Math  /  Data & Statistics

QuestionPoints: 0 of 6
Engineers want to design seats in commercial aircraft so that they are wide enough to fit 99%99 \% of all adults. (Accommodating 100%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 14.6 in. and a standard deviation of 0.8 in. Find P99. That is, find the hip width for adults that separates the smallest 99%99 \% from the largest 1%1 \%.
What is the maximum hip width that is required to satisfy the requirement of fitting 99%99 \% of adults? \square in. (Round to one decimal place as needed.)

Studdy Solution
Convert the z-score to the actual hip width using the formula for a normal distribution:
X=μ+zσ X = \mu + z \cdot \sigma
Substitute the known values:
X=14.6+2.330.8 X = 14.6 + 2.33 \cdot 0.8
Calculate:
X=14.6+1.864=16.464 X = 14.6 + 1.864 = 16.464
Round to one decimal place:
X16.5 X \approx 16.5
The maximum hip width required to fit 99% of adults is:
16.5 \boxed{16.5} inches.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord