STAT 230 Final with Solution, Summer 2009

American University of Beirut STAT 230 Introduction to Probability and Random Variables Summer 2009 Final Exam - solution Exercise 1 Let X and Y be a couple of random variables with pdf f(x, y) = cy 0 < x < y < 1 a. find the value of the constant c � 1 0 � y 0 cydxdy = � 1 0 cy2dy = c 3 = 1 (cause f is a pdf), and hence c = 3 b. find the marginal pdf of X and Y . Are X and Y independent ? g(x) = � 1 x 3ydy = 3 2 � y2�1 y=x = 3 2(1 − x2) , 0 < x < 1 h(y) = � y 0 3ydx = 3y2 , 0 < y < 1 f(x, y) ̸= g(x) × h(y), then X and Y are not independent. c. find P(X + Y > 1/4) (its advised to sketch the region of integration !) P(X + Y > 1/4) = 1 − � 1 8 0 � 1 4−x x 3ydydx = 1 − 3 2 � 1 0 � 1 16 − x 2 � dx = 1 − 3 512 = 509 512 d. find E(X) E(X) = � 1 0 xg(x)dx = 3 2 � 1 0 (x − x3)dx = 3 8 Exercise 2 Let (X, Y ) be a couple of random variables with joint pdf f(x, y) = x + y , 0 < x < 1 , 0 < y < 1 Let U = X and V = XY . Find the joint pdf of the couple (U, V ), then find the marginal pdf of V . solving the system of equations yields x = u and y = v u the Jacobian is J(u, v) = ������ 1 0 − v u2 1 u ������ = 1 u and hence g(u, v) = 1 + v u2 0 ≤ u ≤ 1 , 0 ≤ v ≤ u h(v) = � 1 v (1 + v u2 ) du = � u − v u �1 u=v = 2 − 2v , 0 < v < 1 (its advised to sketch the region of integration) Exercise 3 An electronic device runs until one of its three components fails. The lifetimes (in weeks), X1, X2, X3, of these components are independent, and each has the pdf f(x) = 2x 25 e−(x/5)2 , 0 < x < ∞ Find the probability that the device stop running in the first three weeks. p = 1 − (P(X1 > 3))3 = 1 − �� ∞ 3 2x 25 e−(x/5)2dx �3 = 1 − �� −e−(x/5)2�∞ 3 �3 = 1 − e−27/25 = 0.66 Exercise 4 Let X1, . . . , X18 be a random sample of size 18 with χ2(1) = G(1 2, 2) distribution. Let Y = X1 + X2 + . . . + X18. a. Find P(9.39 < Y ≤ 34.80) Y ⇝ χ2(18) (the sum of independent χ2 distribution is a χ2 distribution), and hence P(9.39 < Y ≤ 34.80) = 0.99 − 0.05 = 0.94 (from the χ2 table with r = 18) b. Approximate P(9.39 < Y ≤ 34.80) E(X1) = 1 and V ar(X1) = 2; by the central limit theorem Y 18−1 √ 2 √ 18 = 3( Y 18 − 1) has a standard normal distribution N(0, 1), hence P(9.39 < Y ≤ 34.80) ≃ P(3(9.39 18 − 1) < Z ≤ 3(34.80 18 − 1)) = P(−1.43 < Z ≤ 2.8) = 0.9974 − (1 − 0.9236) = 0.921 Exercise 5 Each item produced by a certain manufacturer is, independently, of acceptable quality with probability 0.95. Approximate the probability that at most 10 of the next 150 items produced are unacceptable. Let X be the number of unacceptable items among the 150; X has a Binomial distribution b(150; 0.05). X can be approximated by a normal distribution Y ⇝ N(7.5; 7.125). P(X ≤ 10) ≃ P(Y ≤ 10) = P � Y −7.5 √ 7.125 ≤ 10−7.5 √ 7.125 � = P(Z ≤ 0.94) = 0.8264 (from the standard normal table)