1. Let X be a random variable with probability density function. 1 x < f(x) = 0 otherwise

Transcription

1 Name M36K Final. Let X be a random variable with probability density function { /x x < f(x = 0 otherwise Compute the following. You can leave your answers in integral form. (a ( points Find F X (t = P (X t for t <. (b ( points E[ X ] =? (c Find the density function of Y =. Hint: first find the distribution function of X Y. Answer. For t, P (X t = t Of course, F X (t = 0 if t <. [ ] E = (/x(/x dx = X Observe that ] t dx = = /t. x x F Y (t = P (Y t = P (/X t = P (/t X = /x 3 dx = /(x ] = /. /t /x dx = /x] /t = t if /t <. Equivalently, this occurs if 0 < t. Of course, F Y (t = 0 if t 0 and F Y (t = if t. The density function is f Y (t = F Y (t = for 0 t and f Y (t = 0 if t / [0, ]. So Y is uniformly distributed over [0, ].. (0 points You throw a dart at a 0cm circular target. You win 00 dollars if your dart lands within cm of the center; 0 if your darts lands between and 3cm of the center and nothing otherwise. Suppose that your shot is uniformly distributed over the target. Find your expected winnings. Solution. E[X] = 00 π π00 + 0π(9 π00 = + =. 3. Suppose the pair (X, Y is uniformly distributed over the semi-circle defined by x +y, y > 0. (This is the upper half of the disk of radius. (a Find the joint density function of (X, Y. (b Find the density function of Y.

2 (c Find the density function of X conditioned on Y =. In other words, find f X (x Y =. Solution. (a The joint density function is { /(π x f(x, y = + y, y > 0 0 otherwise (b The density of Y is obtained by integrating the joint density function over x. It is if 0 < y < and 0 otherwise. (c y f Y (y = /(π dx = y y π f X (x Y = = f(x, f Y ( = = 6 if 3 x 3 and 0 otherwise. In other words, X conditioned on Y = is uniformly distributed on [ 3, 3].. (0 points You are dealt 3 cards at random (from a standard deck of cards. Let X be the number of aces you receive. Find the expected value and variance of X. Solution. Write X = X + + X 3 where { if the i-th card drawn is an ace X i = 0 otherwise E[X i ] = / = /3 E[X] = (3(/3 =. Now Also if i j then Var(X = Var( i X i = 3 i= Var(X i + i<j Cov(X i, X j. Var(X i = E[X i ] E[X i ] = E[X i ] E[X i ] = (/ (/. Cov(X i, X j = E[X i X j ] E[X i ]E[X j ] = (/(3/ (/. So ( 3 Var(X = 3[(/ (/ ] + [(/(3/ (/ ] = 3(/[ / + ((3/ (/] = (/3 + ((3/ (/3] = 36/.

3 . (0 points An achievement test is given to a certain population of students. The score of a randomly selected student is a random variable with expectated value 00 and standard deviation 00. Let Z be the MEAN (not the sum of the scores of a random sample of 0 students. Use Chebyshev s inequality to find an upper bound for P ( Z (Do not use the Central Limit Theorem or a normal approximation. Solution. Note that Z = X + X 0 0 has expectation 00 and variance equal to 0(00 0 = 000. So by Chebyshev, P ( Z = (0 points You have two friends, Alice and Bob and they both promised to call you sometime after 7pm on Saturday. So you are sitting at home at 7pm on Saturday and you decide to go out with either Alice or Bob, whoever calls you first. Let A be the amount of time after 7pm before Alice calls you, and B the amount of time after 7pm before Bob calls you. Assume A and B are independent exponential random variables; the expectation of A is hour and the expectation of B is 30 minutes. What is the probability that you will go out with Alice? Solution. The question asks for P (B A. Note that the joint density function of (A, B is f(x, y = (/60(/30 exp( x/60 y/30 if x, y > 0 are 0 otherwise. So the answer is 0 x (/60(/30 exp( x/60 y/30 dydx = /3. 7. (0 points If X and Y are independent random variables, X is normal with mean 00 and standard deviation 0 and Y is normal with mean 00 and standard deviation 30 then what is the probability that X < Y? Write your answer in terms of the function Φ. Solution. P (X < Y = P (X Y < 0. Now E[X Y ] = 00 and Var(X Y = Var(X + Var(Y = = 0. So ( X Y + 00 P (X Y < 0 = P (X Y + 00 < 00 = P < = Φ( 0 since X Y is a standard normal random variable. 8. Each time that Jim charges an item to this credit card, he rounds the amount to the nearest dollar in his records. Assuming he has used his credit card 300 times in the last months, use the central limit theorem to approximate the probability that his record differs from the total expenditure by, at most, 0 dollars. You may assume that, for each number Jim records, the difference between the actual number and the recorded number is uniformly distributed over [ /, /]. Hence the standard deviation for one such difference is. Leave your answer in terms of the Φ-function.

