Question

In designing seats for commercial airplanes, engineers need to consider the hip width of passengers. Because men are in general larger than women, we consider their size. Men have hip widths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.0 inches. Suppose you randomly select 16 male passengers. Based on this sample, what is the probability that men have hips wider than 15.2 inches, on average?

Answer 1

0.07% probability that men have hips wider than 15.2 inches, on average

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 = (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 problem, we have that:

`\mu = 14.4, \sigma = 1, n = 16, s = (1)/(√(16)) = 0.25`

What is the probability that men have hips wider than 15.2 inches, on average?

This is 1 subtracted by the pvalue of Z when X = 15.2. So

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

By the Central Limit Theorem

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

`Z = (15.2 - 14.4)/(0.25)`

`Z = 3.2`

`Z = 3.2` has a pvalue of 0.9993

1 - 0.9993 = 0.0007

0.07% probability that men have hips wider than 15.2 inches, on average