Math 408, Spring 2005 Final Exam Solutions

Transcription

1 Math 408, Spring 005 Final Exam Solutions. Assume A and B are independent events with P A) = 0. and P B) = 0.. Let C be the event that neither A nor B occurs, let D be the event that exactly one of A or B occurs a) Find P C). Solution. P C) = P A B ) = P A )P B ) = P A)) P B)) = = 0.56 b) Find P D). Solution. P D) = P A B \ A B) = P A B) P A)P B) = P A) + P B) P A B) = 0.8 c) Find P A D). Solution. P A D) = P A D)/P D) = P A \ A B)/P D) = )/0.8 = 7/9 7 = d) Are C and D independent? Justify your answer! Solution. C and D are not independent since P C D) = 0, but P C)P D) = NO 0.9. [.4-9, Variant] Suppose A, B, and C are mutually independent events with probabilities P A) = 0.5, P B) = 0.8, and P C) = 0.. Find the probability that at least one of these events occurs. Solution. The probability in question is P A B C). Using the formula for the probability of a union of three events, and the independence of the three sets, we get P A B C) = P A) + P B) + P C) P A B) P A C) P B C) + P A B C) = ) ) ) ) = 0.9.

2 840. How many ways are there to seat 0 people, consisting of 5 couples, in a row of seats 0 seats wide) if all couples are to get adjacent seats? Solution. 5! 5 = 840 or = 840) 4. [Actuarial Exam Problem #7] The probability that a randomly chosen male has a circulation problem is 0.5. Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem. What is the conditional probability that a male has a circulation problem, given that he is a smoker? Solution. Let C denote the event has a circulation problem and S the event 0.4 is a smoker. We are given that P C) = 0.5 and P S C) = P S C ), and we need to compute P C S). By Bayes Rule, P C S) = = P S C)P C) P S C)P C) + P S C )P C ) P S C)0.5 P S C)0.5 + /)P S C) 0.5) = = 5 = [Similar to Actuarial Exam Problem #57] Suppose a random variable X has moment generating function ) + e t 9 Mt) =. Calculate the variance of X. Solution. We use the formula Var X = EX ) EX) = M 0) M 0). Computing the derivatives of Mt), we get so VarX) = =. ) + e M t 8 t) = 9 et, ) + e M t 7 ) e t + e t 8 t) = 8 + e t, ) + e M 0 8 0) = e 0 =, ) 7 ) 8 M 0) = 8 + =,

3 0, [Actuarial Exam Problem #7] The lifetime of a printer costing 00 is exponentially distributed with mean years. The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, and a one-half refund if it fails during the second year. If the manufacturer sells 00 printers, how much should it expect to pay in refunds? Solution. Let X denote the lifetime of a printer, and let Y denote the refund paid by the manufacturer on a single printer. Now, 00 if X, Y = 00 if < X, 0 if X >, so EY ) = 00P X ) + 00P < X ) = 00 e / ) + 00 e / ) e / )) = 0.56, since P X x) = e x/θ = e x/ by the given exponential) distribution of X. The expected refund on 00 such printers then is 00EY ) = 0, [Actuarial Exam Problem #56] An insurance policy is written to cover a loss, X, where X has uniform distribution on [0, 000]. At what level must a deductible be set in order for the expected payment to be 5% of what it would be with no deductible? Solution. Let Y be the insurance payment and D the deductible. Then { 0 if X D, Y = X D if D X 000. Since X has density fx) = /000 on the interval [0, 0000], we get EY ) = 000 D 000 D) x D) dx = Now the expected payment with no deductible is EY ) = 500 obtained by setting D = 0 above, or by symmetry arguments), so we need to choose D such that EY ) = 5. This leads to the equation 000 D) = 5 000, and solving for D we get D = 500.

4 8. [Actuarial Exam Problem #46] A device that continuously measures and records seismic activity is placed in a remote region. The time, T, to failure of this device is exponentially distributed with mean years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is X = maxt, ). Determine EX). Solution. First note that since T has exponential distribution with mean, the.54 density of T is ft) = /)e t/ for t 0. Since X = maxt, ) equals if T, and T if T, it follows that using integration by parts for the second integral) EX) = 0 e t/ dt + t e t/ dt = [ e t/] + [ e t/ t ] + 0 e t/ dt = e / ) e / ) + / e / = + e / = [Actuarial Exam Problem #76] Claim amounts for wind damage to insured homes are independent random variables with common density function fx) = x 4 for x >, and fx) = 0 otherwise, where x is the amount of a claim in thousands. Suppose such claims are made. What is the expected value of the largest of the three claims? Solution. Let X, X, X denote the three claims, and Y = maxx, X, X ) 05 the largest of these. We need to compute EY ). This requires first computing the c.d.f. F y) and the p.d.f. fy) = F y) of Y. By the maximum trick, so F y) = P maxx, X, X ) y) = P X y)p X y)p X y) y = x dx) 4 = ), y for y. Therefore fy) = F y) = ) y y = 9 y ) 4 y 4 EY ) = = 9 = 9 9 yfy)dy = ) dy y y y y + ) dy 6 y ) = =.05 Since the units are thousands of dollars, the answer is 05.

