CHAPTER 6

Note: Problem 6.34 uses the exponential. It was incorrectly assigned

6.1 (a) P(Z < 1.57) = 0.9418

(b) P(Z > 1.84) = 1 – 0.9671 = 0.0329

(c) P(1.57 < Z < 1.84) = 0.9671 – 0.9418 = 0.0253

(d) P(Z < 1.57) + P(Z > 1.84) = 0.9418 + (1 – 0.9671) = 0.9747

(e) P(– 1.57 < Z < 1.84) = 0.9671 – 0.0582 = 0.9089

(f) P(Z < – 1.57) + P(Z > 1.84) = 0.0582 + 0.0329 = 0.0911

(g) Since the Z-distribution is symmetric about its mean, half of the area will be above Z = 0.

(h) If P(Z > A) = 0.025, P(Z < A) = 0.975. A = + 1.96.

(i) If P(–A < Z < A) = 0.6826, P(Z < A) = 0.8413. So 68.26% of the area is captured between –A = – 1.00 and A = + 1.00.

6.3 (a) P(Z < 1.08) = 0.8599

(b) P(Z > – 0.21) = 1.0 – 0.4168 = 0.5832

(c) P(0 < Z < 1.08) = 0.8599 – 0.5 = 0.3599

(d) P(Z < 0) + P(Z > 1.08) = 0.5 + (1.0 – 0.8599) = 0.6401

(e) P(– 0.21 < Z < 0) = 0.5 – 0.4168 = 0.0832

(f) P(Z < – 0.21) + P(Z > 0) = 0.4168 + 0.5 = 0.9168

(g) P(– 0.21 < Z < + 1.08) = 0.8599 – 0.4168 = 0.4431

(h) P(Z < – 0.21) + P(Z > 1.08) = 0.4168 + (1 – 0.8599) = 0.5569

6.4 (a) P(Z > 1.08) = 1 – 0.8599 = 0.1401

(b) P(Z < – 0.21) = 0.4168

(c) P(– 1.96 < Z < – 0.21) = 0.4168 – 0.0250 = 0.3918

(d) P(– 1.96 < Z < 1.08) = 0.8599 – 0.0250 = 0.8349

(e) P(1.08 < Z < 1.96) = 0.9750 – 0.8599 = 0.1151

(f) Since the Z-distribution is symmetric about its mean, half of the area will be below Z = 0.

(g) If P(Z < A) = 0.1587, A = – 1.00.

(h) If P(Z > A) = 0.1587, P(Z < A) = 0.8413. So A = + 1.00.

6.5 (a) = – 2.50

P(X > 75) = P(Z > – 2.50) = 1 – P(Z< – 2.50) = 1 – 0.0062 = 0.9938

(b) = – 3.00

P(X < 70) = P(Z < – 3.00) = 0.00135

6.5 (c) = 1.20

cont. P(X > 112) = P(Z > 1.20) = 1 – P(Z < 1.20) = 1.0 – 0.8849 = 0.1151

(d) = – 1.50

P(75 < X < 85) = P(– 2.50 < Z < – 1.50) = 0.0668 – 0.0062 = 0.0606

(e)

P(X < 80) = P(Z < – 2.00) = 0.0228

P(X > 110) = P(Z > 1.00) = 1 – P(Z < 1.00) = 1.0 – 0.8413 = 0.1587

P(X < 80) + P(X > 110) = 0.0228 + 0.1587 = 0.1815

(f) P(X < A) = 0.10,

P(Z < – 1.28) = 0.10 Z =

Solving for A, A = 100 – 1.28(10) = 87.20

(g) P(X_lower < X < X_upper) = 0.80

P(Z < – 1.28) = 0.10 and P(Z < 1.28) = 0.90

X_lower = 100 – 1.28(10) = 87.20 and X_upper = 100 + 1.28(10) = 112.80

(h) P(X > A) = 0.7, so P(X < A) = 0.3

A = 100 – 0.52(10) = 94.8

6.7 (a) P(X < 25) = P(Z < –1.116) = 0.1322

(b) P(X > 50) = P(Z > 1.384) = 0.0832

(c) P(X > 75) = P(Z > 3.884) = 0.0001

(d) P(30 < X < 40) = P(–0.616 < Z < 0.384) = 0.3806

(e) P(X < A) = 0.99 Z = 2.3263 = A = 59.4234

(f) P(X > A) = 0.80 Z = –0.8416 = A = 27.7438

(g) The normality assumption may be invalid if there are some households spending an abnormal amount on gourmet coffee, which will cause the distribution to be skewed to the right.

Note: The above answers are obtained using PHStat. They may be slightly different when Table E.2 is used.

