Some Continuous Probability

Chapter 6 Some Continuous Probability

Distributions

6.1 (a) Area=0.9236. (b) Area=1 − 0.1867 = 0.8133.

(c) Area=0.2578 − 0.0154 = 0.2424. (d) Area=0.0823.

(e) Area=1 − 0.9750 = 0.0250. (f) Area=0.9591 − 0.3156 = 0.6435.

6.2 (a) The area to the left of z is 1 − 0.3622 = 0.6378 which is closer to the tabled value 0.6368 than to 0.6406. Therefore, we choose z = 0.35.

(b) From Table A.3, z = −1.21. (c) The total area to the left of z is 0.5000+0.4838=0.9838. Therefore, from Table

A.3, z = 2.14. (d) The distribution contains an area of 0.025 to the left of −z and therefore a total

area of 0.025+0.95=0.975 to the left of z. From Table A.3, z = 1.96.

6.3 (a) From Table A.3, k = −1.72. (b) Since P (Z > k) = 0.2946, then P (Z < k) = 0.7054/ From Table A.3, we find

k = 0.54. (c) The area to the left of z = −0.93 is found from Table A.3 to be 0.1762. Therefore,

the total area to the left of k is 0.1762+0.7235=0.8997, and hence k = 1.28.

6.4 (a) z = (17 − 30)/6 = −2.17. Area=1 − 0.0150 = 0.9850. (b) z = (22 − 30)/6 = −1.33. Area=0.0918. (c) z 1 = (32−3)/6 = 0.33, z 2 = (41−30)/6 = 1.83. Area = 0.9664−0.6293 = 0.3371. (d) z = 0.84. Therefore, x = 30 + (6)(0.84) = 35.04.

72 Chapter 6 Some Continuous Probability Distributions (e) z 1 = −1.15, z 2 = 1.15. Therefore, x 1 = 30 + (6)(−1.15) = 23.1 and x 2 =

6.5 (a) z = (15 − 18)/2.5 = −1.2; P (X < 15) = P (Z < −1.2) = 0.1151. (b) z = −0.76, k = (2.5)(−0.76) + 18 = 16.1. (c) z = 0.91, k = (2.5)(0.91) + 18 = 20.275.

(d) z 1 = (17 − 18)/2.5 = −0.4, z 2 = (21 − 18)/2.5 = 1.2;

P (17 < X < 21) = P (−0.4 < Z < 1.2) = 0.8849 − 0.3446 = 0.5403.

6.6 z 1 = [(µ − 3σ) − µ]/σ = −3, z 2 = [(µ + 3σ) − µ]/σ = 3;

P (µ − 3σ < Z < µ + 3σ) = P (−3 < Z < 3) = 0.9987 − 0.0013 = 0.9974.

6.7 (a) z = (32 − 40)/6.3 = −1.27; P (X > 32) = P (Z > −1.27) = 1 − 0.1020 = 0.8980. (b) z = (28 − 40)/6.3 = −1.90, P (X < 28) = P (Z < −1.90) = 0.0287.

(c) z 1 = (37 − 40)/6.3 = −0.48, z 2 = (49 − 40)/6.3 = 1.43; So, P (37 < X < 49) = P (−0.48 < Z < 1.43) = 0.9236 − 0.3156 = 0.6080.

6.8 (a) z = (31.7 − 30)/2 = 0.85; P (X > 31.7) = P (Z > 0.85) = 0.1977. Therefore, 19.77% of the loaves are longer than 31.7 centimeters.

(b) z 1 = (29.3 − 30)/2 = −0.35, z 2 = (33.5 − 30)/2 = 1.75; P (29.3 < X < 33.5) = P (−0.35 < Z < 1.75) = 0.9599 − 0.3632 = 0.5967.

Therefore, 59.67% of the loaves are between 29.3 and 33.5 centimeters in length. (c) z = (25.5 − 30)/2 = −2.25; P (X < 25.5) = P (Z < −2.25) = 0.0122.

Therefore, 1.22% of the loaves are shorter than 25.5 centimeters in length.

