Question

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?

Answer 1

`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