Homework 10 (due December 2, 2009)

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1 Homework (due December, 9) Problem. Let X and Y be independent binomial random variables with parameters (n, p) and (n, p) respectively. Prove that X + Y is a binomial random variable with parameters (n + n, p). Solution: This problem is solved in the book, see the section on Sums of Random Variables. Problem. Each day a meteorologist makes a prediction of what time the sun will rise. The difference in minutes between his prediction and the actual time the sun rises is described by a normal random variable with mean and variance 3. What is the probability that in a month of 3 days his prediction is within 5 minutes of the actual sunrise on exactly days? Solution: Let X the number of days among 3 in which his prediction is within 5 minutes. This is a binomial random variable where n 3 and p is the probability that any one day his prediction is off by less than 5 minutes. Since his error, which I will denote as the random variable E, is a normal random variable with mean (µ ), and variance 3 (σ ). Thus, Now since we have p then, p ( 5 P ( 5 < E < 5) P Φ(5/) Φ( 5/) Φ(5/) + Φ(5/) < E Φ(5/) (.7977).5953 P (X ) < 5 ) ( ) 3 (.5953) (.5953). Problem 3. The weight that a bridge can support without structural damage is estimated to be 35, in units of s of pounds. The weight of a single car is on average 3 (again, in s of pounds) with standard deviation.3. How many cars would need to be on the bridge for the probability of structural damage to exceed.?

2 Solution: Suppose that the wight of car i is X i, then E[X i ] 3and V ar(x i ).9. We are interested in finding n so that: P (X + X X n > 35) >. Since we are assuming that n is large we can apply the central limit theorem and we need to subtract off nµ and divide by σ n or: ( ) X + X X n 3n P (X + X X n > 35) P n > n.3 n ( ) 35 3n Φ.3 n If we want this to be greater than. then we have: ( ) 35 3n Φ.3 >. n or ( ) 35 3n.9 > Φ.3 n Now the first value where Φ(x).9 is when x.9. Thus to find the n we seek we set or or n.3 n.387 n 35 3n 3n n 35 Using the quadratic formula we have that: n.387 ± (3)(35).387 ± We must take the + sign since n cannot be negative. Thus n.73, So with cars the probability will exceed.. Problem 4. The length of time to fix a certain machine is known to be an exponential random variable with mean. hours. If the repair man has 3 machines to fix approximate the probability that he can complete the work in under 5.5 hours. State clearly any assumptions you make.

3 Solution: Let X i be the amount of time to fix machine i. Then each X i is an exponential random variable with mean. which is equal to thus the variance is λ /λ is.4 ad the σ.. Now we are interested in P (X + X X 3 < 5.5). We can apply the Central Limit Theorem since we are dealing with the sum of identical independent random variables and n 3 is large enough for the approximation to be valid. So, ( ) X + X X 3 3(.) P (X + X X 3 < 5.5) P. 5.5 (3)(.) < 3. 3 ( ).5 Φ. 3 Φ(.45) Problem 5. Problem 8. on page 4 of the text. (a) Give an upper bound for the probability that a student s test score will exceed 85. Solution: Here we want an upper bound for P (X > 85), P (X > 85) E[X]/85 75/85 (b) Suppose that in addition, that the professor knows that the variance of a student s test score is equal to 5. Solution: What can be said about the probability that a student will score between 5 and 85? Here we know, P (5 < X < 85) P (5 75 < X 75 < 85 75) P ( < X 75 < ) P ( X 75 < ) P ( X 75 ) If we apply Chebyshev s Inequality then we have that Thus, P ( X 75 ) 5.5 P (5 < X < 85) P ( X 75 ).5.75 (c) How many students would have to take the examination to ensure, with probability at least.9 that the class average would be within 5 of 75? Do not use the Central Limit 3

4 Theorem. Solution: The average score of n students is A n (X +X +...+X n )/n. The mean of A n 75 and the variance of A n is 5/n. Now according to Chebyshev s inequality we have that: P ( A n 75 5) 5 n 5 n. So in order to ensure that the average is within 5 with probability of at least.9 then we need the probability that that average is more than 5 from 75 is less than. so we must take /n <. or n >. So we must take at least students to assure that we have the desired situation. Problem. X, the average rainfall in the month of June in Minneapolis, is known to be approximately a normal random variable with mean 8 and variance 4. In Paris, the average rainfall in June is given by a normal random variable, Y, with mean 9 and variance 7. If it is assumed that the rainfall in Minneapolis is independent of that in Paris, what distribution ( and with what parameters) would the represent the average of X and Y? Solution: Since X and Y are normal then the sum of X and Y is also normal with mean equal to the sum of the means and variance equal to the sum of the variances. Thus X + Y is normal with mean 7 and variance. Now, the average is (X + Y ). This makes the average normally distributed with mean 7/ and variance /4. Problem 7. Show that. f(x, y) { x < y < x <, otherwise (a) Show that f(x, y) is a true joint density function for two random variable X and Y. Solution: To show this we compute the integral of the density function over the region in which it is non-zero which gives: x x dydx x x dydx So yes it is indeed a proper density function. (b) Find the marginal densities f X (x) and f Y (y). (x )dx x 4

