mont4e_sm_ch07_supplemental

7 50 assume x is approximately normally distributed p

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Unformatted text preview: 7 −50 ≤ Z ≤ ⎜ ⎝ 12 / 36 53− 50 12 / 36 ⎞ ⎟ ⎠ = P (−1.5 ≤ Z ≤ 1.5) = P ( Z ≤ 1.5) − P ( Z ≤ −1.5) = 0.9332 − 0.0668 = 0.8664 No, because Central Limit Theorem states that with large samples (n ≥ 30), X is approximately normally distributed. 7-50 Assume X is approximately normally distributed. P ( X > 4985) = 1 − P ( X ≤ 4985) = 1 − P ( Z ≤ 4985 − 5500 ) 100 / 9 = 1 − P ( Z ≤ −15.45) = 1 − 0 = 1 7-51 z= X −μ s/ n = 52 − 50 2 / 16 = 5.6569 P(Z > z) ~0. The results are very unusual. 7-52 P( X ≤ 37) = P( Z ≤ −5.36) = 0 7-53 Binomial with p equal to the proportion of defective chips and n = 100. 7-54 E (aX 1 + (1 − a ) X 2 = aμ + (1 − a ) μ = μ 7-16 V ( X ) = V [aX 1 + (1 − a) X 2 ] = a 2V ( X 1 ) + (1 − a) 2 V ( X 2 ) = a 2 ( σ 1 ) + (1 − 2a + a 2 )( σ 2 ) n n 2 = 2 2 a 2σ 2 σ 2 2aσ 2 a 2σ 2 + − + = (n2 a 2 + n1 − 2n1a + n1a 2 )( σ ) n2 n2 n2 n1n2 n1 2 ∂V ( X ) = ( σ )(2n2 a − 2n1 + 2n1a) ≡ 0 n1n2...
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