4 Solution. Let S = X + + X 300 where X i is the difference between the recorded amount of the i-th item and the actual amount. So X i is uniformly distributed over [ /, /]. Observe E[S] = 0, V ar(s = 300/ = and SD(S =. So P ( S 0 = P ( S / P ( χ = Φ( (0 points For a survey, we contact N people at random and ask them whether they plan to vote for candidate A. How large should N be so that there is at least a 9% chance that the sample mean determined by the survey is within % of the true mean? Use the normal approximation. Hint: if χ is a standard normal then P ( χ 0.9. Solution. Let p denote the true mean and S the number of people in our survey that plan to vote for candidate A. Then we want to choose N large enough so that ( S P N p Now S p is approximately normal with mean 0 and variance N ( ( S S Var N p = Var = Var(S/N = N p( p/n = p( p/n. N We don t know what p is, but we do know that p( p /. So ( S Var N p N. Therefore, the standard deviation of S p is at most N ( S/N p P SD(S/N p 0.0 P SD(S/N p = P ( χ 0.0 N N ( χ Remember that P ( χ 0.9. So we want to choose N so that In other words, N = (00 = 0, N =... So we choose N so that 0.0 SD(S/N p 0. (0 points What is the probability that in a group of 30 people at least have the same birthday? Answer. 36! (36 30! Suppose the monthly worldwide average number of commercial airplane crashes is. (a If there are 0000 monthly flights, each flight having the same probability of crashing independently of other flights, then what is the probability that there will be at least crashes in a given month? Solution. This is a binomial random variable with n = 0000 and p = /0000. The answer is ( 0000 ( ( 0000 (

5 (b Suppose that the number of monthly flights is very large. Using Poisson random variables, find an approximation for the probability that there will be at least crashes in a given month. Solution. λ =. So the answer is e e!.. (0 points Suppose that E, E, E 3, E are four events such that P (E i = 0. (for all i, P (E i E j = 0.3 (for all i j, P (E i E j E k = 0. if i, j, k are distinct and P (E E E 3 E = 0. What is P (E E E 3 E? Answer. (0. 6( = (0 points If 7% of the population are women who own dogs and 8% of the population own dogs, then what percentage of all dog-owners are women? Solution. Let W stand for women and D for dogowners. Then P (W D = 0.07, P (D = 0.8. We want to know P (W D. P (W D = P (W D P (D = 0.7/0.8 = /.. There are 3 urns. Urn # contains 3 red balls and blue balls. Urn # contains red balls and blue balls. Urn #3 contains red balls and 6 blue balls. An urn is chosen at random and a ball is selected from the urn. (a (0 points What is the probability that a red ball is selected? Solution. Let E i be the event that the ith urn is selected and R the event that a red ball is selected. Then P (R = 3 P (R E i P (E i = (/3 (3/7 + /9 + /. i= (b (0 points Given that a red ball is selected, what is the probability that Urn # was chosen? Solution. P (E R = P (R E P (E P (R = (3/7(/3 (/3 (3/7 + /9 + /.. A school play has distinct male roles and distinct female roles. Suppose 7 men and 8 women audition for the play. (a (0 points How many possible casts are there? (b ( points What if Bob and Alice refuse to be in the play together? (it is OK for Bob to be in the play without Alice or Alice without Bob.

6 Solution. The number of possible casts is ( ( 7! 8! (choose men, then assign them roles, choose women, then assign them roles. The number of ways that Alice and Bob can both be in the play is ( ( 6 3 3! 7! (choose a role for Bob, then roles for the other males, then a role for Alice, then roles for the other females. So if Alice and Bob refuse to be in the play together the number of possible casts is ( ( 7 8!! ( 6 3 3! ( 7! = ( 6 ( 8!! + ( 7 ( 7!! ( 6! ( 7!. 6. Suppose E, E, E 3 are jointly independent events with P (E = P (E = P (E 3 = 0.. What is the probability of P (E E E 3? Note: this is a union, not an intersection of events. Solution. P (E E E 3 = P (E c E c E c 3 = (0.8 3 = 0. = (0 points Suppose 30% of the population are Democrats. You call people at random until you get a Democrat. What is the probability that you have to make at most 3 calls? Answer. If X is the number of people that you call then X is a geometric random variable with success probability 0.. So P (X 3 = P (X = + P (X = + P (X = 3 =

7 Some helpful formulae (this will be on the exam; no need to memorize it Type of Random Variable Density or Mass Function Expected Value Variance Binomial(n, p P (X = k = ( n k p k ( p n k (0 k n np np( p λ λk Poisson(λ P (X = k = e (k N λ λ k! Geometric(p P (X = k = ( p k p p, (k N /p p Exponential(λ f(x = λe λx, (x 0 λ λ Normal(µ, σ f(x = /σ πσ e (x µ µ σ Uniform over an interval [a, b] f(x = b a Var(X = E[(X µ ] = E[X ] E[X]. (x [a, b] a+b Cov(X, Y = E[(X E[X](Y E[Y ]] = E[XY ] E[X]E[Y ]. (Markov s inequality P (X 0 P (X t E[X]/t. (Chebyshev s inequality P ( X µ t Var(X t. Φ(t = P (χ t (where χ is a standard normal random variable. P ( χ 0.68, P ( χ 0.9, P ( χ (b a