Engineers want to design seats in commercial aircraft so that they are wide enough to fit 98% 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 14.4 in. and a standard deviation of 1.0 in.. Find P98. That is, find the hip breadth for men that separates the smallest 98% from the largest 2

P98 = 16.154in

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Men have hip breadths that are normally distributed with a mean of 14.4 in. and a standard deviation of 1.0 in.

This means that

Find P98

This is the 98th percentile, that is, X when Z has a pvalue of 0.98, so X when Z = 2.054.

So

P98 = 16.154in

is this part of a question or something

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11 -1> 1 -1+ a

-12 > a