STAT/MA 416 Answers Homework 6 November 15, 2007 Solutions by Mark Daniel Ward PROBLEMS

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1 STAT/MA 4 Answers Homework November 5, 27 Solutions by Mark Daniel Ward PROBLEMS Chapter Problems 2a. The mass p, corresponds to neither of the first two balls being white, so p, 8 7 4/39. The mass p, corresponds to the first ball being red and the second ball 3 2 being white, so p, 8 5 /39. The mass p, corresponds to the first ball being 3 2 white and the second ball being red, so p, 5 8 /39. The mass p, corresponds 3 2 to both of the first two balls being white, so p, 5 4 5/ a. The mass p, corresponds to the white balls numbered and 2 not appearing within the three choices, so p, 3 5/2. The mass p, corresponds to the white 3 3 balls numbered not being chosen, and the white ball numbered 2 getting chosen, within the three choices, so p, 2 5/2. The mass p, corresponds to the white balls 3 3 numbered 2 not being chosen, and the white ball numbered getting chosen, within the three choices, so p, 2 5/2. The mass p, corresponds to both of the white 3 3 balls numbered and 2 appearing within the three choices, so p, 2 2 /2. 5. The only modification from problem 3a above is that the balls are replaced after each draw. Thus p, /297; also, p, /297; similarly, p, 397/297; and finally, p, / We first note that X and X 2 are independent, so the joint mass is the product of the masses of X and X 2, i.e., px, x 2 p X x p X2 x 2. For k, the probability that exactly k failures precede the first success is p X k P X k p k p, and similarly, p X2 k P X 2 k p k p. Therefore the joint mass is px, x 2 p x +x 2 p 2. 8a. To find c, we write fx, y dx dy y y cy 2 x 2 e y dx dy 8c 3 3 and thus c /8. 8b. For all x, the marginal density of X is f X x fx, y dy x 8 y2 x 2 e y dy x e x 8

2 2 For y, the marginal density of Y is f Y y fx, y dx y For y <, the marginal density of Y is f Y y. 8c. The expected value of X is E[X] xf X x dx y 8 y2 x 2 e y dx e y y 3 x x e x dx 9a. We verify that this is a joint density by noting that fx, y for all x, y, and also by computing 2 fx, y dx dy x 2 + xy dx dy 7 2 9b. For x, the marginal density of X is 2 f X x fx, y dy x 2 + xy dy x2 + x For x < and for x >, the marginal density of X is f X x. 9c. The desired probability is P X > Y x 2 + xy dx dy 5/5 7 2 We could also have computed this by writing x P X > Y x 2 + xy dy dx 5/ d. The desired probability is P Y > 2 X < P Y > /2 and X < /2 2 P X < /2 9e. The expected value of X is E[X] y xf X x dx 2 /2 /2 /2 7 x 2 + xy 2 dx dy 7 2x2 + x dx x 7 2x2 + x dx 5/7 9/8 9f. First we need the marginal density of Y. For y 2, the marginal density of Y is f Y y fx, y dx x 2 + xy dx y For y < and for y > 2, the marginal density of Y is f Y y. Now we can compute the expected value of Y, which is 2 E[Y ] yf Y y dy y y dy 8/7 a. We first compute P X < Y y e x+y dx dy /2

3 Another method is to compute P X < Y x e x+y dy dx /2 Finally, a third method is to just notice that X and Y are independent and have the same distribution, so half the time we have X < Y and the other half of the time we have Y < X. Thus P X < Y /2. b. For a <, we have P X < a. For a >, we compute P X < a a fx, y dx dy a e x+y dx dy e a Another method is to simply note that the joint density of X and Y shows us that, in this case, X and Y must be independent exponential random variables, each with λ. So P X < a e a for a >, and P X < a otherwise, since this is the cumulative distribution function of an exponential random variable. 2. We let X and Y denote, respectively, the number of men and women who enter the drugstore. We assume that X and Y are independent, and X is Poisson with mean 5, and Y is Poisson with mean 5; this agrees with the problem statement that X + Y is Poisson with mean. Therefore P X 3 Y P X 3 e 5 5! + e 5 5! + e ! + e ! 8 3 e One possibility is to compute the desired area, 3 3 units given in minutes over the entire area of the sample space, 3 8, so the desired probability is 3/8 /. Another possibility is to integrate: 45 x+5 5 x 5 8 dy dx A third possibility is to notice that, regardless of when the man arrives, the woman has a total interval of out of minutes in which she must arrive, so the desired probability is. The woman arrives first half of the time, and the man arrives first half of the time. 4. We write X for the location of the ambulence, and Y for the location of the accident, both in the interval [, L]. The distance in between is D X Y. We know that P D < a for a < and P D < a for a L, since the minimum and maximum distances for D are and L, respectively. So D must be between and L. Perhaps the easiest method for computing P D < a with a L is to draw a picture of the sample space and then divide the desired area over the entire area of the sample space; this method works since the joint distribution of X, Y is uniform. So the desired probability is 2 L a2 L 2 2 L a2 a2l a/l 2. L 2 Another possibility is to integrate, and we need to break the desired integral into three regions: P D < a a x+a dy dx + L2 L a x+a a x a L L dy dx + L2 L a x a 3 dy dx a2l a/l2 L2

