Homework_6_Solutions

Homework_6_Solutions - Homework 5 Solutions 8.65 a. =...

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Homework 5 Solutions 8.65 a. μ = 100(1/2) = 50 and . 5 ) 5 . 1 )( 5 (. 100 = - = σ b. P(X 60) ≈ P(Z (60 - 50)/5) = P(Z 2) = 1 – P(Z < 2) = 1 - .9772 = .0228. c. P(X 59.5) ≈ P(Z (59.5 – 50)/5) = 1 – P(Z < 1.9) = 1 - .9713 = .0287. 8.66 a. Answer = .0571. For a binomial random variable with n = 50 and p = .512, μ = np = 50(.512) = 25.6, and σ = ) 512 . 1 )( 512 (. 50 - = 3.535. Thus, for X = 20, 58 . 1 535 . 3 6 . 25 20 - = - = z . P(X 20) ≈ P(Z - 1.58) = .0571. b. Answer = .0749. With the continuity correction, we find P ( X 20.5). For X = 20.5, 44 . 1 535 . 3 6 . 25 5 . 20 - = - = z . So, P(X 20.5) = P( Z -1.44) = .0749. 8.67 a. Answer = .9015. For a binomial random variable with n = 1000 and p = .60, μ = np = 1000(.60) = 600, and σ = 492 . 15 ) 60 . 1 )( 60 (. 1000 = - . For X = 620, 29 . 1 492 . 15 600 620 = - = z . P ( Z 1.29) = .9015. b. Answer = .0039. For a binomial random variable with n = 2000 and p = .87, μ = np = 2000(.87) = 1740, and σ = 04 . 15 ) 87 . 1 )( 87 (. 2000 = - . For X = 1700, 66 . 2 04 . 15 1740 1700 - = - = z . P(Z - 2.66) = .0039. 8.68 a. X is a binomial random variable. There are a fixed number of trials since n = 100. There are two possible outcomes—picking the correct suit and picking an incorrect suit. The outcomes are independent from one trial to the next because the cards are selected with replacement. The probability of picking the correct suit remains the same from trial to trial and p = 1/4 if the subject is just guessing. b.
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Homework_6_Solutions - Homework 5 Solutions 8.65 a. =...

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