Using the first two columns of Table 7 to select 8 numbers between 1 and 50 by discarding numbers outside of this range as well as any previously drawn number gives 13, 4, 29, 14, 36, 48, 22, 35.

These correspond to Illinois, Arkansas, New Hampshire, Indiana, Oklahoma, West Virginia, Michigan, and Ohio.

131 Copyright © 2017 Pearson Education, Inc.

CHAPTER 6 SAMPLING DISTRIBUTIONS

132

6.5 (a) The number of samples ( given that order does not matter) is ⎛ ⎝

⎞ ⎠ = 9 · 8 = 36.

2·1 2 9

(b) The number of samples (given that order does not matter) is ⎛ ⎝

⎞ ⎠ = 23 · 22 = 253.

2·1 2

23

6.6 For a random sample, each outcome must have equal probability. Thus, the probabilities are: 1 = .02278. (a) 36 1 = .00395. (b) 253 6.7 (a) The probability of each of the numbers is given in the table: Number

−4

−3

−2

−1

0

1

2

3

4

Probability

1/6

2/15

1/10

1/15

1/15

1/15

1/10

2/15

1/6

The mean of the distribution is 1 2 2 1 (−4) + (−3) + · · · + (3) + (4) 6 15 15 6 1 2 1 1 1 = (4 − 4) + (3 − 3) + (2 − 2) + (1 − 1) + (0) = 0.

6 15 10 15 15 The variance is 4 2 2 1 2 (16) + (9) + (4) + (1) + (0) = 8.667.

6 15 10 15 15 (b) The samples and mean are given in Table 6.1 Table 6.1. 50 Samples of Size 10 Taken Without Replacement

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

mean

1

3

-1

4

4

4

-4

-4

-4

-2

0.1

-2

3

3

-2

2

-4

4

-4

-3

-4

-0.7

4

4

-2

-3

-3

2

1

3

4

0

1.0

3

3

-3

4

2

1

-2

0

4

0

1.2

4

0

-1

-2

3

1

2

4

-1

1

1.1

1

-2

-3

-4

2

4

-2

3

4

-4

-0.1

-1

2

1

-1

-4

-3

0

-4

3

1

-0.6

-4

-2

2

1

-4

-3

3

2

-2

1

-0.6

1

-3

2

-4

2

4

3

-4

-3

-2

-0.4

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CHAPTER 6 SAMPLING DISTRIBUTIONS

133

Table 6.1. (continued)

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

mean

1

4

4

4

-4

3

-1

-2

2

-4

0.7

4

0

-1

1

-3

1

4

-3

4

2

0.9

-4

4

2

-4

2

-2

-2

-1

4

3

0.2

2

-1

-2

4

-4

0

-4

1

3

-2

-0.3

-4

-1

-3

4

0

-4

-3

1

2

4

-0.4

-2

4

4

-1

-4

1

4

1

-4

-3

0.0

-3

0

-3

-2

0

-4

4

-1

1

-2

-1.0

4

-4

3

4

-4

-1

3

2

0

1

0.8

3

3

-3

4

-3

1

3

-3

-4

-1

0.0

3

-4

-4

4

4

3

4

-3

2

-2

0.7

4

0

2

-3

-3

-3

-2

3

-4

-4

-1.0

2

-1

4

-3

2

3

4

4

-4

4

1.5

3

2

3

4

4

4

3

1

2

-2

2.4

4

-3

0

-1

-3

-3

0

-4

4

-4

-1.0

-3

3

-4

3

0

1

-4

-4

-3

-4

-1.5

-2

-4

-2

-3

4

4

3

-3

1

-3

-0.5

3

-1

-2

-3

1

-4

-2

3

4

2

0.1

4

2

-4

3

-2

2

-2

1

4

3

1.1

0

-2

4

-2

0

-1

2

3

-3

-4

-0.3

1

-4

4

4

-2

-3

-4

-4

0

3

-0.5

2

4

3

-2

3

1

0

0

-4

-3

0.4

4

0

2

-1

1

-4

-4

-4

-3

4

-0.5

4

-3

-1

-3

-3

0

4

4

3

1

0.6

-4

-1

0

-3

4

-3

2

1

3

4

0.3

0

4

-4

3

-2

2

0

4

2

-4

0.5

3

2

-2

4

-4

4

3

4

0

-3

1.1

-4

2

0

-2

-3

2

-1

-2

-4

-2

-1.4

-3

-1

-2

4

2

4

0

1

2

2

0.9

2

-4

-2

-3

-1

3

1

-2

-4

-3

-1.3

-4

3

1

0

-1

0

-3

-3

2

4

-0.1

Copyright © 2017 Pearson Education, Inc.

CHAPTER 6 SAMPLING DISTRIBUTIONS

134 Table 6.1. (continued)

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

mean

2

2

4

-4

-4

4

4

-3

-3

-2

0.0

-4

1

4

4

3

2

-2

-3

4

-3

0.6

0

-4

-4

0

-4

1

4

-3

1

4

-0.5

-4

4

-3

1

4

-1

-4

-3

-4

-2

-1.2

4

-3

-3

4

4

-3

1

2

3

-1

0.8

-2

3

2

-4

-1

3

-4

-1

-3

1

-0.6

4

2

4

-3

-4

-4

-1

-3

-4

-2

-1.1

-3

-1

1

2

-3

0

-4

-2

4

2

-0.4

0

-1

4

-4

2

-2

3

-3

-2

3

0.0

-4

2

1

-2

-2

-1

-3

-4

3

-3

-1.3

-4

4

4

4

-3

2

3

3

3

4

2.0

(c) The mean of the 50 sample means is .034. The sample variance of the 50 sample means is .8096. (d) According to Theorem 6.1, the distribution of the means has mean 0 and variance σ2 N − n 8.667 20 · = · = .5977. n N −1 10 29 The sample values in part (c) compare well with these theoretical values.

