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STAT/MA 416 Answers Homework 5 October 16, 2007 Solutions by Mark Daniel Ward PROBLEMS We always let “Z” denote a standard normal random variable in the computations below. We simply use the chart from page 222 of the Ross book, when working with a standard normal random variable; of course, a more accurate computation is possible using the computer. 1a. We compute Z 1 2 c(1 − x ) dx = c 1= −1 1 1 3 x− x = c(4/3) 3 x=−1 and thus c = 3/4. 1b. We compute F (a) = P (X ≤ a) = 0 for a ≤ −1, and F (a) = P (X ≤ a) = 1 for a ≥ 1. For −1 < a < 1, we have a Z a 3 3 1 3 3 1 3 2 2 F (a) = P (X ≤ a) = (1 − x ) = x− x = a− a + 4 3 4 3 3 −1 4 x=−1 So the cumulative distribution function of X is if a ≤ −1 0 3 2 1 3 F (a) = 4 a − 3 a + 3 if −1 < a < 1 1 if a ≥ 1 2. We first compute ∞ Z ∞ −x/2 e e−x/2 − dx 1= Cxe dx = C x −1/2 x=0 −1/2 0 0 Z ∞ ∞ = 2C e−x/2 dx = −4Ce−x/2 x=0 = 4C Z ∞ −x/2 0 Thus C = 1/4. Now we compute the probability that the system functions for at least 5 months: −x/2 ∞ Z ∞ Z ∞ −x/2 1 −x/2 1 e e P (X ≥ 5) = xe dx = x − dx 4 4 −1/2 x=5 −1/2 5 5 Z ∞ ∞ 7 1 1 −5/2 −5/2 −x/2 = 10e +2 e dx = 10e − 4 e−x/2 x=5 = e−5/2 4 4 2 5 3a. The function ( C(2x − x3 ) if 0 < x < 5/2 f (x) = 0 otherwise 1 2 √ √ 3 cannot be a probability density function. R ∞ To see this, note 2x − x = −x(x − 2)(x + 2). If C = 0, then f (x) = 0 for all x, so −∞ f (x) = 0, but every probability density function √ R∞ has −∞ f (x) = 1. If C > 0, then f (x) < 0 on the range x ∈ (0, 2), but every probability √ density function has f (x) ≥ 0 for all x. If C < 0, then f (x) < 0 on the range x ∈ ( 2, 5/2), but every probability density function has f (x) ≥ 0 for all x. So no value of C will satisfy all of the properties needed for a probability density function. 3b. The function ( C(2x − x2 ) if 0 < x < 5/2 f (x) = 0 otherwise 2 cannot be a probability R ∞density function. To see this, note 2x−x = −x(x−2). RIf∞C = 0, then f (x) = 0 for all x, so −∞ f (x) = 0, but every probability density function has −∞ f (x) = 1. If C > 0, then f (x) < 0 on the range x ∈ (0, 2), but every probability density function has f (x) ≥ 0 for all x. If C < 0, then f (x) < 0 on the range x ∈ (2, 5/2), but every probability density function has f (x) ≥ 0 for all x. So no value of C will satisfy all of the properties needed for a probability density function. 