Final Examination December 17, 2012 MATH 323 P (Y = 5) =

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1 . The eight-member Human Relations Advisory Board of Montreal, considered the complaint of a woman who claimed discrimination, based on sex, on the part of a local company. The board, composed of five women and three men, voted 5-3 in favor of the plaintiff, the five women voting in favor of the plaintiff, the three men against. The attorney representing the company appealed the board s decision by claiming sex bias on the part of the board members. If there was no sex bias among the board members, it might be reasonable to conjecture that any group of five board members would be as likely to vote for the complainant as any other group of five. If this were the case, what is the probability that the vote would split along sex lines ( five women for, three men against)? ( marks) Solution: If there is no sex bias, then Y the number of women voting for the plaintiff is distributed according to Hypergeometric(N = 8, r = 5, n = 5). Now ( ) P (Y = 5) = 5)( 5 5 ( 8 = 5) 56 < %

2 . In responding to a survey question on a sensitive topic (such as Have you ever tried marijuana? ), many people prefer not to respond in the affirmative. Suppose that 8%of the population have not tried marijuana and all of those individuals will truthfully answer no to your question. The remaining % of the population have tried marijuana and 7% of those individuals will lie. Derive the probability mass function of Y, the number of people you would need to question in order to obtain the first affirmative response. ( marks) Solution: Let A = Affirmative response} and B = Tried marijuana}. Then P (B) =., P (A c B) =.7 and P (A B c ) =. Thus P (A B) =.3. Using the law of total probability we have P (A) = P (A B)P (B) + P (A B c )P (B c ) = (.3)(.) + (.8) =.6 Let Y be the number of individuals to interview in order to obtain the first affirmative response. Having assumed that the subjects we interview are independent, Y Geometric(p =.6), i.e. P (Y = k) = p( p) k = (.6)(.94) k, k =,, 3,

3 3. Wires manufactured for use in a computer system are specified to have resistances between. and.4 ohms. The actual measured resistances of the wires produced by company A have a normal probability distribution with mean.3 ohms and standard deviation.5 ohms. (a) What is the probability that a randomly selected wire from company A s production will meet the specifications? (5 marks) Solution: Let Y be the resistance of a randomly selected wire manufactured by company A. Then Y N(µ =.3, σ =.5). We need to compute P (. < Y <.4). P (. < Y <.4) = F Y (.4) F Y (.) = Φ( ) Φ( ).5.5 = Φ() Φ( ) = [ Φ()] where Φ is the cdf of Z N(, ). (.3) =.954, (b) If four of these wires are used in each computer system and all are selected from company A, what is the probability that all four in a randomly selected system will meet the specifications? (5 marks) Solution: Define X i =, if the i-th wire meets the criterion, and otherwise, for i =,, 3, 4. Then W = 4 i= X i Bin(n = 4, p =.954). Then P (W = 4) = (.954) 4 =.88. 3

4 4. Suppose that plants of a particular species are randomly dispersed over an area, so that the number of plants in a given area follows a Poisson distribution with a mean density of λ plants per unit area. If a plant is randomly selected in this area, find the probability density function of the distance to the nearest neighboring plant. [Hint: If R denotes the distance to the nearest neighbor, then P (R > r) is the same as the probability of seeing no plants in a circle of radius r.] ( marks) Solution: Let X(s) denote the number of plants in a region with surface area s. Then P [X(s) = n] = e λs (λs) n, n =,,, n! Now using the hint, for any r > we have F R (r) = P (R > r) = P [X(a circle with radius r) = ] = e λπr, where πr is the surface area of a circle with radius r. Now differentiating F R (r) with respect r, we obtain, f R (r) = df R(r) πλre λπr, if r >, = dr, Otherwise. 4

5 5. An environmental engineer measures the amount (by weight) of particulate pollution in air samples of a certain volume collected over two smokestacks at a coal-operated power plant. One of the stacks is equipped with a cleaning device. Let Y denote the amount of pollutant per sample collected above the stack that has no cleaning device and Y denote the amount of pollutant per sample collected above the stack that is equipped with the cleaning device. Suppose (Y, Y ) has the joint probability density function (pdf), if y, y, y y, f Y,Y (y, y ) =, oterwise. (a) Consider the stack with the cleaner installed. Find the probability that the amount of pollutant in a given sample from this stack will exceed.5. (5 marks) Solution: y.5 y Using the above graph, There are two ways to find P (Y.5): First approach: Using double integrals We need to integrate the joint density over the area inside the red triangle under the green line. Notice that taking horizontal slices and sweep the area vertically is easier when computing P (Y.5). It then means that the outer integral is with respect to y while the inner integral is with respect to y. P (Y.5) = / y dy dy = = [ z ] / = 4 = 3 4 / ( y )dy = / zdz Second approach: Calculating the area directly Using the fact that the joint density is uniform and equal over the region of interest, P (Y.5) is equal to the area under the green line in the red triangle. This are can be decomposed into rectangle and a triangle. Using the coordinates given on the graph, the area of the rectangle is (.5 ). As for the triangle we have [.5 (.5 )]. Thus the total area is / + /4 = 3/4. 5