6.8 (a) The mean and variance of the distribution are the same as in part (a) of Exercise 6.7. (b) The 50 samples, taken with replacement, are shown in Table 6.2

Table 6.2. 50 Samples of Size 10 Taken with Replacement.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

mean

-3

-4

2

1

0

-3

-1

3

4

-4

-0.5

0

1

-4

-4

0

2

4

3

2

-4

0.0

-3

-1

-3

2

-4

-3

0

4

4

-4

-0.8

4

-3

2

3

4

4

-2

2

-2

-1

1.1

2

3

-1

-3

4

-3

4

-2

-3

3

0.4

4

-2

-4

3

-4

3

2

1

-4

-2

-0.3

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CHAPTER 6 SAMPLING DISTRIBUTIONS

135

Table 6.2 ( continued).

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

mean

0

3

4

4

0

4

0

2

-2

2

1.7

4

-2

-3

-2

3

-4

-3

2

4

0

-0.1

-4

1

-3

4

-3

4

-3

3

4

-2

0.1

-1

-4

-3

4

-4

0

-4

2

4

4

-0.2

4

-3

-4

-2

4

-2

-4

-1

-4

2

-1.0

4

4

3

4

4

-4

-2

-1

-1

-4

0.7

4

3

3

2

3

-2

-4

4

-4

-4

0.5

4

-3

2

4

4

1

-2

3

-1

1

1.3

-1

-2

4

-4

4

-3

-2

-2

-3

-4

-1.3

-3

3

0

-4

4

-4

3

-2

3

4

0.4

-1

-4

0

3

-4

0

3

-3

4

-3

-0.5

3

4

2

-1

-3

0

-3

2

3

-3

0.4

-4

3

-4

-3

-4

-4

-2

-3

1

-4

-2.4

-4

-2

3

-3

-3

4

3

4

-3

-3

-0.4

1

4

1

-4

-2

0

3

3

3

2

1.1

-2

4

2

4

0

2

-4

-1

4

3

1.2

-4

-1

-1

-3

-3

-2

-2

-3

2

-2

-1.9

4

3

3

-1

0

-4

-2

4

0

-3

0.4

-3

1

0

-4

3

3

2

3

-3

-1

0.1

-2

0

2

-2

4

-4

4

-4

2

-4

-0.4

-2

-4

-3

0

-3

4

-3

4

-3

3

-0.7

-4

-1

2

4

-3

-3

2

1

4

-4

-0.2

2

-2

-4

-4

-4

0

-1

2

-4

-4

-1.9

-3

-2

0

-4

-1

1

-4

1

0

3

-0.9

-2

-2

1

3

4

-2

4

2

-4

4

0.8

3

1

-3

4

2

4

-2

-2

-4

-4

-0.1

2

-2

-4

2

1

2

4

-2

2

1

0.6

4

-2

-3

3

4

-2

1

3

4

-1

1.1

-4

4

4

-4

-4

3

-3

4

3

-3

0.0

-2

0

3

-4

1

2

2

1

-2

-1

0.0

-2

4

4

4

3

-2

1

-2

-2

3

1.1

-4

-2

1

3

-4

-2

-4

-1

4

0

-0.9

-3

0

3

4

2

-3

1

-2

4

1

0.7

Copyright © 2017 Pearson Education, Inc.

CHAPTER 6 SAMPLING DISTRIBUTIONS

136 Table 6.2.(continued)

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

obs.

mean

-4

0

-1

4

3

-1

-2

4

3

2

0.8

-4

4

4

0

-3

0

3

4

3

3

1.4

-4

-3

-3

3

2

1

-4

1

-2

-2

-1.1

3

-4

-2

0

3

0

-3

-3

-3

3

-0.6

-3

3

-2

4

-1

2

-1

1

2

4

0.9

-2

-3

-2

4

-1

4

-3

3

-3

4

-0.7

-3

3

4

3

-2

-1

-1

0

3

-4

0.2

-4

-3

-4

4

1

-4

0

-3

4

-4

-1.3

-3

-4

-4

0

-4

4

0

2

4

-4

-0.9

-3

-3

-2

-4

4

4

4

-2

-4

-3

-0.9

3

3

-2

1

-3

0

2

3

-1

0

0.6

(c) The mean of the means is −.048. The sample variance of the means is .8621. (d) According to Theorem 6.1, the distribution of the means has mean 0 and variance σ 2 /n = 8.667/10 = .8667.

The sample values in part (c) compare well with these theoretical values.

6.9 A table of each outcome and the mean follows:

Outcome

Mean

Outcome

Mean

1, 1

1.0

3, 1

2.0

1, 2

1.5

3, 2

2.5

1, 3

2.0

3, 3

3.0

1, 4

2.5

3, 4

3.5

2, 1

1.5

4, 1

2.5

2, 2

2.0

4, 2

3.0

2, 3

2.5

4, 3

3.5

2, 4

3.0

4, 4

4.0

Thus, the distribution of these values is:

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CHAPTER 6 SAMPLING DISTRIBUTIONS

137

Value

No. of Ways Obtained

Probability

1.0

1

.0625

1.5

2

.1250

2.0

3

.1875

2.5

4

.2500

3.0

3

.1875

3.5

2

.1250

4.0

1

.0625

where the probability is ( no. ways )×(.25)2 . Consequently, the distribution of X̄ has mean μX̄ = 1(.0625) + 1.5(.1250) + · · · + 3.5(.1250) + 4(.0625) = 2.5 which may be obtained directly from the symmetry of the distribution. The distribution of X̄ has variance 2 σX̄

=

(1 − 2.5)2 (.0625) + (1.5 − 2.5)2 (.1250) + · · · + (3.5 − 2.5)2 (.1250) + (4 − 2.5)2 (.0625) = .625

Now the mean of the original distribution is also 2.5 and the variance is 1.25. Thus, Theorem 6.1 yields 2.5 and 1.25/2 = .625 as the mean and the variance of the distribution of the sample mean of two observations. These agree exactly as they must.