4a. We compute Z ∞ P (X > 20) = 20 10 −1 ∞ dx = 10(−1)x = 1/2 x=20 x2 4b. We have F (a) = P (X ≤ a) = 0 for a ≤ 10. For a > 10, we compute Z a Z 10 Z a a 10 10 dx = 10(−1)x−1 x=10 = 1 − F (a) = P (X ≤ a) = f (x) dx = 0 dx + 2 a −∞ −∞ 10 x So the cumulative distribution function of X is ( 1 − 10 a F (a) = 0 if a > 10 otherwise 4c. The probability that one such device will function for at least 15 hours is P (X ≥ 15) = 10 1 − P (X < 15) = 1 − F (15) = 1 − 1 − 15 = 10/15 = 2/3. If all 6 devices are independent, and Y denotes the number of the 6 devices that function for at least 15 hours each, then P (Y ≥ 3) = P (Y = 3) + P (Y = 4) + P (Y = 5) + P (Y = 6) 6 6 6 6 3 3 4 2 5 1 (2/3) (1/3) + (2/3) (1/3) + (2/3) (1/3) + (2/3)6 (1/3)0 = 3 4 5 6 ≈ .8999 5. Let X denote the volume of sales in a week, given in thousands of gallons. We have the density of X, and we want a with P (X > a) = .01. Note that we need 0 < a < 1. We have 1 Z ∞ Z 1 Z ∞ (1 − x)5 4 .01 = P (X > a) = f (x) dx = 5(1 − x) dx + 0 dx = 5 (−1) = (1 − a)5 5 a a 1 x=a √ √ 5 5 5 So .01 = (1 − a) , and thus .01 = 1 − a, so a = 1 − .01 ≈ .6019. 3 6a. We compute E[X] = = = = ∞ ∞ Z 1 1 ∞ 2 −x/2 −x/2 (x) (x)(0) dx + xf (x) dx = (x)(e ) dx = xe dx 4 4 0 0 −∞ −∞ Z ∞ Z ∞ −x/2 ∞ −x/2 −x/2 ∞ 1 e 1 e 2e 2 −x/2 x 2x dx = x xe dx − +4 4 −1/2 x=0 −1/2 4 −1/2 x=0 0 0 ∞ Z ∞ −x/2 −x/2 ∞ 1 e e−x/2 2e x dx + 4x −4 4 −1/2 x=0 −1/2 x=0 −1/2 0 −x/2 ∞ −x/2 ∞ −x/2 ∞ e 1 e e 1 x2 + 4x +8 = (16) = 4 4 −1/2 x=0 −1/2 x=0 −1/2 x=0 4 Z Z 0 Z 6b. We compute ∞ Z −1 Z c(1 − x ) dx + −1 −∞ −∞ ∞ Z 2 0 dx + f (x) dx = 1= 1 Z 1 1 x3 0 dx = c x − = c(4/3) 3 x=−1 and thus c = 3/4. Now we compute Z ∞ −1 Z E[X] = Z xf (x) dx = −∞ x2 x − 2 4 = (3/4) x(3/4)(1 − x ) dx + −1 = (3/4) x=−1 ∞ Z 2 (x)(0) dx + −∞ 1 4 1 (x)(0) dx 1 1 1 − 4 4 =0 6c. We compute Z ∞ 5 Z x (x)(0) dx + xf (x) dx = E[X] = ∞ Z −∞ −∞ 5 5 dx = 5 ln x|∞ x=5 = +∞ x2 7. We compute Z ∞ 1= Z 0 Z f (x) dx = 0 dx + −∞ −∞ 1 2 Z ∞ 1 x3 0 dx = ax + b = a + b/3 3 x=0 Z 1 (a + bx ) dx + 0 1 Also Z ∞ 3/5 = E[X] = Z 0 xf (x) dx = −∞ (x)(0) dx + −∞ 2 Z x(a + bx ) dx + 0 ∞ (x)(0) dx 1 1 2 x4 x = a/2 + b/4 = a +b 2 4 x=0 We solve the system 1 = a + b/3 and 3/5 = a/2 + b/4 to obtain a = 3/5 and b = 6/5. 4 8. We compute Z ∞ Z 0 Z ∞ Z ∞ xf (x) dx = (x)(0) dx + (x)(xe ) dx = x2 e−x dx −∞ −∞ 0 0 Z ∞ Z ∞ −x −x ∞ −x ∞ e −x 2e 2e 2x xe dx = x − dx = x +2 −1 x=0 −1 −1 x=0 0 0 Z ∞ −x −x ∞ −x ∞ e e 2e = x + 2x − 2 dx −1 x=0 −1 x=0 −1 0 ∞ ∞ −x ∞ e−x e−x 2e = x + 2x +2 =2 −1 x=0 −1/2 x=0 −1 x=0 E[X] = −x 10a. The times in which the passenger would go to destination A are 7:05–7:15, 7:20–7:30, 7:35–7:45, 7:50–8:00, i.e., a total of 40 out of 60 minutes. So 2/3 of the time, the passenger goes to destination A. 10b. The times in which the passenger would go to destination A are 7:10–7:15, 7:20–7:30, 7:35–7:45, 7:50–8:00, 8:05–8:10, i.e., a total of 40 out of 60 minutes. So 2/3 of the time, the passenger goes to destination A. 11. To interpret this statement, we say that the location of the point is X inches from the left-hand-edge of the line, with 0 ≤ X ≤ L. Since the point is chosen at random on the line, with no further clarification, it is fair to assume that the distribution is