Question

The mean height of women in a country (ages 20-29) is 63.5 inches. A random sample of 50 women in this age group is selected. What is the probability that the mean height for the sample is greater than 64 inches? Assume the standard deviation equals 2.96.

Answer 1

11.70% probability that the mean height for the sample is greater than 64 inches

Explanation:

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

Normal probability distribution

When the distribution is normal, we use the z-score formula.

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

`Z = (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 sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

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

In this question:

`\mu = 63.5, \sigma = 2.96, n = 50, s = (2.96)/(√(50)) = 0.4186`

What is the probability that the mean height for the sample is greater than 64 inches?

This is 1 subtracted by the pvalue of Z when X = 64.

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (64 - 63.5)/(0.4186)`

`Z = 1.19`

`Z = 1.19` has a pvalue of 0.8830

1 - 0.8830 = 0.1170

11.70% probability that the mean height for the sample is greater than 64 inches