6.10 The 25 means are 3.80, 4.25, 4.25, 5.05, 4.85, 4.20, 4.05, 4.15, 4.85, 4.20, 3.00, 5.20, 4.30, 4.45, 3.75, 5.40, 5.60, 5.70, 4.00, 5.10, 3.15, 4.50, 3.40, 5.00, 4.50 The sample mean of these 25 means is 4.428.

The sample standard deviation of these 25 means is .714.

The population mean is: 0

1 1 1 + 1 + ··· + 9 = 4.5.

10 10 10

The population variance is: 02

1 1 1 + 12 + · · · + 92 − (4.5)2 = 8.25, 10 10 10

so the population standard deviation is 2.87228. Theorem 6.1 says that the mean of the distribution

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CHAPTER 6 SAMPLING DISTRIBUTIONS

138

of X̄ is 4.5 and the variance is 8.25/20 = .4125 Thus the theoretical standard deviation is .642. These compare well to the sample results.

6.11 The variance of the sample mean X̄, based on a sample of size n, is σ 2 /n. Thus the standard deviation, √ or standard error of the mean is σ/ n. √ (a) The standard deviation for a sample of size 50 is σ/ 50. The standard deviation for a sample is √ size 200 is σ/ 200. That is, the ratio of standard errors is √ √ σ/ 200 50 1 √ =√ = , 2 σ/ 50 200 so the standard error is halved. (b) The ratio of standard errors is

√ √ 400 2 σ/ 900 √ =√ = , 3 σ/ 400 900

so the standard error for a sample size 900 is 2/3rd’s that for sample size 400. (c) The standard error for a sample size 25 is √ 225 √ =3 25 times as large as that for sample size 225. (d) The standard error for a sample size 40 is √ 640 √ =4 40 times as large as that for sample size 640.

6.12 The finite population correction factor is given by (N − n)/(N − 1).

Thus, when (a) n = 5 and N = 250, the finite population correction factor is 250 − 5 = .9939.

250 − 1 (b) n = 10 and N = 500, the finite population correction factor is 500 − 10 = .9820.

500 − 1

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CHAPTER 6 SAMPLING DISTRIBUTIONS

139

(c) n = 100 and N = 5, 000, the finite population correction factor is 5, 000 − 100 = .9802.

5, 000 − 1 √ √ 6.13 We need to find P (|X̄ − μ| < .6745 · σ/ n). Since the standard deviation of the mean is σ/ n, the √ standardized variable (X̄ − μ)/(σ/ n) is approximately a normal random variable for large n (central limit theorem). Thus, we need to find: P (|

X̄ − μ √ | < .6745). σ/ n

Now, interpolating in Table 3 gives P( Thus, P(

X̄ − μ √ < .6745) = .75. σ/ n

X̄ − μ X̄ − μ √ ≤ −.6745) = P ( √ ≥ .6745) = .25 σ/ n σ/ n

so P (|

X̄ − μ √ | < .6745) = .75 − .25 = .50. σ/ n

The probability that the mean of a random sample of size n , from a population with standard deviation √ σ, will differ from μ by less than (.6745)(σ/ n) is approximately .5 for sufficiently large n. √ 6.14 (a) Chebyshev’s theorem for X̄ ( which has standard deviation σ/ n ) states that: 1 kσ P (|X̄ − μ| < √ ) ≥ 1 − 2 . k n In this case, σ = 2.4 and n = 25 so we take k = (1.2)5/2.4 = 2.5. Then 1 2.5σ P (|X̄ − μ| < 1.2) = P (|X̄ − μ| < √ ) ≥ 1 − = .84.

(2.5)2 n (b) Under the central limit theorem kσ P (|X̄ − μ| < √ ) ≈ F (k) − F (−k), n where F is the standard normal distribution in Table 3. Thus, P (|X̄ − μ| < 1.2)

= =

|X̄ − μ| √ < 2.5) ≈ F (2.5) − F (−2.5) σ/ n .9938 − .0062 = .9876.

P(

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CHAPTER 6 SAMPLING DISTRIBUTIONS

140 which is much larger. √ √ 6.15 Here σ/ n = 0.0042/ 40 = 0.000664 and

P (0.2245 ≤ X ≤ 0.2260) = P (−.781 ≤

X̄ − 0.2250 ≤ 1.563) .0.000664

Since n = 40 is large , the central limit theorem yields the approximation P (−.781 ≤

X̄ − 0.2250 ≤ 1.563) 0.0006664

≈

F (1.563) − F (−.781)

=

.7236.

6.16 Since the population is normal, the distribution of X̄ is exactly normal and we need to find P(

7.75 − 10 X̄ − 10 X̄ − μ √ < √ ) = P( < −3.00) = F (−3.00) = .0013.

1.5/2 σ/ n 1.5/ 4

Thus, if the process is deemed “out of control” by the decision rule ( i.e. X̄ < 7.75 means that the process is termed “out of control”), then it is very unlikely that the process is actually in control.

However, the process may actually be out of control with relatively high probability of not being detected.