uniform, i.e., X has density function f (x) = L1 for 0 ≤ x ≤ L, and f (x) = 0 otherwise. The ratio of the shorter to the longer segment is less than 1/4 if X ≤ 15 L or X ≥ 54 L. So the ratio of the shorter to the longer segment is less than 1/4 with probability Z 1L Z L 5 1 1 dx + dx = 1/5 + 1/5 = 2/5 4 L L 0 L 5 1 13a. Let X denote the time (in minutes) until the bus arrives. Then X has density f (x) = 30 for 0 ≤ x ≤ 30, and f (x) = 0 otherwise. R∞ R 30 1 The probability that we wait longer than 10 minutes is 10 f (x) dx = 10 30 dx = 2/3. 13b. Let E denote the event that the bus arrives after 10:25; let F denote the event that the bus arrives at 10:15. Then the desired probability is R 30 1 R∞ f (x) dx dx P (E ∩ F ) 1/6 30 = R25 = = 1/3 P (E | F ) = = R25 ∞ 30 1 P (F ) 1/2 f (x) dx dx 15 15 30 14. Using proposition 2.1, we compute Z ∞ Z 0 Z 1 Z n n n n E[X ] = x fX (x) dx = (x )(0) dx + (x )(1) dx + −∞ = −∞ 0 1 ∞ n Z (x )(0) dx = 1 xn dx 0 n+1 1 x 1 = n + 1 x=0 n + 1 To use the definition of expectation, we need to find the density of the random variable Y = X n . We first find the cumulative distribution function of Y . We know that P (Y ≤ a) = 0 5 for a ≤ 0, and P (Y ≤ a) = 1 for a ≥ 1. For 0 < a < 1, we have P (Y ≤ a) = P (X n ≤ a) = √ R na √ √ P (X ≤ n a) = 0 1 dx = n a. Therefore, the cumulative distribution function of Y is (√ n a if 0 < a < 1 FY (a) = 0 otherwise So the density of Y is ( fY (x) = 1 1 n a −1 n 0 if 0 < x < 1 otherwise So the expected value of Y is Z ∞ Z 0 Z 1 Z ∞ Z 1 1 1 −1 1 1/n E[Y ] = xfY (x) dx = (x)(0) dx + (x) x n dx + x dx (x)(0) dx = n −∞ −∞ 0 1 0 n 1 1 1 x n +1 1 = 1 = n n + 1 n+1 x=0 15a. We compute 5 − 10 X − 10 > P (X > 5) = P = P (Z > −5/6) = P (Z < 5/6) ≈ Φ(.83) ≈ .7967 6 6 15b. We compute X − 10 16 − 10 4 − 10 < < P (4 < X < 16) = P 6 6 6 = P (−1 < Z < 1) = P (Z < 1) − P (Z < −1) = Φ(1) − (1 − Φ(1)) ≈ .6826 15c. We compute X − 10 8 − 10 P (X < 8) = P < = P (Z < −1/3) = 1−P (Z < 1/3) ≈ 1−Φ(.33) ≈ .3707 6 6 15d. We compute 20 − 10 X − 20 < = P (Z < 5/3) ≈ Φ(1.67) ≈ .9525 P (X < 20) = P 6 6 15e. We compute X − 20 16 − 10 P (X > 16) = P > = P (Z > 1) = 1 − P (Z < 1) = 1 − Φ(1) ≈ .1587 6 6 16. Let X denote the rainfall in a given year. Then X has a uniform (µ = 40, σ = 4) 50−40 distribution, so P (X ≤ 50) = P X−40 ≤ = P (Z ≤ 2.5) = Φ(2.5) ≈ .9938, where Z 4 4 has a standard normal distribution. If the rainfall in each year is independent of all other years, then it follows that the desired probability is (P (X ≤ 50))10 ≈ (.9938)10 ≈ .9397. 