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

Answer:

$P(141.8$

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

$P(-0.639$

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

Step-by-step 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

$P(141.8$

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

$z=\frac{x-\mu}{\sigma}$

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

$P(141.8$

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

$P(-0.639$

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