6.17 We need to find P(

36

Xi > 6, 000)

=

P (X̄ > 166.67) = P (X̄ − 163 > 3.67)

=

P(

i=1

X̄ − 163 > 1.222).

18/6

Since n = 36 is relatively large, we use the central limit theorem to approximate this probability by 1 − F (1.222) = .111.

6.18 (a) If X has density f (x) , and Y = X − μ, then ∞ μY =

−∞

(x − μ)f (x) dx =

∞ −∞

xf (x) dx − μ

∞ f (x) dx. −∞

But, ∞

∞ xf (x) dx = μ,

−∞

and

f (x) dx = 1, −∞

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CHAPTER 6 SAMPLING DISTRIBUTIONS

141

so, μY = μ − μ = 0.

(b) σY2

=

expected value of (Y − μY )2

=

E((X − μ) − 0)2 ∞ (x − μ)2 f (x) dx = σ 2 .

=

−∞

6.19 Let μX = E(X) be the expected value of X. First we will show that E(X + Y ) = E(X) + E(Y ). Let f (x, y) be the joint density function of X and Y . Then, if X takes on discrete values xi and Y takes on discrete values yj , E(X + Y )

∞ ∞

(xi + yj )f (xi , yj ) i=0 j=0 ∞ ∞ ∞

=

=

xi

i=0

But,

f (xi , yj ) +

j=0

yj

j=0

∞

f (xi , yj ).

i=0

∞

j=0 f (xi , yj ) = f1 (x), the marginal density of X. Similarly for Y . Thus

E(X + Y ) =

∞

xi f1 (xi ) +

i=0

∞

yj f2 (yj ) = E(X) + E(Y )

j=0

by definition. By induction E(

n

Xi ) =

i=1

n

E(Xi ).

i=1

Thus, n μX̄

i=1 Xi

=

E(

=

n 1 1 Xi ) = E(Xi ).

E( n i=1 n i=1

n n

)

But, each E(Xi ) = μ so μX̄ = μ.

6.20 Since the sample is from a normal population t=

x̄ − μ 45.3 − 40.5 √ = = 3.038 7.9/5 s/ n

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142

is the value of a t random variable with 24 degrees of freedom. From Table 4, P(

X̄ − μ √ > 2.797) = .005.

S/ n

Before sampling, the probability of getting a value of t > 2.797 is .005. The probability of t being as large or larger than the observed t = 3.214 is even smaller. If, before we sample, we were also concerned about the mean being less than 40.5, we should also include the probability that t would be less than −3.214 which would double the value Even so, the probability is less than .01 so the data tends to refute the claim that the population mean is 40.5.

6.21 The mean of the data is x̄ = 23 and the sample standard deviation is 6.39. Thus, if the data are from a normal population with μ = 20, the statistic t=

23 − 20 x̄ − μ √ = 1.15 √ = s/ n 6.39/ 6

is the value of a t random variable with 5 degrees of freedom. The entry in Table 4 for α = 10 and ν = 5 is 1.476. Before the data are observed, we know that P(

X̄ − μ √ > 1.15) > .10.

S/ n

The data do not give strong evidence against the ambulance service’s claim.

6.22 If the sample is from a normal population with μ = .5000 t=

x̄ − μ .5060 − .5000 √ = √ = 4.74 s/ n .0040/ 10

is the value of a t random variable with 9 degrees of freedom. From Table 4, P(

X̄ − μ √ > 3.25) = .005.

S/ n

Before observing the data, the probability of getting a value of t > 3.25 is .005. The probability of being as large or larger than the observed t = 4.74 is even smaller. Thus it is very unlikely that the process is under control. The data suggest that the true mean is greater then .5000 cm.

6.23 The population of pitch is normal so that the χ2 distribution with n − 1 degrees of freedom applies.

P (S 2 > 0.122)

= =

(n − 1)S 2 (n − 1)0.122 > ) σ2 σ2 (n − 1)S 2 (9)0.122 9S 2 P( > > 16.892). ) = P ( σ2 0.065 σ2

P(

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143

with n − 1 degrees of freedom. From Table 5W, with ν = 9 degrees of freedom, we see that P(

9S 2 > 18.307) = .05. σ2

and a computer calculation shows that P (9S 2 / σ 2 > 16.892) = .0504 Thus, the probability that the claim will be rejected even though σ 2 = 0.065 is about .05 or 5 percent.

6.24 We need to find P (3.14 < S < 8.94) = P (

9(8.94)2 (n − 1)S 2 9(3.14)2 < < ).

2 42.5 σ 42.5

Since the data are from a normal population, the statistic (n − 1)S 2 /σ 2 has a χ2 distribution with ν = n − 1 degrees of freedom. From Table 5W, with ν = 9 degrees of freedom, we see that P (3.14 < S < 8.94)

= =

(n − 1)S 2 (n − 1)S 2 < 16.925) − P ( < 2.0879) σ2 σ2 .95 − .01 = .94.

P(

6.25 We need to find 1 − P(

S2 1 ≤ 12 ≤ 7).

7 S2

Since the samples are independent and from normal populations, the statistic S12 /S22 has an F distribution with n1 − 1 = n2 − 1 = 7 degrees of freedom for both the numerator and the denominator.

From Table 6W(b), we see that P(

S12 > 6.99) = .01.

S22

Using the relation F1−α (ν1 , ν2 ) = we know that P(

1 , Fα (ν2 , ν1 )

S12 1 < ) = .01.

S22 6.99

Thus, approximately P(

1 S2 S2 1 S12 S2 < or 12 > 7) = P ( 12 < ) + P ( 12 > 7) = .01 + .01 = .02.

2 S2 7 S2 S2 7 S2

6.26 (a) F.95 (15, 12) =

1 1 = = .403.