19. We compute .10 = P (X > c) = P Thus Φ c−12 2 = .90. So X − 12 c − 12 > 2 2 c−12 2 =P ≈ 1.28. So c ≈ 14.56. c − 12 Z> 2 =1−Φ c − 12 2 6 20a. We have n = 100 people, each of which is in favor of a proposed rise in school taxes with probability p = .65. The number of the 100 people who are in favor of the rise in taxes is a Binomial (n = 100, p = .65) random variable, which is well-approximated by a normal random variable X with mean np = 65 and variance np(1 − p) = 22.75. So the probability that at least 50 are in favor of the proposition is approximately X − 65 49.5 − 65 P (X > 49.5) = P √ > √ ≈ P (Z > −3.25) = Φ(3.25) ≈ .9994 22.75 22.75 20b. The desired probability is approximately 59.5 − 65 X − 65 70.5 − 65 √ P (59.5 < X < 70.5) = P ≈ P (−1.15 < Z < 1.15) <√ < √ 22.75 22.75 22.75 = P (Z < 1.15) − P (Z < −1.15) = Φ(1.15) − (1 − Φ(1.15)) ≈ .7498 20c. The desired probability is approximately 74.5 − 65 X − 65 < √ ≈ P (Z < 1.99) = P (Z < 1.99) ≈ .9767 P (X < 74.5) = P √ 22.75 22.75 21. We write X for the height of a randomly-selected 25-year-old man. We compute X − 71 74 − 71 P (X > 74) = P √ > √ = P (Z > 1.2) = 1 − Φ(1.2) ≈ 1 − .8849 = .1151 6.25 6.25 Also, P (X > 77 | X > 72) = = X−71 √ 6.25 P > P (X > 77) P (X > 77 and X > 72) = = P (X > 72) P (X > 72) √ P X−71 > 6.25 77−71 √ 6.25 72−71 √ 6.25 P (Z > 2.4) 1 − Φ(2.4) 1 − .9918 = ≈ ≈ .0238 P (Z > .4) 1 − Φ(.4) 1 − .6554 23. Let X denote the number of times “6” shows during 1000 independent rolls of a fair die. Then X is a Binomial random variable with n = 1000 and p = 1/6. So X is approximately normal with mean np = 1000/6 and variance np(1 − p) = 1000(1/6)(5/6). Thus P (150 ≤ X ≤ 200) = P (149.5 ≤ X ≤ 200.5) =P 149.5 − (1000/6) X − (1000/6) 200.5 − (1000/6) p ≤p ≤p 1000(1/6)(5/6) 1000(1/6)(5/6) 1000(1/6)(5/6) ! ≈ P (−1.46 ≤ Z ≤ 2.87) = P (Z ≤ 2.87) − P (Z ≤ −1.46) = P (Z ≤ 2.87) − P (Z ≥ 1.46) = P (Z ≤ 2.87) − (1 − P (Z ≤ 1.46)) = Φ(2.87) − (1 − Φ(1.46)) ≈ .9979 − (1 − .9279) = .9258 Given that “6” shows exactly 200 times, then the remaining 800 rolls are all independent, with possible outcomes 1, 2, 3, 4, 5, each appearing with probability 1/5 on each die. Let Y denote the number of times “5” shows during the 800 rolls. Then Y is a Binomial random variable with n = 800 and p = 1/5. So Y is approximately normal with mean 7 np = 800/5 = 160 and variance np(1 − p) = 800(1/5)(4/5) = 128. Thus Y − 160 149.5 − 160 √ √ P (Y < 150) = P (Y ≤ 149.5) = P ≤ 128 128 ≈ P (Z ≤ −.93) = P (Z ≥ .93) = 1 − P (Z ≤ .93) = 1 − Φ(.93) ≈ 1 − .8238 = .1762 25. Let X denote the number of acceptable items. Then X is a Binomial random variable with n = 150 and p = .95. So X is approximately normal with mean np = (150)(.95) = 142.5 and variance np(1 − p) = (150)(.95)(.05) = 7.125. Thus 139.5 − 142.5 X − 142.5 √ P (150 − X ≤ 10) = P (140 ≤ X) = P (139.5 ≤ X) = P ≤ √ 7.125 7.125 ≈ P (−1.12 ≤ Z) = P (Z ≤ 1.12) = Φ(1.12) = .8686 26. With a fair coin, let X denote the number of heads. Then X is a Binomial random variable with n = 1000 and p = 1/2. So X is approximately normal with mean