1. Let A, B and C are three events such that P(A) = 0.45, P(B) = 0.30, P(C) = 0.35,

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1 1. Let A, B and C are three events such that PA =.4, PB =.3, PC =.3, P A B =.6, P A C =.6, P B C =., P A B C =.7. a Compute P A B, P A C, P B C. b Compute P A B C. c Compute the probability that exactly one of A, B and C happens. Solution. a P A B = P A + P B P A B =.1, P A C = P A + P C P A C =., P B C = P B + P C P B C =.1. b Since P A B C = P A + P B + P C P A B P A C P B C + P A B C we have P A B C = P A B C P A P B P C+P A B+P A C+P B C =.1. c We have P A B C = P A B P A B C =., P A B C = P A C P A B C =.1, P A B C = P B C P A B C =.. Now P A B C = P A P A B C P A B C P A B C =.. P A B C = P B P A B C P A B C P A B C =.1. P A B C = P C P A B C P A B C P A B C =.. So the answer is =.4.. You are dealt a -card hand from a well-shuffled deck of. a What is the probability of the union of the event that both cards are aces with the event that both cards are redhearts or diamons? b That is the probability that both cards have the same denomination e.g. both are s or both are jacks? c What is the conditional probability that the second card is a picture card J, Q, or K given that at least one of the two cards is a picture card? Solution. a Let A be the event that both cards are aces, and R be the event that both 4 6 cards are red. Then P A = since there are 4 aces. Likewise P R = 1. Next

2 P AR = 1 since there is only one good combination A, A. Hence the answer is b After the first card is drawn there are 1 cards left in the desk and 3 cards have the same denomination as the first card. Hence the answer is 3. 1 c Let B 1, B be the events that the first respectively, second card is a picture card. There are 1= 3 4 picture cards. Hence 1 P B 1 = P B = 1.3 and P B 1B =.. Thus if B is an event that at least one card is a picture card, then Hence P B 1 B = P B 1 B P B = P B 1 P B.6. P B = P B 1 + P B P B 1 B In a certain city there are two food stores. 7% of the the population use Cheap FoodStore and 3% use Green FoodStore. Among the shoopers of Cheap FoodStore 7% have household income of less than per year while among the shoopers of Green FoodStore % have household income of less than per year. a Find the probability that a randomly chosen citizen uses Green FoodStore and has household income or more per year. b What proportion of the citizens have household income or more per year. c Given that a person has a household income or more per year how likely he is to shop at Green FoodStore. Solution. Let C be event that a citizen uses Cheap FoodStore and G be event that a citizen uses Green FoodStore. Let R be the event that a citizen has a household income of or more per year. Note that P R C = 1 P R C =.3 and P R G = 1 P R G =.. a P RG = P GP R G =.3. =.1. b P R = P RG + P RC = =.36. c P G R = P RG/P R = /1. 4. Let X have cumulative distribution function F X x = x for x 1 and F X x = for x < and = 1 for x > 1,

3 3 a Find the density of f X x; b find the probability density function f V v of V = 3X + 1; c Compute the median, th and 7th percentiles of X. Solution. a F X x = x = x if x [, 1] and otherwise. b V takes value between = 1 and = 4. If v [1, 4] then P V < v = P 3X + 1 < v = P X < v 1 v 1 3 =. 3 Thus [ v f V v = dv ] 1 = v 1 1 dv = v. 9 c Equation for the k-th percentile is x = k so that x k 1 k =. Thus m =, x 1 7 = 3, x = 1.. The continuous random variable Y has density f Y y = yy + 1 for y 3 and 7 f Y y = otherwise. a Find the cumulative distribution function of Y. b Find E1/Y. c If Y 1, Y... Y 1 are independent random variables with distribution Y find the distribution of M = maxy 1, Y... Y 1. Solution. a f Y y = 7 y + y. Thus b EY = 7 F Y y = 7 3 y y + y dy = y 7 s + sds = 81 y y. 3 y + 1dy = y + 7 c P M < m = P Y 1 < m... Y 1 < m = P Y < m 1 = 3 = [ 81 m3 + 1 ] 1 7 m independent random variables W j, j = 1,..., 1 have been simulated, according to an exponential distribution with parameter λ = 1. a Let N be the number among the variables W j which are larger than ln. What is the exact probability distribution of N? b What is the approximate probability that N is 3 or less? c Answer questions a, b if the number ln is replaced by ln.

