Question

Assume that the heights of men are normally distributed with a mean of "71.3" inches and a standard deviation of 2.1 inches. If 36 men are randomly​ selected, find the probability that they have a mean height greater than 72.3 inches. Round to four decimal places.

Answer 1

Answer:

0.0021 = 0.21% probability that they have a mean height greater than 72.3 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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, 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 sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Assume that the heights of men are normally distributed with a mean of "71.3" inches and a standard deviation of 2.1 inches.

This means that $\mu = 71.3, \sigma = 2.1$

Sample of 36:

This means that $n = 36, s = \frac{2.1}{\sqrt{36}} = 0.35$

Find the probability that they have a mean height greater than 72.3 inches.

This is 1 subtracted by the pvalue of Z when X = 72.3. So

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

By the Central Limit Theorem

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

$Z = \frac{72.3 - 71.3}{0.35}$

$Z = 2.86$

$Z = 2.86$ has a pvalue of 0.9979

1 - 0.9979 = 0.0021

0.0021 = 0.21% probability that they have a mean height greater than 72.3 inches.