F.05 (12, 15) 2.48

F.99 (5, 20) =

1 1 = = .105.

F.01 (20, 5) 9.55

(b)

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CHAPTER 6 SAMPLING DISTRIBUTIONS

144 6.27 We need to find P (S 2 > 180) = P (

(n − 1)S 2 4 · 180 (n − 1)S 2 > > 3.2). ) = P ( σ2 225 σ2

Since the sample is from a normal population, the statistic (n − 1)S 2 / σ 2 has a χ2 distribution with n − 1 = 4 degrees of freedom. The density of the χ2 distribution with 4 degrees of freedom is f (x) = Thus,

1 −x/2 , xe 4

x > 0.

(n − 1)S 2 > 3.2) = P (S > 180) = P ( σ2 2

∞

1 −x/2 dx. xe 3.2 4

Integrating by parts with u = x and dv = e−x/2 gives 2

P (S > 180) = −e

−x/2

∞

1 3.2 ( x + 1) = e−3.2/2 ( + 1) = .5249.

2 2 3.2

Using R software, 1 - pchisq(3.2,4) = .5249 6.28 The value for t.05 for ν = 1 in Table 4 is 6.314. Thus we need to show that ∞ P (t > t.05 ) = P (t > 6.314) = But,

1 1 dt = .05.

2 6.314 π 1 + t

∞

1 1 1 1 π dt = arctan (t) = ( − 1.4137) = .05 2 π π 2 6.314 π 1 + t 6.314

∞

as required.

6.29 The probability that the ratio of the larger to the smaller sample variance exceeds 3 is 1 − P(

S2 1 ≤ 12 ≤ 3) = 1 − 3 S2

3

6x dx.

4 1/3 (1 + x)

Let u = x + 1 or x = u − 1. Then P(

1 S2 ≤ 12 ≤ 3) 3 S2

Thus, 1 − P(

=

4 4 6(u − 1) −3 2 du = + u4 u2 4/3 u3 4/3 4/3

=

24 52 − = .6875.

16 64

4

S2 1 ≤ 12 ≤ 3) = 1 − .6875 = .3125.

3 S2

6.30 (a) Since Z22 + Z32 + Z42 + Z52 is the sum of 4 independent standard normal variables squared, it has Copyright © 2017 Pearson Education, Inc.

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145

the χ2 distribution with 4 degrees of freedom. (b) From Part (a), χ2 = Z22 + Z32 + Z42 + Z52 has a chi square distribution with 4 degrees of freedom, Z22 + Z32 + Z42 + Z52 = 4

χ2 = 4

chi square d.f.

which is independent of the standard normal Z1 because all the Zi ’s are independent. According to the representation of t, their ratio has a t distribution with 4 degrees of freedom. √ 6.31 First Z1 − Z2 has a normal distribution with mean 0 and variance 2 so ( Z1 − Z2 )/ 2 has a standard normal distribution and is independent of Z32 +Z34 +Z52 +Z62 which is the sum of 4 independent standard normal variables squared and so has a chi square distribution with 4 degrees of freedom. According to the representation of t, the ratio √ ( Z1 − Z2 )/ 2 Z22 + Z32 + Z42 + Z52 4

=

Z1 − Z2 Z22 + Z32 + Z42 + Z52 2

has a t distribution with 4 degrees of freedom.

6.32 (a) Since Z12 + Z22 + Z32 + Z42 is the sum of 4 independent standard normal variables squared, it has the χ2 distribution with 4 degrees of freedom. (b) Following the same reasoning Z52 + Z62 + Z72 is the sum of 3 independent standard normal variables squared so it has the χ2 distribution with 3 degrees of freedom. (c) Because Z12 + Z22 + Z32 + Z42 + Z52 + Z62 + Z72 is the sum of 7 independent standard normal variables squared it has the χ2 distribution with 7 degrees of freedom.

6.33 Since χ21 and χ22 are independent, they can be represented as χ21 =

ν1

Zi2

and

ν 1 +ν2

χ22 =

Zi2

i=ν1 +1

i=1

where the Zi ’s are independent standard normal random variables. Then χ21 + χ22 =

ν1

i=1

Zi2 +

ν 1 +ν2

Zi2 , =

ν 1 +ν2

i=ν1 +1

Zi2

i=1

which is the sum of ν1 + ν2 squared standard normal variables. Consequently, it has the χ2 distribution with ν1 + ν2 degrees of freedom.

6.34 We know that Xi has mgf Mi (t) = e−λi + λi e

t

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CHAPTER 6 SAMPLING DISTRIBUTIONS

146 Consequently, M ni=1 Xi (t) =

8 !

e−λi + λi e = e( t

8

i=1 λi ) + (

8

i=1 λi )e

t

i=1

which is the mgf of a Poisson distribution with parameter (a) When λi = 0.5,

8

(b) When λi = 0.04,

i=1 λi

8

i=1 λi .

= 4.0 so the distribution is Poisson with parameter 4.0.

8

i=1 λi

= 0.32 so the distribution is Poisson with parameter 0.32.

6.35 We know that Xi has mgf ( 1 − 2t )− 2i so that M 5i=1 Xi (t) =

n

i=1 Xi has mgf

5 !

1 1 = −2i ( 1 − 2 t ) ( 1 − 2 t )−30 i=1

which is the mgf of a Gamma distribution with parameters α = 2 and β = 30.