5 0. [5.-5] Let X have uniform distribution on the interval [0, ], and given X = x, let Y have uniform distribution on the interval [0, x ]. a) Find the joint density fx, y) of X and Y. Be sure to specify the range!) Solution. We are given that f X x) = /, 0 x uniform distribution on [0, ]), and gy x) = /x, 0 y x uniform distribution on [0, x ]). Therefore a) fx, y) = gy x)f X x) = x = x, 0 x, 0 y x, b) Find the marginal density f Y y) of Y. Be sure to specify the range!) Solution. [ f Y y) = fx, y)dx = x= y x dx = ] x= b) x x= y = 4 + y, 0 y 4 c) Find EXY ). Solution. We have EXY ) = = xyfx, y)dydx = x=0 x x ) dx = 4 x x=0 y=0 0 xy) x dydx x dx = =.

6 . [From Double Integral handout, variant of Actuarial Exam Problem #9] Let X and Y be random variables with joint density fx, y) = x y +, 0 x, y Find the probability that X + Y 0.5. Solution. The probability is given by the double integral over the above density function over the part of the unit square on which x + y 0.5: 0.5 x=0 y=0.5 x 0.5 = = = x= x y + )dydx + xy y + y ) dx + y=0.5 x x=0.5 y=0 x=0.5 x y + )dydx xy ) y + y x + x) x) ) + + x) ) dx 0.5 x + ) dx 8 + x + x ) dx + x + ) x 0.5 = = 7 8 = dx y=0. [Actuarial Exam Problem #87, Variant] A computer generates 48 random real numbers, rounds each number to the nearest integer and then computes the average of these 48 rounded values. Assume that the numbers generated are independent of each other and that the rounding errors are distributed uniformly on the interval [ 0.5, 0.5]. Find the approximate probability that the average of the rounded values is within 0.05 of the average of the exact numbers. Solution. Let X,..., X 48 denote the 48 rounding errors, and X = /48) 48 i= X i 0.77 their average. We need to compute P X 0.05). Since a rounding error is uniformly distributed on [ 0.5, 0.5], its mean is µ = 0 and its variance is σ = x dx = [x /] = /. By the Central Limit Theorem, X has approximate distribution Nµ, σ /n) = N0, /)/48) = N0, /4 ). Thus 4X is approximately standard normal, so P X 0.05) P ) 4X ) = Φ.) Φ.) = Φ.) = 0.77.

7 6. [Actuarial Exam Problem #8] A company manufactures a brand of light bulb with a lifetime in months that is normally distributed with mean and variance. A consumer buys a number of these bulbs with the intention of replacing them successively as they burn out. The light bulbs have independent lifetimes. What is the smallest number of bulbs to be purchased so that the succession of light bulbs produces light for at least 40 months with probability 0.977? Solution. Let n be the unknown) number of light bulbs to be purchased, X,..., X n their respective lifetimes, and S = n i= X i the total lifetime of all n bulbs. We need to choose n minimal so that P S 40) Now, by the CLT, ) ) S n P S 40) P 40 n 40 n n Φ n n This is equal to when 40 n)/ n =, or equivalently, n n 40 = 0. Setting x = n, the latter equation becomes x x 40 = 0. Solving ignoring the negative solution) gives x = /6) = 4, so n = x = 6 is the number sought [6.-] Assume the math scores on the SAT test are normally distributed with mean 500 and standard deviation 60, and the verbal scores are normally distributed with mean 450 and standard deviation 80. If two students who took both tests are chosen at random, what is the probability that the first student s math score exceeds the second student s verbal score? Solution. Let X and Y denote the scores of the two students. Then X Y is N , ) = N50, 00 ), so X Y 50 P X > Y ) = P X Y > 0) = P > 0 50 ) = P Z > /) Φ /) = Φ/) = [Example 6.-] Let X, X, X, X 4 be a random sample of size 4 from the normal distribution N76.4, 8), and let X be the sample mean and S the sample variance. Determine a such that P S a) = Solution. We know that S n )/σ = /8)S = S has χ ) a = 798 distribution, and from the chi-square table the 90-th percentile of a χ ) is 6.5. Thus, 0.90 = P /8)S 6.5) = P S 6.58/)) = P S 798). so a = 798.

8 [6.-b)] Suppose X and Y are independent, each having Poisson distribution with means and, respectively. Let Z = X + Y. Find P X + Y = ). Solution. P X + Y = ) = P X = 0, Y = ) + P X =, Y = 0) = e 0 /0!)e /!) + e 0 /0!)e /!) = 5e 5 = 0.06.