STAT 516 Midterm Exam 3 Friday, April 18, 2008

Transcription

1 STAT 56 Midterm Exam 3 Friday, April 8, 2008 Name Purdue student ID (0 digits). The testing booklet contains 8 questions. 2. Permitted Texas Instruments calculators: BA-35 BA II Plus BA II Plus Professional TI-30Xa TI-30X IIS (solar) TI-30X IIB (battery) The memory of the calculator should be cleared at the start of the exam, if possible. No other calculators are permitted. 3. Simplify the answers and write them in the boxes provided. 4. Some partial credit may be awarded at the discretion of the professor. 5. Show all of your work in the exam booklet. Answers without justification may be viewed as exhibiting academic dishonesty. 6. Extra sheets of paper are available from the professor.

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3 . A point X, Y is chosen uniformly from the unit circle. In other words, X, Y are jointly distributed continuous random variables with joint density { C 0 x 2 + y 2 f(x, y) 0 otherwise where C is constant. Find the probability that the distance from (X, Y ) to the origin (0, 0) is less than /2, i.e., P ( X 2 + Y 2 /2). Answer. Method. The desired event happens if and only if (X, Y ) is found in a circle of radius /2 centered at the origin. Since the random variable have a uniform joint distribution, then the desired probability is the area of the circle centered at the origin with radius /2 (namely, (π)(/2) 2 π/4), divided by the area of the circle centered at the origin with radius (namely, (π)() 2 π). So the desired probability is π/4 /4. π Method 2. We can solve for C by computing x 2 x 2 C dy dx which yields C /π (or just noting that C is the reciprocal of the area of the entire circle of radius, centered at the origin). Then the desired probability is ( /2 ) 4 x2 dy dx /2 π ( 4 x2 x 2 x 2 π dy dx ) /4 4

4 2. Let X and Y denote the number of eggs laid by insects A and B, respectively, in a study. It is known that X and Y are independent Poisson random variables, each with parameter λ. If the total number of eggs laid is 50, what is the chance that insect A laid at least 90 eggs? [Hint: Recall that, if X, Y are independent Poissons with parameters λ, λ 2 respectively, then the conditional distribution of X, given X + Y n, is Binomial with parameters n and p λ λ +λ 2.] Answer. As in the hint, the conditional distribution of X is Binomial with n 50 and p λ /2. So the desired probability is P (X 90). Here are two methods for finding λ+λ P (X 90). Method. Note that X is approximately normal with mean np 75 and variance np( p) 37.5, so the desired probability is P (X 90) P (X 89.5) ( X 75 P 37.5 P (N 2.37) Φ(2.37) ) Method 2. Some students decided to give an expression for the exact answer. Since I did not specify whether an approximation was desired, I gave full credit for this type of answer too: ( 50 P (X 90) P (X i) i i90 i90 ) ( 2 ) i ( ) 50 i i90 ( ) 50 Unfortunately, this is difficult to compute without Maple, but I still gave full credit for this type of answer. i 2

5 3. A fair coin is tossed 300 times. Let H 00 denote the number of heads in the first 00 tosses, and H 300 the total number of heads in all 300 tosses. Find the correlation ρ of H 00, H 300, Cov(H i.e., ρ(h 00, H 300 ) 00,H 300 ). Var(H00 )Var(H 300 ) Answer. Note that H 00 is Binomial with parameters n 00 and p /2, so Var(H 00 ) (00)(/2)(/2) 25. Also, H 300 is Binomial with parameters n 300 and p /2, so Var(H 300 ) (300)(/2)(/2) 75. Now write X i to indicate whether the ith toss shows heads, i.e., X i if the ith toss shows heads, and X i 0 otherwise. So H i X i and H j X j. So Cov(H 00, H 300 ) Cov ( 00 ) X i, X j Cov (X i, X j ) i Note that X i, X j are independent if i j. So Cov(X i, X j ) 0 if i j. So the sum in the formula above simplifies to j Cov(H 00, H 300 ) Cov (X i, X i ) Var(X i ) Var i 25 So we conclude that the correlation is (25)(75) 3 i i j ( 00 ) X i Var(H 00 ) 25 i 3

6 4. Consider independent standard normal random variables W, X, Y, Z. Find P (W + X > Y + Z + ). Answer. We note that P (W + X > Y + Z + ) P (W + X Y Z > ) and also W + X Y Z is a normal random variable with mean and variance Thus, the desired probability is ( W + X Y Z 0 P > 0 ) P (N > /2) Φ(/2) where N is standard normal. 4

