Question

A​ half-century ago, the mean height of women in a particular country in their 20s was 62.1 inches. Assume that the heights of​ today's women in their 20s are approximately normally distributed with a standard deviation of 2.85 inches. If the mean height today is the same as that of a​ half-century ago, what percentage of all samples of 24 of​ today's women in their 20s have mean heights of at least 63.42 ​inches? About nothing​% of all samples have mean heights of at least 63.42 inches.

Answer 1

Answer:

About 1.16% of all samples have mean heights of at least 63.42 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 62.1, \sigma = 2.85, n = 24, s = \frac{2.85}{\sqrt{24}} = 0.58175$

What percentage of all samples of 24 of​ today's women in their 20s have mean heights of at least 63.42 ​inches?

This is 1 subtracted by the pvalue of Z when X = 63.42. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{63.42 - 62.1}{0.58175}$

$Z = 2.27$

$Z = 2.27$ has a pvalue of 0.9884

1 - 0.9884 = 0.0116

About 1.16% of all samples have mean heights of at least 63.42 inches.

Answer 2

The percentage of all samples of 24 of​ today's women in their 20s that  have mean heights of at least 63.42 ​inches is 1.2%.

Percentage of sample mean

Standard deviation=2.85/√24

Standard deviation=0.58175

z=x-mean/standard deviation

z=63.42-62.1/0.58175

z=1.32/0.58175

z=2.269

z=2.27(Approximately)

Since 2.27 has pvalue of 0.9884 using standard normal distribution table

Hence:

Percentage=1-pvalue

Percentage=1-0.9884

Percentage=0.0116×100

Percentage=1.2% (Approximately)

Inconclusion the percentage of all samples of 24 of​ today's women in their 20s that  have mean heights of at least 63.42 ​inches is 1.2%.

Learn more about Percentage of sample mean here:https://brainly.com/question/7597734