6.9 (a) z = (224 − 200)/15 = 1.6. Fraction of the cups containing more than 224 mil- limeters is P (Z > 1.6) = 0.0548.

(b) z 1 = (191 − 200)/15 = −0.6, Z 2 = (209 − 200)/15 = 0.6;

P (191 < X < 209) = P (−0.6 < Z < 0.6) = 0.7257 − 0.2743 = 0.4514. (c) z = (230 − 200)/15 = 2.0; P (X > 230) = P (Z > 2.0) = 0.0228. Therefore,

(1000)(0.0228) = 22.8 or approximately 23 cups will overflow. (d) z = −0.67, x = (15)(−0.67) + 200 = 189.95 millimeters.

6.10 (a) z = (10.075 − 10.000)/0.03 = 2.5; P (X > 10.075) = P (Z > 2.5) = 0.0062.

Therefore, 0.62% of the rings have inside diameters exceeding 10.075 cm. (b) z 1 = (9.97 − 10)/0.03 = −1.0, z 2 = (10.03 − 10)/0.03 = 1.0;

P (9.97 < X < 10.03) = P (−1.0 < Z < 1.0) = 0.8413 − 0.1587 = 0.6826. (c) z = −1.04, x = 10 + (0.03)(−1.04) = 9.969 cm.

6.11 (a) z = (30 − 24)/3.8 = 1.58; P (X > 30) = P (Z > 1.58) = 0.0571. (b) z = (15 − 24)/3.8 = −2.37; P (X > 15) = P (Z > −2.37) = 0.9911. He is late

99.11% of the time.

Solutions for Exercises in Chapter 6

73 (c) z = (25 − 24)/3.8 = 0.26; P (X > 25) = P (Z > 0.26) = 0.3974.

(d) z = 1.04, x = (3.8)(1.04) + 24 = 27.952 minutes. (e) Using the binomial distribution with p = 0.0571, we get

b(2; 3, 0.0571) = 3

6.12 µ = 99.61 and σ = 0.08. (a) P (99.5 < X < 99.7) = P (−1.375 < Z < 1.125) = 0.8697 − 0.08455 = 0.7852.

(b) P (Z > 1.645) = 0.05; x = (1.645)(0.08) + 99.61 = 99.74.

6.13 z = −1.88, x = (2)(−1.88) + 10 = 6.24 years.

6.14 (a) z = (159.75 − 174.5)/6.9 = −2.14; P (X < 159.75) = P (Z < −2.14) = 0.0162. Therefore, (1000)(0.0162) = 16 students.

(b) z 1 = (171.25 − 174.5)/6.9 = −0.47, z 2 = (182.25 − 174.5)/6.9 = 1.12. P (171.25 < X < 182.25) = P (−0.47 < Z < 1.12) = 0.8686 − 0.3192 = 0.5494.

Therefore, (1000)(0.5494) = 549 students. (c) z 1 = (174.75 − 174.5)/6.9 = 0.04, z 2 = (175.25 − 174.5)/6.9 = 0.11.

P (174.75 < X < 175.25) = P (0.04 < Z < 0.11) = 0.5438 − 0.5160 = 0.0278. Therefore, (1000)(0.0278)=28 students.

(d) z = (187.75 − 174.5)/6.9 = 1.92; P (X > 187.75) = P (Z > 1.92) = 0.0274.

Therefore, (1000)(0.0274) = 27 students.

6.15 µ = $15.90 and σ = $1.50. (a) 51%, since P (13.75 < X < 16.22) = P 13.745−15.9 1.5 <Z< 16.225−15.9 1.5

= P (−1.437 < Z < 0.217) = 0.5871 − 0.0749 = 0.5122. (b) $18.36, since P (Z > 1.645) = 0.05; x = (1.645)(1.50) + 15.90 + 0.005 = 18.37.

6.16 (a) z = (9.55 − 8)/0.9 = 1.72. Fraction of poodles weighing over 9.5 kilograms = P (X > 9.55) = P (Z > 1.72) = 0.0427.