7 5. Consider two jointly distributed continuous random variables X and Y with joint density { 3x 0 y x f(x, y) 0 otherwise Find P ( 5 Y 2 5 ). Answer. Method. We compute using just one integral P ( 5 Y 2 ) /5 2/5 2/ y 3x 2 2 3x dx dy xy dy 3 2 ( y2 ) dy ) (y y3 2/5 3 y (( 2 5 ( ) ) Method 2. We compute using two integrals P ( 5 Y 2 ) 5 2/5 x 2/5 2/5 3x dy dx + 3xy x y dx + 2/5 3x(x ) dx + 2/5 2/5 ( 5 3 3x dy dx 3xy 2/5 y dx 2/5 ( 2 3x 5 5 ( 3x ) x 2 2/5 + 3 x x 5 2 x2/5 ( (2 ) 3 ( ) ) ( 3 (2 3 ) ) dx ( ) )) 3 5 ( ) ) ( 2 (2/5) 2) 5 0 5

8 6. A machine has a random supply Y at the beginning of a given day and dispenses a random amount X during the day (with measurements in gallons). It is not resupplied during the day; hence, X Y. It has been observed that X, Y have joint density { /2 0 x y 2 f(x, y) 0 otherwise Find the conditional expectation of X, given that Y y, for some fixed y with 0 y 2. Answer. Method. Since X, Y have a joint uniform distribution, then once we are given Y y, we know that the conditional distribution of X is uniform in the range 0 x y, so the conditional expected value of X in this case is y/2. Method 2. We compute E(X Y y) xf X Y (x y) dx y 0 f(x, y) x f Y (y) dx We already know the joint density of X, Y, but we still need to compute the marginal density of Y evaluated at y. This is y f Y (y) 2 dx y 2 x y/2 Thus E(X Y y) y 0 0 x /2 y y/2 dx x 0 y x0 x2 dx 2y y x0 y2 2y y/2 6

9 7. A complex machine is able to operate effectively as long as at least 3 of its 5 motors are functioning. Suppose that each motor independently functions for a random amount of time with density function { xe x x > 0 f(x) 0 otherwise, For fixed t > 0, compute the probability that the machine functions until at least time t. [Note: For simplicity, you do not need to simplify your answer on this problem. Just circle your final answer.] Answer. For t > 0, the probability that a particular motor functions until at least time t is xe x dx. We use u x and dv e x dx, and thus du dx and v e x, to solve the t integral by parts: t xe x dx xe x e x dx te t e x xt xt te t + e t (t + )e t t So the probability that exactly i of the 5 motors will function until at least time t is ( ) 5 ((t + )e t ) i ( (t + )e t ) 5 i i So the probability that the machine functions until at least time t is 5 i3 ( ) 5 ((t + )e t ) i ( (t + )e t ) 5 i i 7

10 8. A standard deck has 52 cards, with values A, 2, 3, 4, 5, 6, 7, 8, 9, 0, J, Q, K (4 cards of each value). Choose 5 cards randomly from the deck, without replacement. Given that exactly 2 Jack cards (i.e., value J ) appear, find the expected number of Ace cards (i.e., value A ) that appear. [Note: For simplicity, you do not need to simplify your answer on this problem. Just circle your final answer.] Answer. Method. Let X denote the number of Ace cards selected, and let Y denote the number of Jack cards selected. Then the desired conditional expectation is E(X Y 2) 3 xp X Y (x 2) dx x0 3 x0 p(x, 2) x p Y (2) 5 x 2) For 0 x 3, we note that p(x, 2) (4 x)( 4 2)( 44 probability is ( 52 5 ) ( ) x x)( 3 x ) 4 x0 ( Also, p Y (2) (4 2)( 48. So the desired ( 52 5 ) Method 2. Let X denote the number of Ace cards selected, and let Y denote the number of Jack cards selected. Then X X + X 2 + X 3 + X 4 + X 5, where X i if the ith card selected from the deck is an Ace, and X i 0 otherwise. So the desired conditional expectation is 5 E(X + + X 5 Y 2) E(X i Y 2) Each X i is an indicator function, so E(X i Y 2) P (X i Y 2) P (X i and Y 2) We note that P (Y 2) (4 2)( 48 3 ) ) 4 52 ( 4 2)( 47 2 ) ( 5 4 ). Thus i 3 ) P (Y 2). ( 52 5 ). Also P (X i and Y 2) P (X i )P (Y 2 X i E(X i Y 2) 4 52 ( 4 2)( 47 2 ) ( 5 4 ) 3 ) ( 52 5 ) ( 4 2)( So the desired probability is E(X + + X 5 Y 2) 5 i E(X i Y 2) Method 3. Since we are given that exactly two of the five cards are Jacks, then the remaining 3 cards are just chosen from 48 non-jack cards. In this new setting, we let X X + X 2 + X 3, where X i if the ith card is an Ace, and X i 0 otherwise. So E(X i ) P (X i ) 4/48 /2, and thus E(X) E(X + X 2 + X 3 ) E(X ) + E(X 2 ) + E(X 3 )