Stat427: Solutions to Practice Final Exam - Spring 2009, Exams of Statistics

Download Stat427: Solutions to Practice Final Exam - Spring 2009 and more Exams Statistics in PDF only on Docsity! Stat427: Solutions to Practice Final Exam; Spring 2009 Problem 1. P (all 10 bulbs work) = (.95)10 assuming bulbs work or fail independently P (string fails) = 1 − (.95)10 = .40. Problem 2. X = amount win in one game, x = 0, 1. pmf of X: P (x) = { 4/6 x = 0 2/6 x = 1 (a) E(X) = (0)(4/6) + (1)(2/6) = 1/3 Dollars expect to win in 5 games: 5E(X) = 5(1/3) = 1.67 (b) Need to win at least 3 games out of 5; Y ∼ Bin(5, 1/3) P (Y ≥ 3) = ( 5 3 ) (1/3)3(2/3)2 + ( 5 4 ) (1/3)4(2/3)1 + ( 5 5 ) (1/3)5(2/3)0 = .21 (c) Need to win more than 100 games out of 200; Y ∼ Bin(200, 1/3) Use normal approximation with continuity correction; check: np = 200/3 ≥ 10; nq = 400/3 ≥ 10 P (Y > 100) = 1−P (Y ≤ 100) = 1−Φ((100+0.5−np)/√npq) = 1−Φ(5.07) ≈ 0. Problem 3. X̄ ∼ N (µ = 5000, σ = 240/ √ 36 = 40) P (X̄ > 5100) = 1 − P (X̄ ≤ 5100) = 1 − Φ(5100 − 5000/40) = 1 − Φ(2.5) = .0062. Problem 4. (a) X = # fails; X ∼ Bin(500, .002). Number expect to fail: np = 500(.002) = 1. (b) Use Poisson approximation with λ = np = 1; check: n ≥ 100; p ≤ .01; np ≤ 20. P (X ≥ 2) = 1 − P (X ≤ 1) = 1 − F (1; 1) = 1 − .736 = .264. Problem 5. (a) P (A | M) = Φ((19 − 21)/4) = Φ(−.5) = .3085 (b) P (A | M ′) = Φ((19 − 20)/2) = Φ(−.5) = .3085 (c) P (A) = P (A | M)P (M)+P (A | M ′)P (M ′) = (.3085)(2/5)+(.3085)(3/5) = .3085 (d) P (M | A) = P (A | M)P (M)/P (A) = (.3085)(2/5)/.3085 = 2/5. 1