Question

Assume that the heights of adult Caucasian women have a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 women are randomly​ selected, find the probability that they have a mean height greater than 63.0 inches. Round to four decimal places.

Answer 1

Answer:

0.9918 = 99.18% probability that they have a mean height greater than 63.0 inches.

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 63.6, \sigma = 2.5, n = 100, s = \frac{2.5}{\sqrt{100}} = 0.25$

Find the probability that they have a mean height greater than 63.0 inches.

This is 1 subtracted by the pvalue of Z when X = 63. So

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

By the Central Limit Theorem

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

$Z = \frac{63 - 63.6}{0.25}$

$Z = -2.4$

$Z = -2.4$ has a pvalue of 0.0082

1 - 0.0082 = 0.9918

0.9918 = 99.18% probability that they have a mean height greater than 63.0 inches.