YORK UNIVERSITY. Faculty of Science Department of Mathematics and Statistics MATH A Test #2 June 11, Solutions

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1 YORK UNIVERSITY Faculty of Science Department of Mathematics and Statistics MATH 2. A Test #2 June, 2 Solutions. ( pts) The probability of a student in MATH 4 passing a test is.82. Suppose students took the test independently. (a) Find the probability that at least 8% of students passed the test. (Answer correct to 4 decimal places.) We have a binomial distribution with n =, p =.82, q = p =.8, i.e. Bi(,.82). µ = np = 6.6, = npq = % of = 4, so = P (X < 4) = P (, ) Φ( ) Φ(.7) = [ Φ(.7)] = Φ(.7) =.76. (b) Given that at least 8% of students passed the test, find the probability that at most 9% of students passed the test. (Answer correct to decimal places.) 9% of = 7. We need to find P (X 7 X 4) = P (X 7 X 4) = P (4, 7) Φ( ) Φ( ) Φ(2.5) Φ(.7) Φ(2.5) [ Φ(.7)] =.998 (.76) (c) Suppose that the probability that less than m students failed the test is approximately.25. Find m. Now, we consider a binomial distribution with n =, p =.8, q =.82, i.e.

2 Bi(,.8). µ = np = 2.4, = npq = P (X < m) =.25 P (, m ) =.25 (m ) +.5 µ Φ( ) =.25 Φ( m.5 µ ) =.25. Φ(z) =.25 = z <. Hence, using the formula Φ(z) = Φ( z), we obtain Φ( m.5 µ ) =.25, that is Φ( µ +.5 m ) =.75, m Φ( ) =.75 Φ(.675). Whence, 2.9 m.675 and m Therefore, since m is a positive integer, m = = (5 + 5 pts) Let A and B be independent events, with P (A) = p and P (B) = q, and indicator random variables I A and I B, respectively, and let Y = (I A + I B ) 2. (a) Determine the probability distribution of Y in terms of p and q. range(y ) = {,, 4}. P (Y = ) = P (I A = I B = ) = P (A c )P (B c ) = ( P (A))( P (B)) = ( p)( q) = p q + pq, P (Y = ) = P [(I A = I B = ) (I A = I B = )] = P (I A = I B = ) + P (I A = I B = ) = P (A)P (B c ) + P (A c )P (B) = P (A)( P (B)) + P (B)( P (A)) = p( q) + q( q) = p + q 2pq, P (Y = 4) = P (I A = I B = ) = P (A)P (B) = pq. (b) What is the mean value of Y in terms of p and q? E(I A ) = P (A), E(I B ) = P (B), E(IA 2 ) = 2 P (A c ) + 2 P (A) = P (A), E(IB 2 ) = 2 P (B c ) + 2 P (B) = P (B). E(Y ) = E[(I A + I B ) 2 )] = E(IA 2 + 2I AI B + IB 2 ) = E(IA 2 ) + 2E(I A)E(I B ) + E(IB 2 ) = P (A) + 2P (A)P (B) + P (B) = p + 2pq + q.. ( pts) Let X be a random variable, with the c.d.f., if x <, F X (x) = 2 (x + ), if x <,, if x. (a) Find P ( X 5 ). 2

