2 years ago

Question 10, 4.A.27

Mathematics

1 answer:

anygoal [31]2 years ago

3 0

dun dun dunnnnnnnnnn

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

Molodets [167]

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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2 years ago