5 Solution: To find the marginal density for x we hold x fixed and integrate of over y. Namely, x f X (x) x dy x x y, < x < Similarly, f Y (y) y ( ) x dx ln(x) y ln() ln(y) ln, < y < y (c) Find E[X] and E[Y ]. Solution: E[X] E[Y ] xf X (x)dx y ln(y)dy x()dx x y ln(y)dy Now to evaluate the remaining integral you can either look up the integral of y ln(y) or you can find it by integrating by parts which is a bit of a trick letting u ln(y) and dv ydy. Thus du y dy and v y /. This gives, E[Y ] y ln(y)dy y ln(y) + y y dy Problem 8. Let X and Y be continuous random variable with joint density given by: { x + cy, < x <, < y < 5 5 f(x, y) otherwise (a) Find c. 5 x 5 + cydydx 4x 5 [ ] 5 xy 5 + cy dx x + cdx 5 + cx Setting this equal to we have that c 3/5 or c /. (b) Are X and Y independent? Justify your answer. ( x + 5c ) ( x 5 ) + c dx 5 + c One could compute both the marginal densities, but one could also observe that even though the region of definition is not a rectangle, there is no way to factor the function f(x, y) into two functions of x and y. Thus, these can t be independent. 5

6 (c) Find P (X + Y > 3). Solution: P (X + Y > 3) 5 3 x x 5 + y dydx [ ] 5 xy 5 + y 4 3 x ( x + 5 ) ( ) x(3 x) (3 x) + dx ( 4x ) ( 4x 8x 9 x + x ( ) x + 7x + dx 4 4 x + 7x3 + 5 x ) dx 5 (d) Write an expression involving integrals that represents P (X > / X + Y > 3). Solution: Using the definition of conditional probability we have that: P (X > / X + Y > 3) P (X > /, X + Y > 3) P (X + Y > 3) 5 x + y dydx 3 x 5 5 Problem 9. Suppose that the joint probability mass function for two discrete random variable X and Y is given by: y p(x, y) x (a) Find the marginal mass functions for X and Y. Solution: To compute the marginal mass functions we add the rows to get p X (x) and add the columns to get p Y (y). x p X (x) y p Y (y)

7 (b) Are X and Y independent? Justify your answer. Solution: If one checks each of the values it is easy to see that p(x, y) p X (x)p Y (y), thus X and Y are indeed independent. (c) What is the probability P (X + Y > )? Solution: To compute this probability we add up all the cases that give values greater than so, P (X+Y > ) p(4, 7)+p(4, )+p(, 4)+p(, 7)+p(, ) (d) Compute the probability P (X Y ). Solution: To compute this we recall the definition of conditional probability to give: P (X Y ) P (X, Y ) P (Y ) p(, ) p Y ()... (e) Find the expected value E[ X Y ]. Solution: To find the expected value we computer we add up the function with its associated probability over both x and y. Namely, E[ X Y ] x y p(x, y) x y or E[ X Y ] 3p( 4, ) + 8p( 4, 4) + p( 4, 7) + 4p( 4, ) + p(, ) + 4p(, 4) +7p(, 7)+p(, )+5p(4, )+p(4, 4)+3p(4, 7)+p(4, )+p(, ) +7p(, 4) + 4p(, 7) + p(, ) (f) Find the cumulative density for X. Express the function with a piecewise definition and sketch a graph. 7

8 Solution: The cumulative density function can be written as a piecewise function by: x < x < F X (x).4 x < x < x 8

9 Bonus Problem. Let X be the number of s and Y be the number of s that occur in n rolls of a fair die. Compute Cov(X, Y ). Hint: First define { if roll i is a X i otherwise { if roll i is a Y i otherwise and write X and Y in terms of the X i s. Clearly, X X i, Y i Y j. j Consequently, due to the properties of covariance we have that: Cov(X, Y ) i Cov(X i, Y j ). Now we compute Cov(X i, Y j ). First, we assume that the results of any particular roll are independent of any other roll, and because of this X i and Y j are independent if i j. Consequently Cov(X i, Y j ) when i j. In the case when i j we first compute E[X i ] P { roll i is a }. Similarly E[Y j]. Now consider X iy i. In order for this product to be nonzero, roll i would have to be both a and a, which is of course not possible, and thus E[X i Y i ] and thus Cov(X i, Y i ) E[X i Y i ] E[X i ]E[Y i ] /3. 3 Consequently, j Cov(X, Y ) Cov(X i, Y j ) i j Cov(X i, Y i ) n 3. i Bonus Problem. Give a combinatorial proof of the identity: ( ) n k ik ( ) i, n k k Hint: Consider the set of number through n, How many subsets of size k have i as their highest numbered member? Bonus Problem 3.Prove that: ( ) n ( ) i i i 9

10 Hint: Use the Binomial Theorem. Bonus Problem 4.Suppose that P (E F ) P (E). Use the definition of conditional probability to show that P (F E) P (F ). Bonus Problem 5. Jeff s factory produces 3 meter beams. Suppose that the amount of weight that their beams can hold before being damaged is a exponentially distributed random variable with mean 5 pounds. The factory sends a beam periodically to a testing center to have its strength tested. The testing center rates the beams on a scale of to 5. A score of is given to beams that support more than 7 pounds before failure, a score of is given to beams that support between 55 and 7 pounds before failure, a score of 3 is given to beams that can support between 45 and 55 pounds before failure and a score of 4 is given if it can support less than 45 pounds before fail. What is the expected score of a randomly chosen beam from Jeff s factory? What is the standard deviation of the score?

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

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