In a normal sample of women, the average height was 64.9 with an std deviation of 2.7. If 5 women are selected at random, what is the probability that their mean height is between 64.1 and 65.1?
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Let X be the height of women.
X ∼ n(64.9; 2.7)
n = 5
By LCT,
Xmean ∼ n(64.9; 2.7/sqrt(5))
P(64.1 ≤ Xmean ≤ 65.1) =
P((64.1 - 64.9)/1.207476708 ≤ (Xmean - 64.9)/1.207476708 ≤ (65.1 - 64.9)/1.207476708))=
P( - 0.66 ≤ Z ≤ 0.17) = P(Z ≤ 0.17) - P(Z ≤ - 0.66) = 0.5675 - 0.2546 = 0.3129
X ∼ n(64.9; 2.7)
n = 5
By LCT,
Xmean ∼ n(64.9; 2.7/sqrt(5))
P(64.1 ≤ Xmean ≤ 65.1) =
P((64.1 - 64.9)/1.207476708 ≤ (Xmean - 64.9)/1.207476708 ≤ (65.1 - 64.9)/1.207476708))=
P( - 0.66 ≤ Z ≤ 0.17) = P(Z ≤ 0.17) - P(Z ≤ - 0.66) = 0.5675 - 0.2546 = 0.3129