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Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If​ women's weights are normally distributed with a mean of 173.6 lb and a standard deviation of 49.8 ​lb, what percentage of women have weights that are within those​ limits? Are many women excluded with those​ specifications?

User Indichimp
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Answer:


P(141.8<X<218)=P((141.8-\mu)/(\sigma)<(X-\mu)/(\sigma)<(218-\mu)/(\sigma))=P((141.8-173.6)/(49.8)<Z<(218-173.6)/(49.8))=P(-0.639<z<0.892)

And we can find this probability with this difference and using the normal standard table:


P(-0.639<z<0.892)=P(z<0.892)-P(z<-0.639)=0.814-0.261= 0.553

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:


X \sim N(173.6,49.8)

Where
\mu=173.6 and
\sigma=49.8

We are interested on this probability


P(141.8<X<2188)

And the best way to solve this problem is using the normal standard distribution and the z score given by:


z=(x-\mu)/(\sigma)

If we apply this formula to our probability we got this:


P(141.8<X<218)=P((141.8-\mu)/(\sigma)<(X-\mu)/(\sigma)<(218-\mu)/(\sigma))=P((141.8-173.6)/(49.8)<Z<(218-173.6)/(49.8))=P(-0.639<z<0.892)

And we can find this probability with this difference and using the normal standard table:


P(-0.639<z<0.892)=P(z<0.892)-P(z<-0.639)=0.814-0.261= 0.553

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

User Samad
by
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