(b) z = (8.65 − 8)/0.9 = 0.72. Fraction of poodles weighing at most 8.6 kilograms = P (X < 8.65) = P (Z < 0.72) = 0.7642.

(c) z 1 = (7.25 − 8)/0.9 = −0.83 and z 2 = (9.15 − 8)/0.9 = 1.28. Fraction of poodles weighing between 7.3 and 9.1 kilograms inclusive = P (7.25 < X < 9.15) = P (−0.83 < Z < 1.28) = 0.8997 − 0.2033 = 0.6964.

6.17 (a) z = (10, 175 − 10, 000)/100 = 1.75. Proportion of components exceeding 10.150

kilograms in tensile strength= P (X > 10, 175) = P (Z > 1.75) = 0.0401.

(b) z 1 = (9, 775 − 10, 000)/100 = −2.25 and z 2 = (10, 225 − 10, 000)/100 = 2.25. Proportion of components scrapped= P (X < 9, 775) + P (X > 10, 225) = P (Z < −2.25) + P (Z > 2.25) = 2P (Z < −2.25) = 0.0244.

74 Chapter 6 Some Continuous Probability Distributions

6.18 (a) x 1 = µ + 1.3σ and x 2 = µ − 1.3σ. Then z 1 = 1.3 and z 2 = −1.3. P (X > µ+1.3σ)+P (X < 1.3σ) = P (Z > 1.3)+P (Z < −1.3) = 2P (Z < −1.3) = 0.1936. Therefore, 19.36%.

(b) x 1 = µ+0.52σ and x 2 = µ−0.52σ. Then z 1 = 0.52 and z 2 = −0.52. P (µ−0.52σ <

X < µ + 0.52σ) = P (−0.52 < Z < 0.52) = 0.6985 − 0.3015 = 0.3970. Therefore, 39.70%.

6.19 z = (94.5 − 115)/12 = −1.71; P (X < 94.5) = P (Z < −1.71) = 0.0436. Therefore, (0.0436)(600) = 26 students will be rejected.

6.20 f (x) = 1 B−A for A ≤ x ≤ B. R B x

B 2 2 (a) µ = A+B −A

A B−A dx = 2(B−A) = 2 .

(b) E(X 2

dx = B 3 −A A 3 B−A 3(B−A) .

B 2 −2AB+A 2 (B−A) So, σ 2 =

3(B−A) − 2 =

2 B 3 −A 3 A+B 2 4(B 2 +AB+A 2 )−3(B 2 +2AB+A 2 )

6.21 A = 7 and B = 10. (a) P (X ≤ 8.8) = 8.8−7

(b) P (7.4 < X < 9.5) = 9.5−7.4 3 = 0.70.

(c) P (X ≥ 8.5) = 10−8.5

6.22 (a) P (X > 7) = 10−7 10 = 0.3. (b) P (2 < X < 7) = 7−2 10 = 0.5.

6.23 (a) From Table A.1 with n = 15 and p = 0.2 we have

P (1 ≤ X ≤ 4) = b(x; 15, 0.2) − b(0; 15, 0.2) = 0.8358 − 0.0352 = 0.8006.

x=0

(b) By the normal-curve approximation we first find µ = np = 3 and then σ 2 = npq = (15)(0.2)(0.8) = 2.4. Then σ = 1.549. Now, z 1 = (0.5 − 3)/1.549 = −1.61 and z 2 = (4.5 − 3)/1.549 = 0.97. Therefore, P (1 ≤ X ≤ 4) = P (−1.61 ≤ Z ≤ 0.97) = 0.8340 − 0.0537 = 0.7803.

6.24 µ = np = (400)(1/2) = 200, σ = √npq = (400)(1/2)(1/2) = 10. (a) z 1 = (184.5 − 200)/10 = −1.55 and z 2 = (210.5 − 200)/10 = 1.05.

P (184.5 < X < 210.5) = P (−1.55 < Z < 1.05) = 0.8531 − 0.0606 = 0.7925.

(b) z 1 = (204.5 − 200)/10 = 0.45 and z 2 = (205.5 − 200)/10 = 0.55.

