Question

A​ half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of​ today's women in their 20s are approximately normally distributed with a standard deviation of 2.07 inches. If the mean height today is the same as that of a​ half-century ago, what percentage of all samples of 21 of​ today's women in their 20s have mean heights of at least 65.86 ​inches?

Answer 1

99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

Explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of​ today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of​ today's women in their 20's have been taken.

Let
`\bar X` = sample mean heights

The z-score probability distribution for sample mean is given by;

Z =
`(\bar X -\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = population mean height of women = 64.7 inches

`\sigma` = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches is given by = P(
`\bar X`
`\geq` 65.86 inches)

P(
`\bar X`
`\geq` 65.86 inches) = P(
`(\bar X -\mu)/((\sigma)/(√(n) ) )`
`\geq`
`(65.86-64.7)/((2.07)/(√(21) ) )` ) = P(Z
`\geq` -2.57) = P(Z
`\leq` 2.57)

= 0.99492 or 99.5%

The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.

Therefore, 99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.