Question

A half-century ago, the mean height of women in a particular country in their 20s was 64.5 inches. Assume that the heights of today's women in their 20s are approximately normally distributed with a standard deviation of 3.18 inches. If the mean height today is the same as that of a half-century ago, what percentage of all samples of 22 of today's women in their 20s have mean heights of at least 66.27 inches? About % of all samples have mean heights of at least 66.27 inches. (Round to one decimal place as needed.)

Answer 1

u = 64.5, 0 = 3.18

sample size (n) = 22< 30..but as the sample is from a normal population ..the sample mean will follow normal distribution with :-

3.18 01 0.6780 n 22

the percentage of all samples of 22 of today's women in their 20s have mean heights of at least 66.27 inches be:-

$=P(\bar{X}\geq66.27)$

$=P(\frac{\bar{X}-64.5}{0.6780}\geq\frac{66.27-64.5}{0.6780})$

= P(Z > 2.61

= 1- P(Z < 2.61)

$=1-\Phi( 2.61)$

[ using standard normal table ]

=l-0.995

=0.005

= 5%

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