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irakobra [83]
3 years ago
11

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 9090​% of all males.​ (Accommodating 1

00% 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.)
Mathematics
1 answer:
Molodets [167]3 years ago
4 0

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%.

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