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Unformatted text preview: IE 300 / GE 331 Discussion Problem Set 3 • 376 Let X denote the number of patients with heart failures due to outside factors. Then it has a binomial distribution with n= 20 and p= 0.13. Note that the pmf for a binomial random variable is as follows: P(X=k) = ? ¡ ¢£ ¡ (1 − £ ) ?−¡ (a) P(X=3)= 20 3 ¢£ 3 (1 − £ ) 17 = 0.235 (b) P(X≥3)= 1P(X<3) = 1P(X=0)P(X=1)P(X=2) = 0.492 (c) µ= E(X)= np= (20)(0.13)= 2.6 Var(X)= E[(X µ) 2 ]= np(1p)= 2.262 σ= ¤ Var(X) = 1.504 • 378 (a) X is a binomial random variable with n = 20 and p = 0.01, so E(X) = np= 20 (0.01) = 0.2 Var(X) = np(1p)= 20 (0.01) (0.99) = 0.198 E(X)+3 ¤ Var(X) = 0.2 + 3 √ 0.198 =1.53 P(X>1.53) = P(X≥2) = 1P(X<2) = 1 ¥ 20 ¢ (0.01 )(0.99 20 ) + 20 1 ¢ (0.01 1 )(0.99 19 ) ¦ = 0.0169 (b) X is a binomial random variable with n = 20 and p = 0.04, so P(X>1) = 1 P(X≤1) = 1 ¥ 20 ¢ (0.04 )(0.96 20 ) + 20 1 ¢ (0.04 1 )(0.96 19 ) ¦ = 0.189 (c) Let Y denote the number of times that X exceeds 1 in the next five samples. Note that Y is a Let Y denote the number of times that X exceeds 1 in the next five samples....
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This note was uploaded on 03/16/2010 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 NegarKayavash

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