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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.
750 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 = 1 751 z= X −μ
s/ n = 52 − 50
2 / 16 = 5.6569 P(Z > z) ~0. The results are very unusual.
752 P( X ≤ 37) = P( Z ≤ −5.36) = 0 753 Binomial with p equal to the proportion of defective chips and n = 100. 754 E (aX 1 + (1 − a ) X 2 = aμ + (1 − a ) μ = μ 716 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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 Spring '14

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