6.36 (a) We know that Xi has mgf

1 2 2

Mi (t) = etμi + 2 t σi so, by independence,

MX1 +2X2 −3X3 (t) = E[ etX1 +2tX2 −3tX3 ] = MX1 (t)MX2 (2t)MX3 (−3t) This last product is equal to 1 2 2

1

1

2 2

2 2

etμ1 + 2 t σ1 · e(2t)μ2 + 2 (2t) σ2 · e(−3t)μ3 + 2 (−3t) σ3

1 2

2

2

2

= et(μ1 + 2μ2 − 3μ3 ) + 2 t (σ1 + 4σ2 + 9σ3 ) This is the form of the moment generating function of a normal distribution. (b) From the form of MX1 +2X2 −3X3 (t) we see that the mean is the coefficient of t or μ1 + 2μ2 − 3μ3 = 2+2(1)−3(−1) = 7. The variance is the coefficient of t2 /2 or σ12 +4σ22 +9σ32 = 4+4(9)+9(1) = 49.

6.37 (a) We know that Xi has mgf

1 2 2

Mi (t) = etμi + 2 t σi so, by independence,

MX1 −3X2 +2X3 −5 (t) = E[ etX1 −3tX2 +2tX3 −5 ] = MX1 (t)MX2 (−3t)MX3 (2t)e−5t This last product is equal to 1 2 2

1

2 2

1

2 2

etμ1 + 2 t σ1 · e(−3t)μ2 + 2 (−3t) σ2 · e(2t)μ3 + 2 (2t) σ3 · e−5t

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CHAPTER 6 SAMPLING DISTRIBUTIONS

147 1 2

2

2

2

= et(μ1 − 3μ2 + 2μ3 − 5) + 2 t (σ1 + 9σ2 + 4σ3 ) This is the form of the moment generating function of a normal distribution. (b) From the form of MX1 −3X2 +2X3 −5 (t) we see that the mean is the coefficient of t or μ1 − 3μ2 + 2μ3 − 5 = 2 − 3(1) + 2(−1) − 5 = −8. The variance is the coefficient of t2 /2 or σ12 + 9σ22 + 4σ32 = 4 + 9(9) + 4(1) = 89.

6.38 (a) We know that Xi has mgf

1 2 2

Mi (t) = etμi + 2 t σi so

M2X1 −X2 +5X3 (t) = E[ et2X1 −tX2 +5tX3 ] = MX1 (2t)MX2 (−t)MX3 (5t)

( by independence )

This last product is equal to 1

1

2 2

1

2 2

2 2

e(2t)μ1 + 2 (2t) σ1 ·e(−t)μ2 + 2 (−t) σ2 ·e(5t)μ3 + 2 (5t) σ3

1 2

2

2

2

= et(4μ1 − μ2 5μ3 ) + 2 t (4σ1 + σ2 + 25σ3 ) This is the form of the moment generating function of a normal distribution. (b) From the form of M2X1 −X2 +5X3 (t) we see that the mean is the coefficient of t or 2μ1 − μ2 + 5μ3 = 2(−4)−0+5(3) = 7. The variance is the coefficient of t2 /2 or 4σ12 +σ22 +25σ32 = 2(1)+4+25(1) = 31.

6.39 (a) We know that Xi has mgf

1 2 2

Mi (t) = etμi + 2 t σi so, by independence,

M7X1 +X2 −2X3 +7 (t) = E[ et7X1 +tX2 −2tX3 +7 ] = MX1 (7t)MX2 (t)MX3 (−2t)e7t This last product is equal to 1

2 2

1

2 2

1

2 2

e(7t)μ1 + 2 (7t) σ1 · etμ2 + 2 (t) σ2 · e(−2t)μ3 + 2 (−2t) σ3 et7 1 2

2

2

2

= et(7μ1 + μ2 − 2μ3 − 7) + 2 t (49σ1 + σ2 + 4σ3 ) This is the form of the moment generating function of a normal distribution. (b) From the form of M7X1 +X2 −2X3 −7 (t) we see that the mean is the coefficient of t or 7μ1 + μ2 − 2μ3 + 7 = 7(−4) + 0 − 2(3) + 7 = −27. The variance is the coefficient of t2 /2 or 49σ12 + σ22 + 4σ32 = 49(1) + 4 + 4(1) = 57.

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CHAPTER 6 SAMPLING DISTRIBUTIONS

148 6.40 (a) From Exercise 5.100, we know that Xi has mgf

Mi (t) = [ pet / ( 1 − ( 1 − p ) et ]

for

pet < 1

so MX1 +X2 +···+Xr (t) = MX1 (t)MX2 (t) · · · MXr (t)

by independence

This last product is equal to [ pet / ( 1 − ( 1 − p ) et ]r

for

pet < 1

(b) When X is the number of trials until the r-th success occurs, it is the sum of trials to get the first success, plus the number to get the second success, and so on. That is, the negative binomial variable X = X1 +X2 +· · ·+Xr . Furthermore, the Xi depend of different sets of trials and so they are independent. Consequently, the mgf in Part (a) is that of the negative binomial distribution.

(c) From the solution to Exercise 5.100, the first two derivatives of the common geometric moment generating function M ( t ) satisfy M (0) = 1 / p and M (0) = ( 2 − p ) /p2 . Because M X ( t ) = M r ( t ), i

r−1 M ( t ) M ( t ) Xi ( t ) = r M

and the mean

M Xi ( 0 ) = r · 1 ·

1 r = p p

Next,

r−1 ( t ) M ( t ) + r M ( t ) ( r − 1) M r − 2 ( t ) M ( t ) M Xi ( t ) = r M

so that

M Xi ( 0 ) = r

2−p 1 (1 − p + r ) + r(r − 1) 2 = r p2 p p2

The variance is then

2 (1 − p + r ) r r(1 − p)

M ( 0 ) = r − ( )2 = Xi ( 0 ) − M Xi p2 p p2

6.41 (a) We know from Exercise 6.40 that Xi has mgf [ pet / ( 1 − ( 1 − p ) et ]ri

for

pet < 1

so MX1 +X2 +···+Xn (t) = MX1 (t)MX2 (t) · · · MXn (t) Copyright © 2017 Pearson Education, Inc.