P (204.5 < X < 205.5) = P (0.45 < Z < 0.55) = 0.7088 − 0.6736 = 0.0352.

(c) z 1 = (175.5 − 200)/10 = −2.45 and z 2 = (227.5 − 200)/10 = 2.75. P (X < 175.5) + P (X > 227.5) = P (Z < −2.45) + P (Z > 2.75) = P (Z < −2.45) + 1 − P (Z < 2.75) = 0.0071 + 1 − 0.9970 = 0.0101.

Solutions for Exercises in Chapter 6

6.25 n = 100.

(a) p = 0.01 with µ = (100)(0.01) = 1 and σ = (100)(0.01)(0.99) = 0.995. So, z = (0.5 − 1)/0.995 = −0.503. P (X ≤ 0) ≈ P (Z ≤ −0.503) = 0.3085.

p (b) p = 0.05 with µ = (100)(0.05) = 5 and σ = (100)(0.05)(0.95) = 2.1794.

So, z = (0.5 − 5)/2.1794 = −2.06. P (X ≤ 0) ≈ P (X ≤ −2.06) = 0.0197.

6.26 µ = np = (100)(0.1) = 10 and σ = (100)(0.1)(0.9) = 3. (a) z = (13.5 − 10)/3 = 1.17; P (X > 13.5) = P (Z > 1.17) = 0.1210.

(b) z = (7.5 − 10)/3 = −0.83; P (X < 7.5) = P (Z < −0.83) = 0.2033.

6.27 µ = (100)(0.9) = 90 and σ = (100)(0.9)(0.1) = 3.

(a) z 1 = (83.5 − 90)/3 = −2.17 and z 2 = (95.5 − 90)/3 = 1.83.

P (83.5 < X < 95.5) = P (−2.17 < Z < 1.83) = 0.9664 − 0.0150 = 0.9514. (b) z = (85.5 − 90)/3 = −1.50; P (X < 85.5) = P (Z < −1.50) = 0.0668.

6.28 µ = (80)(3/4) = 60 and σ = (80)(3/4)(1/4) = 3.873. (a) z = (49.5 − 60)/3.873 = −2.71; P (X > 49.5) = P (Z > −2.71) = 1 − 0.0034 =

0.9966. (b) z = (56.5 − 60)/3.873 = −0.90; P (X < 56.5) = P (Z < −0.90) = 0.1841.

6.29 µ = (1000)(0.2) = 200 and σ = (1000)(0.2)(0.8) = 12.649. (a) z 1 = (169.5 − 200)/12.649 = −2.41 and z 2 = (185.5 − 200)/12.649 = −1.15.

P (169.5 < X < 185.5) = P (−2.41 < Z < −1.15) = 0.1251 − 0.0080 = 0.1171. (b) z 1 = (209.5 − 200)/12.649 = 0.75 and z 2 = (225.5 − 200)/12.649 = 2.02.

P (209.5 < X < 225.5) = P (0.75 < Z < 2.02) = 0.9783 − 0.7734 = 0.2049.

6.30 (a) µ = (100)(0.8) = 80 and σ = (100)(0.8)(0.2) = 4 with z = (74.5 − 80)/4 = −1.38.

P (Claim is rejected when p = 0.8) = P (Z < −1.38) = 0.0838.

(b) µ = (100)(0.7) = 70 and σ = (100)(0.7)(0.3) = 4.583 with z = (74.5 − 70)/4.583 = 0.98.

P (Claim is accepted when p = 0.7) = P (Z > 0.98) = 1 − 0.8365 = 0.1635.

6.31 µ = (180)(1/6) = 30 and σ = (180)(1/6)(5/6) = 5 with z = (35.5 − 30)/5 = 1.1.

P (X > 35.5) = P (Z > 1.1) = 1 − 0.8643 = 0.1357.

6.32 µ = (200)(0.05) = 10 and σ = (200)(0.05)(0.95) = 3.082 with z = (9.5 − 10)/3.082 = −0.16. P (X < 10) = P (Z < −0.16) = 0.4364.