6.8 (a) P(34 < X < 50) = P(– 1.33 < Z < 0) = 0.4082

(b) P(34 < X < 38) = P(– 1.33 < Z < – 1.00) = 0.1587 – 0.0918 = 0.0669

(c) P(X < 30) + P(X > 60) = P(Z < – 1.67) + P(Z > 0.83)

= 0.0475 + (1.0 – 0.7967) = 0.2508

(d) 1000(1 – 0.2508) = 749.2 749 trucks

(e) P(X > A) = 0.80 P(Z < – 0.84) 0.20

A = 50 – 0.84(12) = 39.92 thousand miles or 39,920 miles

(f) The smaller standard deviation makes the Z-values larger.

(a) P(34 < X < 50) = P(– 1.60 < Z < 0) = 0.4452

(b) P(34 < X < 38) = P(– 1.60 < Z < – 1.20) = 0.1151 – 0.0548

= 0.0603

(c) P(X < 30) + P(X > 60) = P(Z < – 2.00) + P(Z > 1.00)

= 0.0228 + (1.0 – 0.8413) = 0.1815

(d) 1000(1 – 0.1815) = 818.5 819 trucks

(e) A = 50 – 0.84(10) = 41.6 thousand miles or 41,600 miles

6.11 (a) P(X < 180) = P(Z < – 1.50) = 0.0668

(b) P(180 < X < 300) = P(– 1.50 < Z < 1.50) = 0.9332 – 0.0668 = 0.8664

(c) P(X < 180) + P(X > 300) = P(Z < – 1.50) + P(Z > 1.50)

= 1 – P(– 1.50 < Z < 1.50) = 1 – 0.8664 = 0.1336

1000(0.1336) = 133.6 134 calls

(d) P(110 < X < 180) = P(– 3.25 < Z < – 1.50) = 0.0668 – 0.00058 = 0.06622

(e) P(X < A) = 0.01 P(Z < – 2.33) = 0.01

A = 240 – 2.33(40) = 146.80 seconds

6.13 (a) P(21.9 < X < 22.00) = P(– 20.4 < Z < – 0.4) = 0.3446

(b) P(21.9 < X < 22.01) = P(– 20.4 < Z < 1.6) = 0.9452

(c) P(X > A) = 0.02 Z = 2.05 A = 22.0123

(d) (a) P(21.9 < X < 22.00) = P(– 25.5 < Z < – 0.5) = 0.3085

(b) P(21.9 < X < 22.01) = P(– 25.5 < Z < 2) = 0.9772

(c) P(X > A) = 0.02 Z = 2.05 A = 22.0102

6.38 (a) P( < 95) = P(Z < – 2.50) = 0.0062

(b) P(95 < < 97.5) = P(– 2.50 < Z < – 1.25) = 0.1056 – 0.0062 = 0.0994

(c) P( > 102.2) = P(Z > 1.10) = 1.0 – 0.8643 = 0.1357

(d) P(99 < < 101) = P(– 0.50 < Z < 0.50) = 0.6915 – 0.3085 = 0.3830

(e) P( > A) = P(Z > – 0.39) = 0.65 = 100 – 0.39() = 99.22

(f) (a) P( < 95) = P(Z < – 2.00) = 0.0228

(b) P(95 < < 97.5) = P(– 2.00 < Z < – 1.00)

= 0.1587 – 0.0228 = 0.1359

(c) P( > 102.2) = P(Z > 0.88) = 1.0 – 0.8106 = 0.1894

(d) P(99 < < 101) = P(– 0.40 < Z < 0.40)

= 0.6554 – 0.3446 = 0.3108

(e) P( > A) = P(Z > – 0.39) = 0.65 = 100 – 0.39() = 99.025

6.41 (a) Sampling Distribution of the Mean for n = 2 (without replacement)

Sample Number Outcomes Sample Means _i

1 1, 3 ₁ = 2

2 1, 6 ₂ = 3.5

3 1, 7 ₃ = 4

4 1, 7 ₄ = 4

5 1, 12 ₅ = 6.5

6 3, 6 ₆ = 4.5

7 3, 7 ₇ = 5

8 3, 7 ₈ = 5

9 3, 12 ₉ = 7.5

10 6, 7 ₁₀ = 6.5

11 6, 7 ₁₁ = 6.5

12 6, 12 ₁₂ = 9

13 7, 7 ₁₃ = 7

14 7, 12 ₁₄ = 9.5

15 7, 12 ₁₅ = 9.5

6.41 (a)

cont.

Mean of All Possible Mean of All

Sample Means: Population Elements:

Both means are equal to 6. This property is called unbiasedness.