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CHAPTER 6 SAMPLING DISTRIBUTIONS

149

This last product is equal to [ pet / ( 1 − ( 1 − p ) et ]

n

i=1

ri

for

pet < 1

(b) Note that this is of the form of negative binomial mgf but with the number of successes given by n r = i=1 ri . The sum of independent negative binomial random variables, all having the same success probability p, is again a negative binomial random variable.

6.42 From the example, g(y) = f (

√

y)

y −1/2 √ y −1/2 + f( − y ) 2 2

where

2 1 f (z) = √ e−z /2 2π

Consequently, we verify that 1 y −1/2 y −1/2 1 1 g(y) = √ e−y/2 + √ e−y/2 = √ e−y/2 y −1/2 2 2 2π 2π 2π 6.43 Let Y = Z 3 where Z is standard normal with density function 2 1 f (z) = √ e−z /2 2π

Then, since the transformation is monotone increasing, G(y) = P ( Y ≤ y ) = P ( Z 3 ≤ y ) = P ( Z ≤ y 1/3 ) = F (y 1/3 ) Differentiating with respect to y, we obtain the density 2/3 1 1 2 g(y) = f (y 1/3 ) | y −2/3 | + f (y 1/3 )| y −2/3 | = √ y −2/3 e− y /2 3 3 3 2π

6.44 Let Y = 1 − e−X where X has the density function f (x) = e−x for x > 0. Since the transformation is monotone increasing, for any 0 < y < 1, we have G(y) = P ( Y ≤ y ) = P ( 1 − e−X ≤ y ) = P ( X ≤ − ln(1 − y)) = F (− ln(1 − y)) = 1 − eln(1−y) = 1 − (1 − y) = y Differentiating with respect to y, we obtain the density function g(y) = 1, for 0 < y < 1. This is the uniform distribution.

6.45 Let Y = ln(X) where X has the density function f (x) = e−x for x > 0. Since the transformation is

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CHAPTER 6 SAMPLING DISTRIBUTIONS

150 monotone increasing, for any −∞ < y < −∞ we have

G(y) = P ( Y ≤ y ) = P ( ln(X) ≤ y ) = P ( X ≤ ey ) = F (ey ) = 1 − e−e

y

Differentiating with respect to y, we obtain the density function g(y) = ey e−e . y

6.46 Let y = h(x) = x2 which is monotone increasing on 0 < x < 2 where f (x) = 0.5. Also, on the range, √ the inverse function w(y) = y so 1 g(y) = f ( w(y) ) | w (y) | = 0.5 y −1/2 2

for

0<y<4

and g(y) = 0 elsewhere.

6.47 Let y = h(x) = − ln(x) which is monotone decreasing on 0 < x < 1 where f (x) = 1. Also, on the range, the inverse function w(y) = e−y so g(y) = f ( w(y) ) | w (y) | = 1 e−y

0<y<∞

for

and g(y) = 0 elsewhere. This is the exponential distribution with β = 1.

6.48 Let X and Y be independent and each have a exponential distribution with β = 1. For any z > 0, ∞ fX+Y (z) =

−∞

fX (x) fY (z − x ) dx =

z

e−x e−(z − x ) , dx = e−z

0

z

1 dx = z e−z

0

which is density function of a gamma distribution with α = 2 and β = 1.

6.49 When X and Y are independent and each has the same gamma distribution, ∞ fY / X (z) =

−∞

| x | fX (x) fY (xz) dx =

∞ x 0

1 1 α−1 −x / β 1 1 x e (xz)α−1 e−(xz) / β dx Γ(α) β α Γ(α) β α

This last integral becomes 1 Γ(α)2

∞ z 0

α−1

1 2α−2 −x(1+z) / β 1 x e dx = β 2α Γ(α)2

∞

z α−1 x2α−1

0

1 −x(1+z) / β e dx β 2α

Under the change of variable v = (1 + z)x/β, and multiplying and dividing by Γ(2α) and (1 + z)2α we obtain z α−1 1 Γ(2α) Γ(α)2 (1 + z)2α

∞ 0

1 Γ(2α) z α−1 v 2α−1 e−v dx = Γ(2α) Γ(α)2 (1 + z)2α

for 0 < z < 1

6.50 When X and Y are independent and each has a discrete uniform distribution on 0, 1, 2 the convolution

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151

formula becomes

fX+Y (z) =

z

0

⎧ ⎨ z 1 1 = z+1 z = 0, 1, 2 0 3 3 9 fX (x) fY (z − x ) =

2 1 1 5−z ⎩ z = 3, 4 z−2 3 3 = 9

6.51 (a) Number the states from 1 to 50 according to their alphabetical order. Using the last two columns of the second page of Table 7, starting at row 11, to select 10 numbers between 1 and 50 by discarding numbers outside of this range as well as any previously drawn number gives 17, 2, 12, 1, 44, 18, 33, 39, 40, 41.

These correspond to Kentucky, Alaska, Idaho, Alabama, Utah, Louisiana, North Carolina , Rhode Island, South Carolina, and South Dakota. (b) No. High population states like California would have many participants but could send only two.

A student from a small state would have a better chance than a student in California.