6.33 µ = (400)(1/10) = 40 and σ = (400)(1/10)(9/10) = 6.

76 Chapter 6 Some Continuous Probability Distributions (a) z = (31.5 − 40)/6 = −1.42; P (X < 31.5) = P (Z < −1.42) = 0.0778.

(b) z = (49.5 − 40)/6 = 1.58; P (X > 49.5) = P (Z > 1.58) = 1 − 0.9429 = 0.0571. (c) z 1 = (34.5 − 40)/6 = −0.92 and z 2 = (46.5 − 40)/6 = 1.08;

P (34.5 < X < 46.5) = P (−0.92 < Z < 1.08) = 0.8599 − 0.1788 = 0.6811.

6.34 µ = (180)(1/6) = 30 and σ = (180)(1/6)(5/6) = 5. (a) z = (24.5 − 30)/5 = −1.1; P (X > 24.5) = P (Z > −1.1) = 1 − 0.1357 = 0.8643.

(b) z 1 = (32.5 − 30)/5 = 0.5 and z 2 = (41.5 − 30)/5 = 2.3.

P (32.5 < X < 41.5) = P (0.5 < Z < 2.3) = 0.9893 − 0.6915 = 0.2978.

(c) z 1 = (29.5 − 30)/5 = −0.1 and z 2 = (30.5 − 30)/5 = 0.1.

P (29.5 < X < 30.5) = P (−0.1 < Z < 0.1) = 0.5398 − 0.4602 = 0.0796.

6.35 (a) p = 0.05, n = 100 with µ = 5 and σ = (100)(0.05)(0.95) = 2.1794.

So, z = (2.5 − 5)/2.1794 = −1.147; P (X ≥ 2) ≈ P (Z ≥ −1.147) = 0.8749. (b) z = (10.5 − 5)/2.1794 = 2.524; P (X ≥ 10) ≈ P (Z > 2.52) = 0.0059.

6.36 n = 200; X = The number of no shows with p = 0.02. z = √ 3−0.5−4 (200)(0.02)(0.98) = −0.76. Therefore, P (airline overbooks the flight) = 1 − P (X ≥ 3) ≈ 1 − P (Z > −0.76) =

6.37 (a) P (X ≥ 230) = P Z > 230−170

(b) Denote by Y the number of students whose serum cholesterol level exceed 230 among the 300. Then Y ∼ b(y; 300, 0.0228 with µ = (300)(0.0228) = 6.84 and p

σ= (300)(0.0228)(1 − 0.0228) = 2.5854. So, z = 8−0.5−6.84

= 0.26 and P (X ≥ 8) ≈ P (Z > 0.26) = 0.3974.

6.38 (a) Denote by X the number of failures among the 20. X ∼ b(x; 20, 0.01) and P (X >

20 0 20 20 1) = 1−b(0; 20, 0.01)−b(1; 20, 0.01) = 1− 19

(b) n = 500 and p = 0.01 with µ = (500)(0.01) = 5 and σ = (500)(0.01)(0.99) = 2.2249. So, P (more than 8 failures) ≈ P (Z > (8.5 − 5)/2.2249) = P (Z > 1.57) =

R 2.4

6.39 P (1.8 < X < 2.4) =

− 3.4e = 0.1545. R ∞

−x

−x

1.8 xe dx = [−xe −e ]| 1.8 = 2.8e

−x 2.4 −1.8

6.40 P (X > 9) = ∞ 1

9 9 x −x/3 dx =

3 e −x/3

−e −x/3

9 = 4e −3 = 0.1992.

6.41 Setting α = 1/2 in the gamma distribution and integrating, we have

x −1/2 e −x/β dx = 1.

Solutions for Exercises in Chapter 6

77 Substitute x = y 2 /2, dx = y dy, to give

since the quantity in parentheses represents one-half of the area under the normal curve √ n(y; 0, β).

6.42 (a) P (X < 1) = 4

0 xe −2x

−2x dx = [−2xe 1 −e ]|

dx = [−2xe

−2x

−e ]| 2 = 5e −4 = 0.0916.