(b) Sampling Distribution of the Mean for n = 3 (without replacement)

Sample Number Outcomes Sample Means _i

1 1, 3, 6 ₁ = 3 1/3

2 1, 3, 7 ₂ = 3 2/3

3 1, 3, 7 ₃ = 3 2/3

4 1, 3, 12 ₄ = 5 1/3

5 1, 6, 7 ₅ = 4 2/3

6 1, 6, 7 ₆ = 4 2/3

7 1, 6, 12 ₇ = 6 1/3

8 3, 6, 7 ₈ = 5 1/3

9 3, 6, 7 ₉ = 5 1/3

10 3, 6, 12 ₁₀ = 7

11 6, 7, 7 ₁₁ = 6 2/3

12 6, 7, 12 ₁₂ = 8 1/3

13 6, 7, 12 ₁₃ = 8 1/3

14 7, 7, 12 ₁₄ = 8 2/3

15 1, 7, 7 ₁₅ = 5

16 1, 7, 12 ₁₆ = 6 2/3

17 1, 7, 12 ₁₇ = 6 2/3

18 3, 7, 7 ₁₈ = 5 2/3

19 3, 7, 12 ₁₉ = 7 1/3

20 3, 7, 12 ₂₀ = 7 1/3

This is equal to , the population mean.

(c) The distribution for n = 3 has less variability. The larger sample size has resulted in more sample means being close to .

(d) (a) Sampling Distribution of the Mean for n = 2 (with replacement)

.

Sample Number Outcomes Sample Means _i

1 1, 1 ₁ = 1

2 1, 3 ₂ = 2

3 1, 6 ₃ = 3.5

4 1, 7 ₄ = 4

5 1, 7 ₅ = 4

6 1, 12 ₆ = 6.5

7 3, 1 ₇ = 2

8 3, 3 ₈ = 3

9 3, 6 ₉ = 4.5

10 3, 7 ₁₀ = 5

11 3, 7 ₁₁ = 5

12 3, 12 ₁₂ = 7.5

13 6, 1 ₁₃ = 3.5

14 6, 3 ₁₄ = 4.5

15 6, 6 ₁₅ = 6

16 6, 7 ₁₆ = 6.5

17 6, 7 ₁₇ = 6.5

18 6, 12 ₁₈ = 9

19 7, 1 ₁₉ = 4

20 7, 3 ₂₀ = 5

21 7, 6 ₂₁ = 6.5

22 7, 7 ₂₂ = 7

23 7, 7 ₂₃ = 7

24 7, 12 ₂₄ = 9.5

25 7, 1 ₂₅ = 4

26 7, 3 ₂₆ = 5

27 7, 6 ₂₇ = 6.5

28 7, 7 ₂₈ = 7

29 7, 7 ₂₉ = 7

30 7, 12 ₃₀ = 9.5

31 12, 1 ₃₁ = 6.5

32 12, 3 ₃₂ = 7.5

33 12, 6 33 = 9

34 12, 7 ₃₄ = 9.5

35 12, 7 ₃₅ = 9.5

36 12, 12 ₃₆ = 12

(a) Mean of All Possible Mean of All

. Sample Means: Population Elements:

Both means are equal to 6. This property is called unbiasedness.

(b) Repeat the same process for the sampling distribution of the mean for n = 3 (with replacement). There will be different samples.

This is equal to , the population mean.

(c) The distribution for n = 3 has less variability. The larger sample size has resulted in more sample means being close to .

6.47 (a) P( < 55000) = P(Z < – 1.227) = 0.1099

(b) P( > 60000) = P(Z > 1.773) = 0.0381

(c) P( > 111600) = P(Z > 32.733) = 0.0000

(d) This indicates that the household income distribution is not normally distributed and is skewed to the right.

(e) Since the distribution is not symmetrical about the mean, a sample size of 20 will not be large enough for the central limit theorem to take effect so that the sampling distribution of the sample mean can be approximated by a normal distribution and, hence, the methods used in (a)-(c) will not be appropriate.

Note: The above answers are obtained using PHStat. They may be slightly different when Table E.2 is used.

6.51 (a) p_s = 14/40 = 0.35 (b) = = 0.0725

6.52 (a) ,

P(p_s > 0.55) = P (Z > 0.98) = 1.0 – 0.8365 = 0.1635

(b) ,

P(p_s > 0.55) = P (Z > – 1.021) = 1.0 – 0.1539 = 0.8461

(c) ,

P(p_s > 0.55) = P (Z > 1.20) = 1.0 – 0.8849 = 0.1151

(d) Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2.

(a) P(p_s > 0.55) = P (Z > 1.96) = 1.0 – 0.9750 = 0.0250

(b) P(p_s > 0.55) = P (Z > – 2.04) = 1.0 – 0.0207 = 0.9793

(c) P(p_s > 0.55) = P (Z > 2.40) = 1.0 – 0.9918 = 0.0082