4 4 9a. The marginal density of X is f X x for x and also for x. For < x <, the marginal density of X is x f X x x dy 9b. The marginal density of Y is f Y y for y and also for y. For < y <, the marginal density of Y is 9c. The expected value of X is E[X] 9c. The expected value of Y is E[Y ] f Y y y xf X x dx yf Y y dy dx ln/y x x dx /2 y ln/y dy /4 To see this, use integration by parts, with u ln/y and dv y dy. 2a. Yes, X and Y are independent, because we can factor the joint density as follows: fx, y f X xf Y y, where xe x x > e y y > f X x f Y y else else 2b. No; in this case, X and Y are not independent. To see this, we note that the density is nonzero when < x < y <. So the domain does not allow us to factor the joint density into two separate regions. For instance, P < X < > since X can be in the range 4 between /4 and. On the other hand, P < X < 4 Y 8, since X cannot be in the range between /4 and when Y /8; instead, X must always be smaller than Y. 23a. Yes, X and Y are independent, because we can factor the joint density as follows: fx, y f X xf Y y, where x x < x < 2y < y < f X x f Y y else else 23b. We compute E[X] xf Xx dx xx x dx /2. 23c. We compute E[Y ] yf Y y dy y2y dy 2/3. 23d. We compute E[X 2 ] x2 f X x dx x2 x x dx 3/. Thus VarX /2. 23e. We compute E[Y 2 ] y2 f Y y dy y2 2y dy /2. Thus VarY /8. 27a. The joint density of X, Y is fx, y e y for < x < and y >. The cumulative distribution function of Z X + Y is P Z a for a, since Z is never negative in

5 this problem. For < a <, we compute For a, we compute P Z a P Z a a a x a x e y dy dx e a + a e y dy dx e a + e a So the cumulative distribution function of Z is a F Z a P Z a e a + a < a < e a + e a a 27b. The cumulative distribution function of Z X/Y is P Z a for a, since Z is never negative in this problem. For < a, we compute X /a ay P Z a P Y a P X ay e y dxdy+ e y dxdy a e /a An alternate method of computing is to write X P Z a P Y a P a X Y So the cumulative distribution function of Z is a F Z a P Z a a e /a < a x/a /a e y dy dx a e /a 28. The cumulative distribution function of Z X /X 2 is P Z a for a, since Z is never negative in this problem. For < a, we compute ax2 X P Z a P a P X ax 2 λ λ 2 e λ x +λ 2 x 2 dx dx 2 λ a X 2 λ a + λ 2 An alternative method of computing is to write X P Z a P a P X 2 a X X 2 x /a λ λ 2 e λ x +λ 2 x 2 dx 2 dx λ a 5 λ a + λ 2 3a. The number of crashes X in a month is roughly Poisson with mean λ 2.2, so the probability that X is more than 2 is P X > 2 P X 2 e e e !! 2! 3b. The number of crashes X in two months is roughly Poisson with mean λ , so the probability that X is more than 4 is P X > 4 P X 4 e e e e e !! 2! 3! 4! 32a. We assume that the weekly sales in separate weeks is independent. Thus, the number of the mean sales in two weeks is by independence simply The variance of sales in one week is 23 2, so that variance of sales in two weeks is by independence simply