6.43 (a) µ = αβ = (2)(3) = 6 million liters; σ 2 = αβ 2 = (2)(9) = 18.

(b) Water consumption on any given day has a probability of at least 3/4 of falling √ in the interval µ ± 2σ = 6 ± 2

18 or from −2.485 to 14.485. That is from 0 to 14.485 million liters.

2 6.44 (a) µ = αβ = 6 and σ 2 = αβ = 12. Substituting α = 6/β into the variance formula we find 6β = 12 or β = 2 and then α = 3. (b) P (X > 12) = 1

16 12 x 2 e −x/2 dx. Integrating by parts twice gives

P (X > 12) =

−2x ∞ − 8xe − 16e

e −x/2

R 3 −x/4

6.45 P (X < 3) = 1

Let Y be the number of days a person is served in less than 3 minutes. Then

P 6 −3/4

6 4 2 6 P (Y ≥ 4) = 5 b(y; 6, 1 − e )=

6.46 P (X < 1) =

= 0.3935. Let Y be the number of switches that fail during the first year. Using the normal approximation we find p µ = (100)(0.3935) = 39.35, σ =

e −x/2

(100)(0.3935)(0.6065) = 4.885, and z = (30.5 − 39.35)/4.885 = −1.81. Therefore, P (Y ≤ 30) = P (Z < −1.81) = 0.0352.

R ∞ −x 2 /2

6.47 (a) E(X) =

6.48 µ = E(T ) = αβ 1/β

0 t e −αt dt. Let y = αt , then dy = αβt β−1 dt and t = (y/α) . Then

µ= 1/β (y/α) e −y dy = α −1/β

y (1+1/β)−1 e −y dy = α −1/β Γ(1 + 1/β).

78 Chapter 6 Some Continuous Probability Distributions Z ∞

E(T 2/β ) = αβ t e −αt dt = (y/α) e −y dy = α −2/β

y (1+2/β)−1 e −y dy

=α −2/β Γ(1 + 2/β). So, σ 2 = E(T 2 )−µ 2 =α −2/β

6.49 R(t) = ce − R 1/ t dt = ce −2 t . However, R(0) = 1 and hence c = 1. Now

√ / t,

f (t) = Z(t)R(t) = e −2 t

t dt = −e −4 =e = 0.0183.

6.50 f (x) = 12x 2 (1 − x), 0 < x < 1. Therefore,

Z 1 P (X > 0.8) = 12 2 x (1 − x) dx = 0.1808.

6.51 α = 5; β = 10; (a) αβ = 50.

(b) σ 2 = αβ 2 = 500; so σ = 500 = 22.36.

(c) P (X > 30) = β α Γ(α) 30 x α−1 e −x/β dx. Using the incomplete gamma with y = x/β, then

Z 3 y 4 e −y

1 − P (X ≤ 30) = 1 − P (Y ≤ 3) = 1 − dy = 1 − 0.185 = 0.815.

(a) Using integration by parts,

x β α−1 0 e Γ(α) −x/β dx. Using the incomplete gamma with y = x/β, we have

P (X < 10) = P (Y < 2) =

ye −y dy = 0.5940.

6.53 µ = 3 seconds with f (x) = 1

3 e −x/3 for x > 0.

Solutions for Exercises in Chapter 6

5 =e −5/3 = 0.1889. (b) P (X > 10) = e −10/3 = 0.0357.

(a) P (X > 5) = 5 3 e −x/3

6.54 P (X > 270) = 1 − Φ ln 270−4

4 12 6.55 µ = E(X) = e 4 =e ;σ =e (e − 1) = e (e − 1).

6.56 β = 1/5 and α = 10. (a) P (X > 10) = 1 − P (X ≤ 10) = 1 − 0.9863 = 0.0137.

(b) P (X > 2) before 10 cars arrive.

1 x α−1 e −x/β

Given y = x/β, then

Z 10

Z 10

y 10−1 e −y P (X ≤ 2) = P (Y ≤ 10) =

with P (X > 2) = 1 − P (X ≤ 2) = 1 − 0.542 = 0.458.

e 0 −10x dx = e 6.57 (a) P (X > 1) = 1 − P (X ≤ 1) = 1 − 10 −10 = 0.000045. (b) µ = β = 1/10 = 0.1.