6.52 (a) The number of samples ( given that order does not matter) is ⎛ ⎝

⎞ ⎠ = 12 · 11 = 66.

2·1 2

12

(b) The number of samples (given that order does not matter) is ⎛ ⎝

⎞ ⎠ = 20 · 19 = 190.

2·1 2

20

6.53 For a random sample, each outcome must have equal probability. The probabilities are: 1 = .01515. (a) 66 1 = .00526. (b) 190 6.54 The finite population correction factor is (N − n)/(N − 1). Thus (a)

(b)

12 − 2 N −n = = .909.

N −1 12 − 1 20 − 2 N −n = = .947.

N −1 20 − 1

√ 6.55 (a) Here σ/ n = 9/6 = 1.5 and we will bound the one-tailed probability P (X̄ > 66.75) = P (X̄ − 63 > 66.75 − 63).

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152

Using Chebyshev’s theorem for the two-tailed probability P (|X̄ − 63| > 66.75 − 63)

=

P (|X̄ − 63| >

≤

(

66.75 − 63 · 9/6) 9/6

9/6 )2 = .16.

66.75 − 63

Thus, P (X̄ > 66.75) ≤ .16. (b) We need to find P(

66.75 − 63 X̄ − 63 X̄ − 63 √ > √ > 2.5) ) = P( 9/6 σ/ n σ/ n

which by the central limit theorem is the much smaller probability P (Z > 2.5) = 1 − F (2.5) = 1 − .9938 = .0062.

6.56 Let X̄ be the mean number of pieces of mail in a random sample of size 35.

(a) By Chebyshev’s theorem, P (X̄ < 40 or X̄ > 48) = P (|X̄ − 44| > 4) ≤

1 k2

where 8 4 = kσX̄ = k √ 35

or

1 = .114 k2

Thus, P (X̄ < 40 or X̄ > 48) ≤ .114 (b) By the central limit theorem, we get the much smaller value P (X̄ < 40 or X̄ > 48)

≈

1 − P (40 ≤ X̄ ≤ 48) X̄ − μ 48 − 44 40 − 44 √ ) √ ≤ ≤ 1 − P( √ σ/ n 8/ 35 8/ 35 1 − P (−2.96 ≤ Z ≤ 2.96) = 2P (Z > 2.96)

=

2(1 − .9985) = .0030

= =

6.57 Since the population is normal, the distribution of X̄ is exactly normal and we need to find P (|X̄ − μ| ≤ .02)

= =

X̄ − μ .02 |≤ ) = F (2.5) − F (−2.5) .04/5 .04/5 .9938 − .0062) = .9876.

P (|

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153

6.58 By the central limit theorem, P (|X̄ − μ| > .06)

= ≈

.06 |X̄ − μ| √ ) √ > σ/ n .16/ 36 P (|Z| > 2.25) = 2(1 − .9878) = .0244

P(

6.59 We need to find P(

S12 > 4.0).

S22

Since the samples are independent and from normal populations, the statistic S12 /S22 has an F distribution with n1 − 1 = 8 and n2 − 1 = 15 degrees of freedom for the numerator and the denominator, respectively . From Table 6W(b), we see that P(

S12 > 4.0) = .01, S22

so the probability is .01 or 1 percent.

6.60 By Theorem 6.5, F = S22 /S12 has F distribution with 7 and 25 degrees of freedom. Hence , from Table 6W(a), we have P(

S22 > 2.4) = P (F > 2.4) = .05 S12

6.61 The variance of the sample mean X̄, based on a sample of size n, is σ 2 /n. Thus the standard deviation, √ or standard error, of the mean is σ/ n. √ (a) The standard deviation for a sample of size 100 is σ/ 100. The standard deviation for a sample √ is size 200 is σ/ 200. That is, the ratio of standard errors is √ √ 1 σ/ 200 100 √ =√ = √ = .707, σ/ 100 200 2 so the standard error for sample size 200 is .707 times the standard error for sample size 100. (b) The ratio of standard errors is √ √ σ/ 300 200 √ =√ = σ/ 200 300

2 = .816, 3

so the standard error for a sample size 300 is .816 times that for sample size 200. (c) The ratio of standard errors is

√ σ/ 90 √ = 2, σ/ 360

so the standard error doubles.

6.62 Under this sampling scheme, the observations will not satisfy the independence assumption required for a random sample. The longest lines are likely to occur together when the same cars must wait

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CHAPTER 6 SAMPLING DISTRIBUTIONS

154 several light changes during the rush hour.

6.63 (a) The production process may not be stabilized when the first shaft is made each morning. Alternatively, the first shaft could be given extra attention and may not be representative. (b) Buses and large trucks would take longer to pass the fixed point and thus would be more likely to be included in the sample than a small car.

6.64 (a) No, we would expect the weight of the bags to vary because the weights of the individual apples 30

Xi is the total weight of the vary. With Xi the weight of the i−th apple placed in the bag, i=1

bag. Then, P(

30

Xi ≤ y) = P (X ≤ y/30)

i=1

so the probability concerning total weight can be expressed as a probability concerning X. (b) Further, P (X ≤ y/30) = P (

y/30 − μ X −μ √ ≤ √ ) σ/ n σ/ n

X̄ − μ √ is nearly standard normal when n = 30. Thus, for any σ/ n 30

specified value y, we can approximate the probability P ( Xi ≤ y) by the standard normal

By the central limit theorem

y/30 − μ √ ) probability P ( Z ≤ σ/ n

i=1

(c) With μ = .2 and σ = .03 pound

P(

30

i=1

Xi ≥ 6.2) = P ( Z ≥

6.2/30 − .2 √ ) = 1 − F (1.217) = .1118 .03/ 30

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