np = 500 and variance np(1 − p) = 250. Thus the probability we reach a false conclusion is 524.5 − 500 X − 500 √ P (525 ≤ X) = P (524.5 ≤ X) = P ≤ √ ≈ P (1.55 ≤ Z) 250 250 = 1 − P (Z ≤ 1.55) = 1 − Φ(1.55) ≈ 1 − .9394 = .0606 With a biased coin, let Y denote the number of heads. Then Y is a Binomial random variable with n = 1000 and p = .55. So Y is approximately normal with mean np = 550 and variance np(1 − p) = 247.5. Thus the probability we reach a false conclusion is 524.5 − 550 Y − 550 < √ P (Y < 525) = P (Y < 524.5) = P √ ≈ P (Z ≤ −1.62) 247.5 247.5 = P (Z ≥ 1.62) = 1 − P (Z ≤ 1.62) = 1 − Φ(1.62) ≈ 1 − .9474 = .0526 28. We assume that each person is left-handed, independent of all the other people. Let X denote the number of the 200 people who are lefthanded. Then X is a Binomial random variable with n = 200 and p = .12. So X is approximately normal with mean np = 200(.12) = 24 and variance np(1 − p) = 200(.12)(.88) = 21.12. Thus the probability that at least 20 of the 200 are lefthanded is 19.5 − 24 X − 24 √ P (20 ≤ X) = P (19.5 ≤ X) = P <√ ≈ P (−.98 < Z) 21.12 21.12 = P (Z < .98) = Φ(.98) ≈ .8365 32a. Let X denote the time (in hours) neededR to repair a machine. Then X is exponentially ∞ distributed with λ = 1/2. Then P (X > 2) = 2 12 e−(1/2)x dx = e−1 ≈ .368. 32b. The conditional probability is R ∞ 1 −(1/2)x e dx P (X > 10 & X > 9) P (X > 10) e−5 2 P (X > 10 | X > 9) = = = R10 = = e−1/2 ∞ 1 −(1/2)x −9/2 P (X > 9) P (X > 9) e e dx 9 2 We could also have simply computed the line above by writing P (X > 10) 1 − P (X ≤ 10) 1 − F (10) 1 − (1 − e−(1/2)10 ) e−5 = = = = −9/2 = e−1/2 −(1/2)9 P (X > 9) 1 − P (X ≤ 9) 1 − F (9) 1 − (1 − e ) e An alternative method is to simply compute that probability that the waiting time would R∞ be at least one additional hour, which is P (X > 1) = 1 12 e−(1/2)x dx = e−1/2 (or equivalently, 8 P (X > 1) = 1 − P (X ≤ 1) = 1 − F (1) = 1 − (1 − e−(1/2)1 ) = e−1/2 ). Either way, the answer is e−1/2 ≈ .6065. 33. Let X denote the time in years that the radio continues to functino for Jones. The desired probability is ∞ Z ∞ 1 −(1/8)x 1 e−(1/8)x P (X > 8) = e dx = = e−1 ≈ .368 8 8 −1/8 x=8 8 We could also have written P (X > 8) = 1 − P (X ≤ 8) = 1 − (1 − e−(1/8)8 ) = e−1 ≈ .368. 34a. We write X for the total mileage of the car, given in thousands of miles. If X is 1 exponentially distributed with λ = 20 , then the desired conditional probability is R ∞ 1 −(1/20)x e dx P (X > 30) e−3/2 P (X > 30 & X > 10) 20 = = R30 = −1/2 = e−1 P (X > 30 | X > 10) = ∞ 1 −(1/20)x P (X > 10) P (X > 10) e e dx 10 20 We could also have simply computed the line above by writing P (X > 30) 1 − P (X ≤ 30) 1 − F (30) 1 − (1 − e−(1/20)30 ) e−3/2 = = = = = e−1 P (X > 10) 1 − P (X ≤ 10) 1 − F (10) 1 − (1 − e−(1/20)10 ) e−1/2 An alternative method is to simply compute that that the lifetime has at R ∞ 1probability least 20,000 more miles, which is P (X > 20) = 20 20 e−(1/20)x dx = e−1 (or equivalently, P (X > 20) = 1 − P (X ≤ 20) = 1 − F (20) = 1 − (1 − e−(1/20)20 ) = e−1 ). Either way, the answer is e−1 ≈ .368. 