6.58 Assume that Z(t) = αβt β−1 , for t > 0. Then we can write f (t) = Z(t)R(t), where − R Z(t) dt = ce − R αβt β−1 R(t) = ce dt

= ce −αt β . From the condition that R(0) = 1, we find β that c = 1. Hence R(t) = e αt and f (t) = αβt β−1 e −αt , for t > 0. Since

f (t)

Z(t) =

−αt R(t) = 1 − F (t) = 1 − β dx = 1 + de =e ,

then

αβt β β−1 e −αt

Z(t) =

6.59 µ = np = (1000)(0.49) = 490, σ = √npq = (1000)(0.49)(0.51) = 15.808.

P (481.5 < X < 510.5) = P (−0.54 < Z < 1.3) = 0.9032 − 0.2946 = 0.6086.

80 Chapter 6 Some Continuous Probability Distributions R ∞

6.60 P (X > 1/4) = 1/4 6e −6x

6.61 P (X < 1/2) = 108 0 x e −6x dx. Letting y = 6x and using Table A.24 we have

Z 3 P (X < 1/2) = P (Y < 3) = 2 y e −y dy = 0.577.

6.62 Manufacturer A:

P (X ≥ 10000) = P Z≥ = P (Z ≥ −2) = 0.9772.

Manufacturer B:

P (X ≥ 10000) = P Z≥ = P (Z ≥ −3) = 0.9987.

Manufacturer B will produce the fewest number of defective rivets.

6.63 Using the normal approximation to the binomial with µ = np = 650 and σ = √npq = 15.0831. So,

P (590 ≤ X ≤ 625) = P (−10.64 < Z < −8.92) ≈ 0.

6.64 (a) µ = β = 100 hours.

R ∞ (b) P (X ≥ 200) = 0.01 200 e −0.01x dx = e −2 = 0.1353.

6.65 (a) µ = 85 and σ = 4. So, P (X < 80) = P (Z < −1.25) = 0.1056. (b) µ = 79 and σ = 4. So, P (X ≥ 80) = P (Z > 0.25) = 0.4013.

6.66 1/β = 1/5 hours with α = 2 failures and β = 5 hours. (a) αβ = (2)(5) = 10.

(b) P (X ≥ 12) = ∞

R ∞ 1 −x/5

12 5 2 Γ(2) xe

dx = 25 12 xe −x/5 dx = − 5 e −x/5 −e −x/5

6.67 Denote by X the elongation. We have µ = 0.05 and σ = 0.01. (a) P (X ≥ 0.1) = P Z ≥ 0.1−0.05

0.01 = P (Z ≥ 5) ≈ 0.

(b) P (X ≤ 0.04) = P Z ≤ 0.04−0.05

0.01 = P (Z ≤ −1) = 0.1587.

(c) P (0.025 ≤ X ≤ 0.065) = P (−2.5 ≤ Z ≤ 1.5) = 0.9332 − 0.0062 = 0.9270.

6.68 Let X be the error. X ∼ n(x; 0, 4). So,

P (fails) = 1 − P (−10 < X < 10) = 1 − P (−2.25 < Z < 2.25) = 2(0.0122) = 0.0244.

Solutions for Exercises in Chapter 6

6.69 Let X be the time to bombing with µ = 3 and σ = 0.5. Then

= P (−4 ≤ Z ≤ 2) = 0.9772.

P (of an undesirable product) is 1 − 0.9772 = 0.0228. Hence a product is undesirable is 2.28% of the time.

1 R 200

6.70 α = 2 and β = 100. P (X ≤ 200) = β 2 0 xe −x/β dx. Using the incomplete gamma

table and let y = x/β,

0 ye dy = 0.594.

−y

6.71 µ = αβ = 200 hours and σ 2 = αβ 2 = 20, 000 hours.

