Question

Suppose that an airline uses a seat width of 16.2 in. Assume men have hip breadths that are normally distributed with a mean of 14 in. and a standard deviation of 1 in. Complete parts (a) through (c) below.

Answer 1

Given:

population mean (μ) = 14 inches

population standard deviation (σ) = 1 inch

sample size (n) = 126

Find: the probability that a sample mean > 16.2 inches

Solution:

To determine the probability, first, let's convert x = 16.2 to a z-value using the formula below.

`x=\frac{\bar{x}-\mu}{\sigma/√(n)}`

Let's plug into the formula above the given information.

`z=(16.2-14)/(1/√(126))`

Then, solve.

`z=(2.2)/(0.089087)`
`z=24.6949`

The equivalent z-value of x = 16.2 is z = 24.6949

Since we are looking for the probability of greater than 16.2 inches, let's find the area under the normal curve to the right of z = 24.6949.

Based on the standard normal distribution table, the area from the center to z = 24.6949 is 0.5

Since we want the area to the right, let's subtract 0.5 from 0.5.

`0.5-0.5=0`

Therefore, the probability that a sample mean of 126 men is greater than 16.2 inches is 0.