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Bingel [31]
3 years ago
5

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.
Mathematics
1 answer:
Luba_88 [7]3 years ago
8 0

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.

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