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Homework 4

# Homework 4 - HW 4 Solutions 1 Solve 6.18(a Let x1 = 1.3 and...

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HW 4 Solutions 1. Solve 6.18. (a) Let x 1 = μ + 1.3 σ and x 2 = μ 1.3 σ . Then z 1 = 1.3 and z 2 = 1.3. P( X > μ + 1.3 σ )+P(X < μ - 1.3 σ ) = P(Z > 1.3)+P(Z < 1.3) = 2P(Z < 1.3) = 0.1936. Therefore, 19.36%. (b) Let x 1 = μ + 0.52 σ and x 2 = μ− 0.52 σ . Then z 1 = 0.52 and z 2 = 0.52. P( μ + 0.52 σ < X < μ− 0.52 σ ) = P( 0.52 < Z < 0.52) = 0.6985 0.3015 = 0.3970. Therefore, 39.70%. 2. Solve 6.60. Let X: time, in hours, between successive calls. Here X~Exp(1/6) P(X > 1/4) = 6 1/4 6 x e dx e 1.5 = 0.223. 3. Solve 8.17. We need to find ) 4 . 0 9 . 1 ( X X X X X P . Notice X X X Z follows N(0,1) distribution. ) 4 . 0 9 . 1 ( X X X X X P =P(-1.9 Z -0.4) =P(Z -0.4)-P(Z -1.9) = 0.3446-0.0287 =0.3159 4. Solve 8.20. Given n=36, 2 n X . Then 2 36 , so σ =12.

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So, for 2 . 1 new X n , 2 . 1 12 new n . So, n new =100. 5. Solve 8.21. Here μ =240 mL, σ =15mL. So, 240 X and . 37 . 2 40 15 n X So, X X 2 =(235.26,244.74). Since, X =236 , it lies in the range (235.26,244.74). So, they can conclude that no adjustments are needed.
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