3 P ( X 5 ) = F X( 5 ) F X( ) = 2 [( 5 ) + ] 2 [( ) + ] = 2 ( 25 + ) = 9 2. (b) Let Y = X 2. Derive the p.d.f. of Y. The c.d.f. of Y F Y (y) = P (Y y) = P (X 2 y) = P ( y X y) = P (X y) P (X y) = F X ( y) F X ( y) = 2 [( y) + ] 2 [( y) + ] = y 2, d dx F Y (y) = d dx (y 2 ) = y. 2 X (, ) = (X + ) [, 2) = 2 (X + ) [, ). So, Y [, ), i.e. { range(y ) = [, ). Hence, f Y (y) = 2 y, when y <,, otherwise. (c) Find the mean and variance of the random variable W = 5Y. E(Y ) = yf Y (y) dy = y y dy = y 2 dy = 2 2 5, E(Y 2 ) = y 2 f Y (y) dy = y 2 y dy = y 5 2 dy = 2 2 7, V ar(y ) = E(Y 2 ) [E(Y )] 2 = 7 ( 5 )2 = 2 75, E(W ) = E(5Y ) = 5E(Y ) = 5 5 = 2, V ar(w ) = V ar(5y ) = V ar(y ) = = ( + + pts) An urn contains red balls and black balls. Balls are drawn out at random with replacement until at least one ball of each color has been drawn out. Let random variable D be the number of draws. Find: (a) the probability distribution of D; Range(D) = {2,, 4, }, with P (success) = p = P (failure) = q = 2. Then P (D = ) =, P (D = 2) = p = 2, P (D = ) = qp = ( 2 )2 = 4, P (D = 4) = q 2 p = ( 2 ) = 8,, P (D = n) = qn 2 p = ( 2 )n,.

4 (b) E(D); From part (a), D = T +, where T is a geometric random variable with parameter p = 2. Hence, E(D) = E(T + ) = E(T ) + = p + =.5 + = 2 + =. (c) SD(D). V ar(d) = V ar(t + ) = V ar(t ) = q p =.5 2 (.5) = 2. 2 Hence, SD(D) = V ar(d) = 2. Or equivalently (but longer), E(D 2 ) = E[(T + ) 2 ] = E(T 2 + 2T + ) = E(T 2 ) + 2E(T ) + = + q + 2 p 2 p + = +.5 (.5) = =. Therefore, V ar(d) = E(D 2 ) [E(D)] 2 = 2 = 2, and SD(D) = (7 pts) A typist typing a 45 word passage at a constant speed of 6 words per minute makes a mistake in any particular word with probability.4, independently from word to word. Each incorrect word must be corrected; a task which takes 5 seconds per word. Would it be less time consuming, on average, to type the passage at 45 words per minute if this reduces the probability of an error to.2? Justify your answer. Let the r.v. X be the number of words needing correction at speed 6 words per minute, and the r.v. T be the time to type the passage. Then T = 5X + 45, where the r.v. X Bi(45,.4). E(X) = np = 45(.4) = 8. So, E(T ) = E(5X + 45) = 5E(X) + 45 = = 72(seconds). Now, let the r.v. Y be the number of words needing correction at speed 45 words per minute. Then printing each word takes 6 45 = 4 (seconds). We have Y Bi(45,.2), with E(X) = np = 45(.2) = 9, and T = 5Y = 5Y + 6. Hence, E(T ) = E(6 + 5Y ) = 6 + 5E(Y ) = = 75(seconds). Therefore, the typing the passage at 45 words per minute, on avarage, takes longer time. 6. (7 pts) Suppose that your score on your Final Exam on June 2 is a random variable, with the values between and, mean 66, and variance 6. (a) Can you conclude from this information that the probability that your score will be between 55 and 75 is smaller than.95? Explain your reasoning carefully. 4

5 Let your Final Exam score be a random variable X. Given X, E(X) = µ = 66, V ar(x) = 2 = 6. We need to evaluate P (55 X 75). Assume X N(µ, 2 ) = N(66, 6). Then P (µ 2 X µ + 2).9545 P ( X ).9545, that is P (54 X 78) But P (55 X 75) P (54 X 78) Therefore, we can make such kind of conclusion, provided that X N(µ, 2 ). (b) Can the chance of receiving a score, smaller than 45, be equal to 5%? Explain your reasoning carefully. Hint: You may need to use Chebyshev s inequality. X < 45 X µ < 45 µ x 66 < X 66 < 2. Then x 66 > 2 > 8 = 6 =. Hence, using Chebyshev s inequality P ( x µ k), for µ = 66, = 6 and k2 k =, we obtain P ( x 66 ) 2 = 9 <.5. Therefore, the probability of receiving a score, smaller than 45, is less than or equal to 9 <.5, and consequently, cannot be equal to 5%. The end. 5