4 4 Solution. a P W > ln = e ln = 1. Accordingly N Bin1, 1 b Since 1 1 and 1 1 = is not too large we have N Pois. Thus P N 3 = P N = +P N = 1+P N = +P N = 3 e = c P W > ln = e ln = 1. Accordingly N Bin1, 1. 1.! + 1 1! +! + 3 3! P N 3 = P N = + P N = 1 + P N = + P N = ! ! 1 Since the last term is much larger than all other terms we get P N ! Let X, Y have joint density f X,Y x, y = xy + 8x 3 y 3 if x 1 and y 1 and f X,Y x, y = otherwise. a Compute the marginal density of X. b Compute EX and V X. c Find the CovX, Y. Solution. b EX = Likewise EY = EX = a p X x = x + x 3 xdx = x + x 3 x dx = Hence V X = = = c EXY = = = + 7 xydy + x dx + x 3 dx + xy + 8x 3 y 3 xydxdy = 8x 3 y 3 dy = x + x 3. = 1 1. Thus CovX, Y = x 4 dx = = x dx = = 7 1. x y dxdy + 11 = 1. 8x 4 y 4 dxdy

5 8. Toll rates at a certain bridge is $ for axis vehicles, $ 4 for 3 axis vehicles and $ 1 for vehicles with 4 or more axis. Suppose that 8% of cars passing the bridge have axis, 1% have 3 axis and 1% have 4 or more axis. Let S be the toll collected from next 1 million cars passing the bridge. a Compute ES. b Compute V S. c Compute approximately P S > 33. Solution. Let X j be the toll paid by j-th car. Then S = X 1 + X +... X 1. Next, EX = = 3 so ES = 1EX = 3. Likewise V X = =.8, so V S = 1.8 = 8 and σ S = 8 = 48. By Central Limit Theorem S N3, 48, that is S Z where Z N, 1. Thus P S > 33 P Z > 33 = P 48Z > 3 = P Z > 1. = 1 P Z < = Suppose that X and Y are independent random variables with EX = 1, σx =, EY = 4, σy = 3. Let U = X 4Y. a Find the mean and variance of U. b Compute CovX, U. c If X and Y are each normally distributed find P U 1. Solution. a EU = EX 4EY = 16 = 11. V X = V X +4 V Y = = 98. b CovX, U = CovX, X CovX, 4Y = V X = 1. c By part a U N 11, 98. Hence U = 98Z 11 where Z N, 1. Accoridngly P U 1 = P 98Z 11 1 = P 98Z 1 = P Z 1 98 P Z 1.1 = 1 P Z < Let X 1, X,... X be a sample of some unknown distribution X. Let ˆµ = X 1 + X 4 + X X 4 + X be an estimator of the population mean a Is ˆµ unbiased or not? b Express V ˆµ in terms of V X. c Compare ˆµ with the sample mean. Solution. a Eˆµ = E X1 +E X 4 +E X3 8 +E X4 +E 16 X 1 = EX = EX 16

6 6 so ˆµ is unbiased. X1 X X3 X4 b V ˆµ = V + V + V + V + V = V X =.33937V X 16 c V X = V X estimator. X 16 =.V X. So the sampe mean has a smalller variance and hence is a better 11. Suppose that W is a discrete random variable such that P W = = 1 a/4, P W = 1 = 1 + a/, and P W = = 1 3a/4, where 1 < a < 1 is an unknown parameter. 3 a Compute EW. Let,, 1, 1, 1, be the sample from the distribution W. b Find the method of moment estimator for a. c Find the maximum likelyhood estimator of a. Solution. a EW = 1 a a + 1 3a = a. 4 4 b From part a we have X = â so â MOM = X = 7 = a 1 + a 1 3a c P,, 1, 1,, = 4 4 so the likelyhood function is Thus L a = if La = ln1 a + 3 ln1 + a + ln1 3a ln 4. L a = 1 1 a a 6 1 3a. 1 + a1 3a + 61 a1 3a 61 a1 + a =. That is 36a 9a 1 = Roots have form 9 ± 98. Only one of those roots belongs to the parameter interval, so 7 that â MLE = Let X be a random variable with density f X x = 1 θ + θx if x 1 and f X x = otherwise where < θ < 1 is an unknown parameter. a Compute EX and V X. Let X 1, X... X n be a sample of this distribution. b Find the method of moment estimator for θ. c Give an estimate for the variance of the estimator from part b.

