Question

Aircraft seat width engineers want to design seats in commercial aircraft so that they are wide enough to fit 99% of all adults. (accommodating 100% of adults would require very wide seats that would be much too expensive.) assume adults have hip widths that are normally distributed with a mean of 14.3 in. and a standard deviation of 0.9 in. (based on data from applied ergonomics). find p99. that is, find the hip width for adults that separates the smallest 99% from the largest 1%.

Answer 1

The hip width for adults that separates the smallest 99% from the largest 1% is approximately 16.38 inches.

Step-by-step explanation:

To find the hip width for adults that separates the smallest 99% from the largest 1%, we need to find the z-score associated with the 99th percentile of the standard normal distribution, and then use that z-score to find the corresponding hip width.

First, we calculate the z-score using the formula: z = (x - μ) / σ, where x is the value we want to find the z-score for, μ is the mean, and σ is the standard deviation.

For this problem, we want to find the z-score that corresponds to the 99th percentile, which is the value that separates the smallest 99% from the largest 1%. This is denoted as z0.99.

Using a standard normal distribution table or a calculator, we can find that z0.99 is approximately 2.33.

Now, we can use the formula z = (x - μ) / σ to find the hip width (x) that corresponds to the z-score of 2.33. Rearranging the formula, we have x = z * σ + μ.

Plugging in the values, we get:

x = 2.33 * 0.9 + 14.3 = 16.38 inches

Therefore, the hip width for adults that separates the smallest 99% from the largest 1% is approximately 16.38 inches.