STAT 400 Lecture AL1 Answers for 2.1, 2.2, 2.3 Spring 2015 Dalpiaz 1. A balanced (fair) coin is tossed twice. Outcomes x f ( x ) Let X denote the number of H's. Construct the probability distribution of X. S = { HH , HT , TH , TT } X = 2 1 1 0 TT 0 1/4 HT TH 1 1/2 HH 2 1/4 1.00 Just for fun: Outcomes x f ( x ) Suppose P(H) = 0.60, TT 0 0.40 0.40 = 0.16 P(T) = 0.40. HT TH 1 0.60 0.40 + 0.40 0.60 = 0.48 HH 2 0.60 0.60 = 0.36 1.00 1 ½ . Suppose Homer Simpson has five coins: 2 nickels, 2 dimes and 1 quarter. Let X denote the amount Bart gets if he steals two coins at random. a) Construct the probability distribution of X. Outcomes x f (x) N N 0.10 2/5 1/4 = 2/20 = 0.1 N D D N 0.15 2/5 2/4 + 2/5 2/4 = 8/20 = 0.4 D D 0.20 2/5 1/4 = 2/20 = 0.1 N Q Q N 0.30 2/5 1/4 + 1/5 2/4 = 4/20 = 0.2 D Q Q D 0.35 2/5 1/4 + 1/5 2/4 = 4/20 = 0.2 1.0 ORN1 N2 D1 D2 Q x f (x) N1 * 0.10 0.15 0.15 0.30 0.10 2/20 = 0.1 N2 0.10 * 0.15 0.15 0.30 0.15 8/20 = 0.4 D1 0.15 0.15 * 0.20 0.35 0.20 2/20 = 0.1 D2 0.15 0.15 0.20 * 0.35 0.30 4/20 = 0.2 Q 0.30 0.30 0.35 0.35 * 0.35 4/20 = 0.2 * – do not steal the same coin twice. 1.0 OR Outcomes x f (x) N N 0.10 25010222 CCCC = 0.1 N D 0.15 25011212 CCCC = 0.4 D D 0.20 25012202 CCCC = 0.1 N Q 0.30 25110212 CCCC = 0.2 D Q 0.35 25111202 CCCC = 0.2 1.0 x f (x) x f (x) (x X) 2 f (x) x 2 f (x) 0.10 0.15 0.20 0.30 0.35 0.1 0.4 0.1 0.2 0.2 0.01 0.06 0.02 0.06 0.07 0.00144 0.00196 0.00004 0.00128 0.00338 0.0010 0.0090 0.0040 0.0180 0.0245 1.0 0.22 0.00810 0.0565b) Find the expected value of the amount that Bart gets, E( X ). X = E(X) = xall(x) x f = $0.22. c) Find the standard deviation SD( X ). xall2X2X(x) x Var(X) μσ f = 0.0081. OR 2X xall22Xμσ (x) xVar(X) f = 0.0565 (0.22) 2 = 0.0565 0.0484 = 0.0081. X = SD(X) = 0081.0 = $0.09. 2. Suppose a random variable X has the following probability distribution: x f ( x ) 10 0.20 11 0.40 12 0.30 13 0.10 a) Find the expected value of X, E(X). x f ( x ) x f ( x ) 10 0.2 2.0 11 0.4 4.4 12 0.3 3.6 13 0.1 1.3 1.0 11.3 X = E(X) = x xfx all)( = 11.3.b) Find the variance of X, Var(X). x f ( x ) ( x X ) ( x X ) 2 f ( x ) 10 0.2 – 1.3 1.69 0.2 = 0.338 11 0.4 – 0.3 0.09 0.4 = 0.036 12 0.3 0.7 0.49 0.3 = 0.147 13 0.1 1.7 2.89 0.1 = 0.289 0.810 2Xσ = Var(X) = x xfx all2X)(μ = 0.81. OR x f ( x ) x 2 f ( x ) 10 0.2 20.0 11 0.4 48.4 12 0.3 43.2 13 0.1 16.9 128.5 2Xσ = Var(X) = 2all2E(X))( x xfx = 128.5 – (11.3) 2 = 0.81. c) Find the standard deviation of X, SD(X). Xσ = SD(X) = 2Xσ = 0.9. d) Find the cumulative distribution function of X, F ( x ) = P ( X x ). x f ( x ) F ( x ) F ( x ) = 13113129.012116.011102.0100 xxxxx 10 0.2 0.2 11 0.4 0.6 12 0.3 0.9 13 0.1 1.03. Suppose E(X) = 7, SD(X) = 3. a) Y = 2 X + 3. Find E(Y) and SD(Y). E(Y) = 2 E(X) + 3 = 17. SD(Y) = | 2 | SD(X) = 6. b) W = 5 – 2 X. Find E(W) and SD(W). E(W) = 5 – 2 E(X) = – 9. SD(Y) = | – 2 | SD(X) = 6. 3 ½ . Suppose a discrete random variable X has the following probability distribution: f ( x ) = x21, x = 1, 2, 3, … . a) Verify that this is a valid probability distribution. - f ( x ) 0 x - xxf all = 1 1 21xx = basetermfirst1 = 21121 = 1. b) Find P ( X is divisible by 3 ). P ( X is divisible by 3 ) = P( 3 ) + P( 6 ) + P( 9 ) + P( 12 ) + … = ...2121212112963 = 1 81 kk = 811181 = 71. c) Find P ( X is divisible by 3 | X is divisible by 2 ). P ( X is divisible by 3 | X is divisible by 2 ) = 2by divisible is XP2by divisible is X3by divisible is XP = 2by divisible is XP6by divisible is XP. P ( X is divisible by 2 ) = P( 2 ) + P( 4 ) + P( 6 ) + P( 8 ) + … = ...212121218642 = 1 41 kk = 411141 = 31. P ( X is divisible by 6 ) = P( 6 ) + P( 12 ) + P( 18 ) + P( 24 ) + … = ...212121212418126 = 1 641 kk = 64111641 = 631. P ( X is divisible by 3 | X is divisible by 2 ) = 2by divisible is XP2by divisible is X3by divisible is XP = 2by divisible is XP6by divisible is XP = 31631 = 211.d) Find E ( X ). E ( X ) = x xfx all)( = 1 21xxx = ...2142132122114321 21 E ( X ) = ...213212211432 21 E ( X ) = E ( X ) – 21 E ( X ) = ...2112112112114321 = 1. E ( X ) = 2. e) Find the cumulative distribution function of X, F ( x ) = P ( X x ). For k = 1, 2, 3, … , P ( X > k ) = f ( k + 1 ) + f ( k + 2 ) + f ( k + 3 ) + f ( k + 4 ) + … = ...212121214321 kkkk = basetermfirst1 = 211211 k =k 21. P ( X > k ) = 1 – P ( X ≤ k ). P ( X ≤ k ) = 1 – k 21. …
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