7 Solution. a EX = EX = 1 θ + θxxdx = 1 θ + θ 3 = 1 + θ 6. 1 θ + θxx dx = 1 θ + θ 3 = θ 6. V X = θ θ = θ 36. b From part a we have X = 1 + ˆθ so that ˆθ = 6 X 3. 6 c V ˆθ = 6 V X = 36V X. So the estimate ˆV ˆθ = 36 V X. To estimate the variance of n n X we can use either S = 1 n n 1 j=1 X j X which is always a reasonable estimate for the population variance or a model specific estimate 1 ˆθ After examining bags of rice in a warehouse an inspector came with the following 9% confidence interval for the average weight of a bag 9.9 ±. lbs. a Compute 99% confidence interval for the average weight of a bag; b Compute 9% lower confidence bound for the average weight of a bag; c Estimate how many bags need to be examined so that the width of the 9% confidence interval is.1 lbs. Solution. 9% confidence interval has form X s ± z.. Thus X = 9.9 and 1.96s =.. Accordingly s =. =.7 Next, 99% confidence interval has form 1.96 s.8.7 x ± z.99 = 9.9 ± = 9.9 ±.6 = [9.64, 1.16]. n b Lower confidence bound is s X z.9 = 9.9 n = = s c The width of the 9% confidence interval is z N. N. So from equation 1.96s N N =.1 we get N = 1.96s N. Plugging our estimate for ŝ.1 N =.7 we get N = Researchers investigated the physiological changes that accompany laughter. Ninety 9 subjects years old watched film clips desighed to evoke laughter. During the laughing period, the researchers measured the heart rate beats per minute of subject, with the following summary results: x = 73.1, s = 3. It is well known that the mean resting heart rate of adults is 71 beats per minute. a Test whether the true mean heart rate during laughter exceeds 71 beats per minute using α =.. b Find the power of the test at µ = 7 7

8 8 c Estimate true mean heart rate during laughter in a way that conveys information about precision and reliability. Calculate 9% CI. Solution. a We have to test H = {µ = 71} vs H a = {µ > 71}. Test statistics is z = X 71 s/. 9 In our case z =.1 3/ = 6.6. Since z > z 9. = 1.6 we have sufficient evidence to reject H. That is we have a strong evidence that laughter causes the the average heart rate to exceed 71 beats per minute. b using the formula on page 314 of the book we get 71 7 β = Φ z. + 3/ = Φ Hence the power is 1 β =.93. c The confidence interval has the form s X ± z. = 73.1 ± = 73.1 ±.6 = [7.48, 73.7]. n 9 The upper confidence bound has the form s X ± z. = 73.1 ± = 73.1 ±.6 = [7.48, 73.7]. n 9 1. A business journal investigation of the performance and timing of corporate acquisitions discovered that in a random sample of, 863 firms, 848 announced one or more acquisitions during the year. Let p be the proportion of firms which made one or more acquisitions during the year. a Give a point estimate for p; b Construct 9% confidence interval for p; c Construct 9% upper confidence bound for p. Solution. a ˆp = X n = =.96. b The confidence interval takes form ˆp + z./ 863 ˆp1 ˆp/863 + z ±z. / z./ z./863 b The upper confidence bound takes form ˆp + z.1/ z.1/863 + z.1 ˆp1 ˆp/863 + z.1 / z.1/863 =.96±.14 = [.8,.31]. =.96 ±.11 = Strategic placement of lobster traps is one of keys for a successful lobster fisherman. A study was conducted of the average distance separating traps by lobster fishermen. The trap-spacing measurements in meters for a sample of seven teams from Blue Sea fishing

9 cooperative and came with the following data: x = 81. and s = 1. Suppose that the trap-spacing has normal distribution. a Test the hypotheis that the average distance between the traps is at least 9m at the significance level α =.. b What is the problem with using the normal z statistic to find a confidence interval for the average distance between the traps? c Construct 9% confidence interval for the average distance between the traps. d One team from Blue Sea cooperative was not working during the day of test. Given 9% prediction interval for the distance between the traps used by the missing team. Solution. a We have to test H = {µ = 9} vs H a = {µ < 9}. The test statistics is t = s x 9 which in our case equals to.13. The ctritiacal value t.,6 =.447. Since /7 t < t.,6 we do not reject H. That is we do not have sufficient evidence that the average distance between the traps is in fact less that 9 m. b We do not know the standard deviation and since the number of observations is small we can not estimate it accurately enough to use z statistic. c The confidence interval takes form 81 ±.447 s /7 = [7.69, 91.31]. Note that 9 is inside the confidence interval which is in agreement with the fact that we did not reject H. d The prediction interval takes form 81 ±.447 s = [1.84, 11.16]. 7 9

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