Question

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 99% of all males (Accommodating 100% of males would require very wide seats that would be much too expensive ) Men have hip breadths that are normally distributed with a mean of 14.2 in. and a standard deviation of 0.9 in. Find P

99
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. That is, find the hip breadth for men that separates the smallest 99% from the largest 1% The hip breadth for men that separates the smallest 99% from the largest 1% is P99= in. (Round to one decimal place as needed.)

Answer 1

To find the hip breadth for men that separates the smallest 99% from the largest 1%, we can use the concept of standard deviations in a normal distribution.

Given:
Mean (μ) = 14.2 inches
Standard Deviation (σ) = 0.9 inches

We want to find the hip breadth that corresponds to the 99th percentile, denoted as P99.

First, we need to determine the z-score corresponding to the 99th percentile. The z-score represents the number of standard deviations a value is away from the mean.

To find the z-score corresponding to the 99th percentile, we can use a standard normal distribution table or a calculator. The z-score for the 99th percentile is approximately 2.326.

Next, we can calculate the hip breadth for P99 using the formula:

P99 = μ + (z * σ)

P99 = 14.2 + (2.326 * 0.9)

P99 ≈ 14.2 + 2.0934

P99 ≈ 16.2934 inches

Therefore, the hip breadth for men that separates the smallest 99% from the largest 1% is approximately 16.3 inches (rounded to one decimal place).