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hw6_soln6 - Homeworks 17-19 Solutions 5.46 x = diameter of...

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Homeworks 17-19 Solutions 5.46 x = diameter of a piston ring cm cm 04 . 12 = = σ μ (a) cm x 12 = = μ μ = = n n x 04 . σ σ (b) When n = 64 005 . 64 04 . 12 = = = x x cm σ μ (c) The mean of a random sample of size 64 is more likely to lie within .01 cm of μ , since x σ is smaller. 5.50. (a) x = the weight of a bag of fertilizer μ = 50 lbs σ = 1 lb n = 100 ( 29 ( 29 ( 29 ( 29 9876 . 0062 . 9938 . 5 . 2 5 . 2 5 . 2 5 . 2 100 1 50 25 . 50 100 1 50 75 . 49 25 . 50 75 . 49 = - = - - = - = - - z P z P z P z P x P (b) μ = 49.8 lbs σ = 1 lb n = 100 ( 29 ( 29 ( 29 ( 29 6915 . 3085 . 1 5 . 0 5 . 4 5 . 4 5 . 0 25 . 50 75 . 49 = - = - - = - z P z P z P x P 5.54. (a) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 35 . 05 . 3 05 . 2 1 . 1 8 . 0 ) ( = + + + = = x xp μ ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 7921 . 05 . 35 . 3 05 . 35 . 2 1 . 35 . 1 8 . 35 . 0 ) ( 2 2 2 2 2 = - + - + - + - = - = x p x μ σ (b) 35 . 64 = = = μ μ x n
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( 29 099 . 64 7921 . = = = n x σ σ (c) ( 29 ( 29 0 57 . 6 099 . 35 . 1 1 = - z P z P x P 5.55 (a) Let p = 'proportion of resistors exceeding 105 '. Then the sampling distribution of p is approximately normal with μ p = π = 0.02 and σ p = n ) 1 ( π π - = 100 ) 02 . 1 ( 02 . - = 0.014. (b) P(p < .03) = P(z < 014 . 02 . 03 . - ) = P(z < .71) = 0.7611. 5.15. The probabilities are shown on the tree diagram of problem 5.6: Bit sent Bit received 0 1 0 1 0 1 .01 .99 .99 .01 .60 .40 (a) The proportion/percentage of 1s received is the sum of the proportion of 0s that were sent and mistakenly received as 1s, (.40)
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