6.72 X follows a lognormal distribution.

ln 50, 000 − 5

P (X ≥ 50, 000) = 1 − Φ = 1 − Φ(2.9099) = 1 − 0.9982 = 0.0018.

6.73 The mean of X, which follows a lognormal distribution is µ = E(X) = e µ+σ 2 /2 =e 7 .

6.74 µ = 10 and σ = 50. (a) P (X ≤ 50) = P (Z ≤ 5.66) ≈ 1.

(b) P (X ≤ 10) = 0.5. (c) The results are very similar.

R 1 9 10 1

6.75 (a) Since f (y) ≥ 0 and 0 10(1 − y) dy = − (1 − y) | 0 = 1, it is a density function.

10 (b) P (Y > 0.6) = − (1 − y) 1 |

ze −z/10

(b) Using integral by parts twice, we get

E(Z )=

z e −z/10 dz = 200.

10 0 So, σ 2 = E(Z 2 2 )−µ 2 = 200 − (10) = 100.

z/10 (c) P (Z > 10) = − e ∞

10 =e −1 = 0.3679.

6.77 This is an exponential distribution with β = 10. (a) µ = β = 10.

82 Chapter 6 Some Continuous Probability Distributions

2 (b) σ 2 =β = 100.

6.78 µ = 0.5 seconds and σ = 0.4 seconds.

(a) P (X > 0.3) = P Z > 0.3−0.5 0.4 = P (Z > −0.5) = 0.6915.

(b) P (Z > −1.645) = 0.95. So, −1.645 = x−0.5

0.4 yields x = −0.158 seconds. The negative number in reaction time is not reasonable. So, it means that the normal

model may not be accurate enough.

6.79 (a) For an exponential distribution with parameter β,

P (X > a + b)

e −a−b

P (X > a + b | X > a) = = −a =e −b = P (X > b).

P (X > a)

So, P (it will breakdown in the next 21 days | it just broke down) = P (X > 21) =

e −21/15 =e −1.4 = 0.2466. (b) P (X > 30) = e −30/15 =e −2 = 0.1353.

6.80 α = 2 and β = 50. So,

Z 10 P (X ≤ 10) = 100 49 x e −2x 50 dx.

Let y = 2x 1 with x = (y/2) and dx =

2 1/50 (50) y −49/50 dy.

Z (2)10 50 Z (2)10 50

y −49/50 e −y dy =

6.81 The density function of a Weibull distribution is

Let z = t 1 which yields t = z and dt = 1/β−1

dz. Hence,

F (y) = αβ

z 1−1/β z 1/β−1 e −αz dz = α

e −αz

−αy dz = 1 − e β .

On the other hand, since de −αy β = −αβy β−1 e −αy β , the above result follows immedi- ately.

83

Solutions for Exercises in Chapter 6

6.82 One of the basic assumptions for the exponential distribution centers around the “lack- of-memory” property for the associated Poisson distribution. Thus the drill bit of problem 6.80 is assumed to have no punishment through wear if the exponential dis- tribution applies. A drill bit is a mechanical part that certainly will have significant wear over time. Hence the exponential distribution would not apply.

6.83 The chi-squared distribution is a special case of the gamma distribution when α = v/2 and β = 2, where v is the degrees of the freedom of the chi-squared distribution. So, the mean of the chi-squared distribution, using the property from the gamma distribution, is µ = αβ = (v/2)(2) = v, and the variance of the chi-squared distribution

2 2 is σ 2 = αβ = (v/2)(2) = 2v.

6.84 Let X be the length of time in seconds. Then Y = ln(X) follows a normal distribution with µ = 1.8 and σ = 2.

(a) P (X > 20) = P (Y > ln 20) = P (Z > (ln 20 − 1.8)/2) = P (Z > 0.60) = 0.2743. P (X > 60) = P (Y > ln 60) = P (Z > (ln 60 − 1.8)/2) = P (Z > 1.15) = 0.1251.

(b) The mean of the underlying normal distribution is e 1.8+4/2 = 44.70 seconds. So, P (X < 44.70) = P (Z < (ln 44.70 − 1.8)/2) = P (Z < 1) = 0.8413.