6 ,8. So the sales in two weeks, denoted by X, has normal distribution with mean 44 and variance 5,8. So P X > 5 P X 44 5,8 > P Z >.84 P Z.84 Φ ,8 32b. The weekly sales Y has normal distribution with mean 22 and variance ,9. So, in a given week, the probability p that the weekly sales Y exceeds 2 is p P Y > 2 Y 22 P > 52,9 P Z >.87 P Z <.87 Φ ,9 The probability that weekly sales exceeds 2 in at least 2 out of 3 weeks is approximately 3 2 p 2 p p a. Write X for Jill s bowling scores, so X is normal with mean 7 and variance Write Y for Jack s bowling scores, so Y is normal with mean and variance So X is normal with mean 7 and variance Thus, Y X is nomal with mean 7 and variance So the desired probability is approximately Y X P Y X > P 25 P Z > 2 5 P Z 2 5 > 25 Φ

7 Since the bowling scores are actually discrete integer values, we get an even better approximation by using continuity correction 7 P Y X > P Y X.5 Y X P 25 P Z >.42 P Z.42 Φ > b. The total of their scores, X + Y, is nomal with mean and variance So the desired probability is approximately X + Y 33 P X + Y > 35 P > 25 P Z > 4 5 P Z.8 Φ Since the bowling scores are actually discrete integer values, we get an even better approximation by using continuity correction P X + Y 35.5 P X + Y > P Z >.82 P Z.82 Φ a. The number of males X who never eat breakfast is Binomial with n 2 and p.252; the number of females who never eat breakfast is Binomial with n 2 and p.23. Thus X is approximately normal with mean np and variance npq , and Y is approximately normal with mean np and variance npq 3.8. So X +Y is approximately normal with mean and variance So the desired probability, using continuity correction,

8 8 is approximately P X + Y P X + Y 9.5 X + Y 97. P 73.7 P Z.39 P Z.39 Φ b. We note that X is approximately normal with mean 5.4 and variance So Y X is approximately normal with mean and variance So the desired probability, using continuity correction, is approximately P Y X P Y X P Y X.5 Y X 3.2 P 73.7 P Z.3 P Z.3 Φ a. The conditional mass of Y given Y 2 is and p Y Y 2 p Y Y 2 p, p Y2 p, p Y2 38b. The conditional mass of Y given Y 2 is and p Y Y 2 p Y Y p, p Y2 397/ p, p Y2 72/ p, p Y2 33/ p, p Y2 397/ a. The conditional mass function of X given Y is p X Y, p, p Y /2 and p X Y 2, p2, p Y /2

9 The conditional mass function of X given Y 2 is p X Y, 2 p, 2 p Y /3 and p X Y 2, 2 p2, 2 p Y /3 4b. Since the conditional mass of X changes depending on the value of Y, then the value of Y affects the various probabilities for X, so X and Y are not independent. 4c. We compute 9 and and P XY 3 p, + p2, + p, P X + Y > 2 p2, + p, 2 + p2, P X/Y > p2, /8 43. The density of Y given X x, for x <, is f Y X y x fx, y f X x cx 2 y 2 e x x x cx2 y 2 e x dy cx2 y 2 e x c 4 3 e x x x y2 x 3 49a. We see that minx,..., X 5 a if and only if at least one of the X i s has X i a. So P minx,..., X 5 a P minx,..., X 5 > a P X > a,..., X 5 > a P X > a P X 5 > a e λa 5 e 5λa 49b. We see that maxx,..., X 5 a if and only if all X i s have X i a. So P maxx,..., X 5 a P X a,..., X 5 a P X a P X 5 a e λa 5 54a. We see that u g x, y xy and v g 2 x, y x/y. Thus x h u, v uv and y h 2 u, v u. The Jacobian is v Jx, y y x x y x y 2x y so Jx, y y. Therefore the joint density of U, V is 2x f U,V u, v f X,Y x, y Jx, y y x 2 y 2 2x 2x 3 y 2 uv 3 u v y x y 2 2u 2 v 54b. We did not discuss these marginal densities in class, but just in case you wanted to know, we can calculate: The marginal density of U is f U u for u < ; for u, the marginal density of U is u /u lnu dv 2u 2 v u 2