34b. If X is uniformly distributed on the interval (0, 40), then the desired probability is R 40 1 dx P (X > 30 & X > 10) 1/4 40 P (X > 30 | X > 10) = = R30 = = 1/3 40 1 P (X > 10) 3/4 dx 10 40 As an alternate method, we could think of the reduced lifetime (given that the lifetime is over 10,000 miles) as between 20,000 and 40,000 miles, so the probability of the lifetime being greater than 30,000 miles is 10/30 = 1/3. 37a. We compute P (|X| > 1/2) = P (X > 1/2 or X < −1/2) = P (X > 1/2) + P (X < −1/2) = 1/2 + 1/2 = 1/2. 2 2 37b. We first compute the cumulative distribution function of Y = |X|. For a ≤ 0, we have P (Y ≤ a) = 0, since |X| is never less than a in this case. For a ≥ 1, we have P (Y ≤ a) = 1, since |X| is always less than a in this case. For 0 < a < 1, we compute 2a =a 2 or equivalently, P (Y ≤ x) = x for 0 < x < 1. In summary, the cumulative distribution function of Y = |X| is 0 x ≤ 0 FY (x) = x 0 < x < 1 1 x ≥ 1 P (Y ≤ a) = P (|X| ≤ a) = P (−a ≤ X ≤ a) = 9 Differentiating throughout with respect to x yields the density of Y = |X| is ( 1 0<x<1 fY (x) = 0 otherwise Intuitively, if X is uniformly distributed on (−1, 1), then Y = |X| is uniformly distribution on [0, 1). ( e−x 39. Since X is exponentially distributed with λ = 1, then X has density fX (x) = 0 and cumulative distribution function ( 1 − e−x x ≥ 0 FX (x) = 0 otherwise x≥0 otherwise Next, we compute the cumulative distribution function of Y = ln X. We have a P (Y ≤ a) = P (log X ≤ a) = P (X ≤ ea ) = FX (ea ) = 1 − e−e x or equivalently, P (Y ≤ x) = 1 − e−e . Differentiating throughout with respect to x yields the density of Y = ln X is x fY (x) = ex e−e 40. Since X is uniformly distributed on (0, 1), then X has density ( 1 0<x<1 fX (x) = 0 otherwise and cumulative distribution function 0 x ≤ 0 FX (x) = x 0 < x < 1 1 x ≥ 1 Next, we compute the cumulative distribution function of Y = eX . For a ≤ 1, we have P (Y ≤ a) = 0, since Y = eX is never less than a in this case (i.e., X is never less than 0 in this case). For a ≥ e, we have P (Y ≤ a) = 1, since Y = eX is always less than a in this case (i.e., X is always less than 0 in this case). For 1 < a < e (and thus 0 < ln a < 1), we compute P (Y ≤ a) = P (eX ≤ a) = P (X ≤ ln a) = FX (ln a) = ln a or equivalently, P (Y ≤ x) = ln x for 1 < x < e. In summary, the cumulative distribution function of Y = eX is x≤1 0 FY (x) = ln x 1 < x < e 1 x≥e Differentiating throughout with respect to x yields the density of Y = eX is ( 1 1<x<e fY (x) = x 0 otherwise