Homework 9 (due November 24, 2009)

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1 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 <, < y <, where c is a constant. (a) Find c so that this is a proper joint density. Solution: To do this we integrate over x and y and set it equal to Then solve for c: = c x + xy ( dydx = c x + xdx = c 3 + ) = 7 6 c. Thus c = 6 7. (b) Find the marginal density function for X Solution: To find the marginal density we integrate over all y. f X (x) = 6 7 x + xy dy = 6 ( x + x ). 7 (c) Find P {X > Y } Solution: To do this we compute: P {X > Y } = 6 7 x x + xy dydx = x3 dx = = (d) Find P {Y < X < } Solution P {Y < X < } = P {Y <, X < } P {X < } = = x + x 6 dx 5/4 = /8 x + xy dydx x + xdx

2 (e) Find E[X] Solution: (g) Find E[Y ]. E[Y ] = 6 7 y E[X] = 6 7 x 3 + x dx = 5 7. x + xy dxdy = 6 y y 4 dy = = 8 7 Problem. A man and a woman agree to meet at a certain location about :3 P.M. If the man arrives at a time uniformly distributed between :5 and :45 and if the woman independently arrives at a time uniformly distributed between and, find the probability that the first to arrive waits no longer then 5 minutes. What is the probability that the man arrives first? Solution: If we let X be the amount of time after : that the man arrives and Y be the amount of time after : that the woman arrives then X and Y are both uniform random variables on the intervals (5, 45) and (, 6) respectively. Since they are independent their joint density function for X and Y is given by: { 5 < x < 45, < y < 6 8 f(x, y) = otherwise To compute the probability that the person who arrives first waits no longer then 5 minutes we need to compute that: P { Y X 5} = P { 5 Y X 5} = P { 5 + X Y 5 + X} 45 x+5 = 8 dydx = 6 5 x 5 To compute the probability that the man arrives first we must compute P {X > Y } P {X > Y } = P {X Y > } = 45 x 5 8 dydx =

3 Problem 3: Jill s bowling scores are approximately normally distributed with mean 7 and standard deviation, while Jack s scores are approximately normally distributed with mean 6 and standard deviation 5. If Jack and Jill each bowl one game, then assuming that their scores are independent random variables, approximate the probability that: (a) Jack s score is higher; (b) the total of their scores is above 35. Solution: If we let X be jacks score and Y be Jill s score then X and Y are independent normal random variables with means µ = 6 and µ = 7 and variances σ = 4 and σ = 5 respectively. Then the answer to (a) is given by: P {X Y > }. However, since Y is normal, so is Y with the same variance, and µ as its expected value. Also the sum of two normal random variables is normal with mean the sum of the means and variance the sum of the variances. Consequently Z = X Y µ +µ IS a standard normal random variable, thus: σ +σ P {X Y > } = P {Z > µ ( ) ( ) + µ } = Φ = Φ.6554 = σ + σ 65 5 For (b) we solve it in a similar way to (a) except here we are interested in P {X +Y > 35}. Since both X and Y are independent normal random variables, thenn X + Y is normal with mean µ + µ = 33 and variance σ + σ X+Y 33 = 65. Thus S = 65 is a standard normal random variable and we have: P {X + Y > 35} = P {Z > ( ) } = Φ.788 =

4 Problem 4. According to the U.S. National Center for Health Statistics, 35. percent of males and 6 percent of females never eat breakfast. Suppose that random samples of men and women are chosen. Approximate the probability that: (a) at least of these 4 people never eat breakfast; Let X =the number of men among the sample of that never eat breakfast. and let Y =the number of women among the sample of that never eat breakfast. Let S = X + Y. Assume that the likelihood of any one person never eating breakfast is independent of any other person never eating breakfast, the E[X] = 7.4 and E[Y ] = 5. Consequently E[S] =.4. Using Markov s Inequality we can obtain that: P (S ) E[S]/ =.4/ of course this is pretty useless information here since this number is bigger than and any probability is less than. However we can do better by assuming that X and Y are binomial with n = p =.35 and n = and p =.6 respectively. Since (.35)(.648) > and (.6)(.74) > so both of these binomial random variables can be approximated by normal random variable with mean np and variance np( p). Thus X is approximately normal with mean µ X = 7.4 and σx = and Y is approximately normal with mean µ Y = 5 and variance σy = Since the sum of normal random variables is normal then X + Y is normal with mean µ S = µ X + µ Y =.4 and Variance σs = σ X + σ Y = Thus we have: P (X + Y ) = P ( X + Y P (Z.35) = Φ(.47) = ( Φ(.47)) = Φ(.47).97 ) Note, that since the sum of X and Y is a discrete random variable I have included a continuity correction to compensate for this. (b) the number of the women who never eat breakfast is at least as large as the number of the men who never eat breakfast. Hint: see example 3f in chapter 6. 4

5 Using X and Y from above then we are interested in the probability P (Y X) = P (Y X ). The difference D = Y X is also a sum of approximately normal random variables and since E[Y ] = 5, V ar(y ) = 38.48, E[ X] = 7. and V ar( X) = (recall V ar(ax) = a V ar(x)). Thus we have that D is approximately normal with µ D = = 8.4 and σ D = So, ( ) D ( 8.4).5 ( 8.4) P (D ) = P P (Z.95) = Φ(.95).9744 =.56 Notice that since D is a discrete random variable being approximated by a continuous random variable I have included a continuity correction. 5

6 Problem 5. Suppose that 3 balls are chosen without replacement from an urn consisiting of 6 white balls and 9 red balls. Let X i be equal to if the ith ball sleected is white, and let it equal otherwise. Give the joint probability mass function (write out all the possible values) for : (a) X, X ; (b) X, X, X 3. Solution: For (a) we compute the values at the four possible pairs (i, j) p X,X (, ) = 6 5 p X,X (, ) = 6 9 p X,X (, ) = 9 6 p X,X (, ) = 9 8 p X,X,X 3 (,, ) = p X,X,X 3 (,, ) = p X,X,X 3 (,, ) = p X,X,X 3 (,, ) = p X,X,X 3 (,, ) = p X,X,X 3 (,, ) = p X,X,X 3 (,, ) = p X,X,X 3 (,, ) = () 6

7 Problem 6: The joint density function of X and Y is { c(9x + 3y) < x < 3, < y < f(x, y) = otherwise The constant c =, but you may leave your answers in terms of c. 45 (a) Are X and Y independent? No it can not be factorized into the product of two. Also you could compute the marginal densities and see that their product is not the joint density. (b) Find the density function of X, (c) Find P {X + Y < }. f X (x) = c 9x + 3ydy = c(9x + 3 ), < x < 3. P {X + Y < } = c y 9x + 3ydxdy = c. 7

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