Question

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 14.8 in. and a standard deviation of 1.1 in. find p99. that is, find the hip width for adults that separates the smallest 99% from the largest 1%.
what is the maximum hip width that is required to satisfy the requirement of fitting 99% of adults?
□ in. (round to one decimal place as needed.)

Explanation:

Step1: Find the z - score

For a 99% confidence level (to separate the smallest 99% from the largest 1%), the z - score \(z\) can be found using the standard normal distribution table. The z - score corresponding to an area of 0.99 is \(z = 2.33\).

Step2: Use the formula for the value in a normal distribution

The formula for a value \(x\) in a normal distribution is \(x=\mu+z\sigma\), where \(\mu = 14.8\) (mean) and \(\sigma=1.1\) (standard deviation).
Substitute the values into the formula: \(x = 14.8+2.33\times1.1\).
First, calculate \(2.33\times1.1 = 2.563\).
Then, \(x=14.8 + 2.563=17.363\).

Answer:

\(17.4\)