Question

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 9090​% 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 14.5 in. and a standard deviation of 1.2 in. Find Upper P90. That​ is, find the hip breadth for men that separates the smallest 90​% from the largest 10​%.
The hip breadth for men that separates the smallest 90​% from the largest 10% is P90__in.


​(Round to one decimal place as​ needed.)

Answer 1

Answer:

Hip breadths less than or equal to 16.1 in. includes 90% of the males.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.5

Standard Deviation, σ = 1.2

We are given that the distribution of hip breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.10.

P(X > x)  

$P( X > x) = P( z > \displaystyle\frac{x - 14.5}{1.2})=0.10$  

$= 1 -P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.10$  

$=P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.90$  

Calculation the value from standard normal z table, we have,  

$P(z < 1.282) = 0.90$

$\displaystyle\frac{x - 14.5}{1.2} = 1.282\\x = 16.0384 \approx 16.1$  

Hence, hip breadth of 16.1 in. separates the smallest 90​% from the largest 10%.

That is hip breaths greater than 16.1 in. lies in the larger 10%.