6 (b) Given that the amount of pollutant in a sample taken above the stack with the cleaner is observed to be.5, find the probability that the amount of pollutant exceeds.5 above the other stack (without the cleaner) (5 marks). Solution: We should find P (Y.5 Y =.5). This requires the conditional density of Y given Y. First step in this calculation is to find marginal density of Y. For any y we have Then f Y (y ) = f (y ) = = f (y ) = We therefore find that y f Y,Y (y, y )dy y dy = ( y ). ( y ) if y, otherwise. f Y Y (y y ) = f(y, y ) f (y ) = ( y ) if y y, otherwise. Now P (Y a Y = y ) = maxa,y } dy ( y ) = maxa, y } ( y ) Substituting for a and y by.5 and.5 we obtain = maxa/, y } ( y ) P (Y a Y =.5) = max.5/,.5} (.5) =.5.5 =. 6

7 6. Suppose the number of defects per yard in a certain fabric, Y, has a Poisson distribution with parameter λ. The parameter λ is assumed to be a random variable with a density function given by e λ, if λ, f(λ) =, otherwise. (a) Find the mean and the variance of Y. ( 5 marks) Solution: We know that Y λ P o(λ) and λ Exp(β = ). Then and E[Y ] = E [E (Y λ)] = E[λ] = V[Y ] = V [E (Y λ)] + E [V (Y λ)] = V[λ] + E[λ] = + = (b) How likely is it that Y exceeds 9? (5 marks) Solution: Define W =, if Y > 9,, otherwise. Note that E(W ) = P (Y > 9). We then have P (Y > 9) = E(W ) = E [E (W λ)] = E [P (Y > 9 λ)] = E [ P (Y 9 λ)] [ 9 ] e λ λ k 9 = E = k! k! E [ e λ λ k] = = = ( ) k= k= 9 k! 9 k= k= e λ/(/) λ (k+) dλ = ( ) k+ = (/) [ (/) ] [ (/)] 9 k= k! ( ) k+ Γ(k + ) 7

8 7. The speed of a molecule in a uniform gas at equilibrium is a random variable V whose density function is given by f(v) = av e bv, v >, where b = m/kt, and k, T, and m denote Boltzmann s constant, the absolute temperature, and the mass of the molecule, respectively. (a) Derive the distribution of W = mv /, the kinetic energy of the molecule. (5 marks) Solution: Given that V >, there is W is a - map of V. The simplest approach seems to be the method of transformation. First note that V = W/m and hence dv/dw = (w/m 3 ) /. Then f W (w) = f V ( ( ) / ( ) w w w/m) = a e b( w m 3 m ) m Simplifying the above expression we have a mw e w m/b, if w >, f W (w) =, otherwise. Using that w α e w/β dw = β α Γ(α), we find that ( w / e w (m/b) m ) 3 dw = Γ( 3 b ). ( w m 3 ). Applying Γ(α + ) = αγ(α) and Γ(/) = π, and using the fact that b = m/kt, we find that a = mπ(kt ). 3 (b) Find E(W ). (5 marks) Solution: E(W ) = a m = a m = 3 kt, w 3/ e w (m/b) dw ( m b)5 Γ( 5 ) = a m ( m b)5 after replacing for a and b and some simplification ( ) ( ) 3 π 8

9 8. The manager of a supermarket wants to obtain information about the proportion of customers who dislike a new policy on cashing checks. (a) How many customers should he sample if he wants the sample proportion to be within.5 of the true proportion, with probability.95? Use the Normal approximation to Binomial distribution? (5 marks) Solution: We want to find n such that P ( ˆp n p <.5) =.95. Using the normal approximation ˆp n p p( p)/n app N(, ), for large enough n. Using the table for normal distribution, we then obtain.5 p( p)/n =.96, and hence n (3 )p( p). Given that p is unknown, we can take a conservative approach and maximize p( p), the variance, that is /4. We should then choose n = 43. (b) How many customers should he sample if he wants the sample proportion to be within.5 of the true proportion, with probability at least.95? Use Tchebysheff s inequality? (5 marks) Solution: Using Tchebysheff s inequality we have P ( ˆp n p < kσ) k. Having noticed that σ = p( p)/n, we need to have k p( p)/n =.5 and k =.95. We therefore find k = and hence n = p( p).5. Taking the conservative approach and maximizing p( p), we fine n = 3. 9

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11 STUDENT NAME: STUDENT ID# McGILL UNIVERSITY FACULTY OF SCIENCE FINAL EXAMINATION MATH33 PROBABILITY THEORY Examiner: Professor M. Asgharian Date: Monday, December 7, Associate Examiner: Professor A. Khalili Time: 8: - : INSTRUCTIONS Answer all 8 questions. Calculators are permitted. Answer directly on the exam. Dictionaries are allowed. This is a closed book exam. Questions Marks This exam comprises the cover page, 9 pages of questions, and page which is blank, pages of formulas and one table.