10 The marginal density of V is f V v for v ; for < v, the marginal density of V is du /2 /v 2u 2 v For v >, the marginal density of V is 2u 2 v du 2v 2 v 57. We see that y g x, x 2 x + x 2 and y 2 g 2 x, x 2 e x. Thus x h y, y 2 lny 2 and x 2 y lny 2. The Jacobian is Jx, y e x ex so Jx, y e x. Therefore the joint density of Y, Y 2 is f Y,Y 2 y, y 2 f X,X 2 x, x 2 Jx, x 2 λ λ 2 e λ x λ 2 x 2 e x λ λ 2 e λ +x +λ 2 x 2 λ λ 2 e λ + lny 2 +λ 2 y lny 2 λ λ 2 y λ +λ 2 2 e y λ 2 Chapter 7 Problems 5. Since the accident occurs at a point uniformly distributed in the square, then the accident occurs at the point X, Y, where X and Y are each uniform on.5,.5, and also X and Y are independent. So the expected travel distance to the hospital is E[ X + Y ] E[ X ] + E[ Y ] x.5 3 dx + y 3 dy x/3 dx /8 3/2.5.5 x/3 dx +.5 y/3 dy +.5 y/3 dy 7a. The expected number of objects chosen by both A and B is X X + + X where if A and B both choose the ith object X i otherwise So E[X i ] P X i 3/3/ 9/. Thus E[X] 9/.9. 7b. The number of objects not chosen by both A nor B is X X + + X where if neither A nor B choose the ith object X i otherwise

11 So E[X i ] P X i 7/7/ 49/. Thus E[X] 49/ c. The number of objects chosen by exactly one of A or B is X X + + X where if exactly one of A or B choose the ith object X i otherwise So E[X i ] P X i 7/3/+3/7/ 42/. Thus E[X] 42/ a. The number of empty urns is X X + + X n where if urn i is empty X i otherwise Only the balls numbered i, i +, i + 2,..., n can possibly go into urn i. The probability that ball i goes into urn i is /i; the probability that ball i + goes into urn i is /i + ; the probability that ball i + 2 goes into urn i is /i + 2; etc., etc. So the probability that urn i is empty is i i i i+ i+2 i i + i + i + 2 n, or more simply n i n n So E[X i ] P X i i. Thus E[X] n i n i n n n i i n n n 2 n 2. 9b. There is only one way that none of the urns can be empty: Namely, the ith ball must go into the ith urn for each i. To see this, first note that the nth ball is the only ball that can go into the nth urn. Next, there are two balls, the n and n ball that can go in urn n, but the nth ball is already committed to the nth urn, so ball n must go into urn n. Next, there are three balls, the n 2, n, and n ball that can go in urn n 2, but balls n and n are already committed to urns n and n, respectively, so ball n 2 must go into urn n 2. Similar reasoning continues. So the probability that none of the urns are empty is n n n 2. n!. The number of changeovers is X X X n where if a changeover occurs from the i st flip to the ith flip X i otherwise So E[X i ] P X i p p + pp 2p p. Thus E[X] n 2p p. 2a. The number of men who have a woman sitting next to them is X X + + X n where if the ith man has a woman sitting next to him X i otherwise So E[X i ] P X i. There are two seats on the end of the aisle where a man can sit; for each such seat, the probability that he is sitting there is /2n, and the probability that a woman is sitting next to him afterwards is n/2n. So the probability a specific man sits on the left or right end, with a woman next to him, is 2 n. There are 2n 2 2n 2n 2n sets in the middle of the row; for each such seat, the probability that he is sitting there is /2n, and the probabilty that a woman is sitting next to him afterwards only on his n n n left; only on his right; both sides; respectively is n n + n + 2n 2n 2 2n 2n 2 2n 2n 2 3n. 22n

12 2 So the probability that a specific man sits in the middle, with a woman next to him, is 2n 2 3n 32n 2. So the total probability is E[X 2n 22n 42n i] P X i + 32n 2 2n 42n 3n. Thus E[X] n 3n n3n. 22n 22n 22n 2b. If the group is randomly seated at a round table, then we write X X + + X n where if the ith man has a woman sitting next to him X i otherwise So E[X i ] P X i. Regardless of where the ith man sits, the probability that a woman is sitting next to him afterwards only on his left; only on his right; both sides; respectively n n is + n n + n n 3n. So E[X 2n 2n 2 2n 2n 2 2n 2n 2 22n i] 3n. Thus E[X] n 3n 22n 22n 3n 2 22n. 3. The number of people whose age matches their card is X X + + X where if the ith person s age matches his card X i otherwise So E[X i ] P X i. Thus E[X] /.

Homework 9 (due November 24, 2009)

Homework 9 (due November 24, 2009) Homework 9 (due November 4, 9) Problem. The join probability density function of X and Y is given by: ( f(x, y) = c x + xy ) < x