An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 191 lb. The new population of pilots has normally distributed weights with a mean of 136 lb and a standard deviation of 28.5 lb. a. If a pilot is randomly selected, find the probability that his weight is between 130 lb and 191 lb. The probability is approximately nothing. (Round to four decimal places as needed.) b. If 36 different pilots are randomly selected, find the probability that their mean weight is between 130 lb and 191 lb. The probability is approximately nothing. (Round to four decimal places as needed.) c. When redesigning the ejection seat, which probability is more relevant? A. Part (a) because the seat performance for a sample of pilots is more important. B. Part (b) because the seat performance for a sample of pilots is more important. C. Part (b) because the seat performance for a single pilot is more important. D. Part (a) because the seat performance for a single pilot is more important.

We will find the z score for each value and then subtract the probabilities for each to give us the area between them. We use the z score for an individual value:

[tex]z=\frac{X-\mu}{\sigma}[/tex]

Our first X is 130 and our second X is 191. Our mean, μ, is 136 and our standard deviation, σ, is 28.5:

[tex]z=\frac{130-136}{28.5}=\frac{-6}{28.5}\approx -0.21\\\\z=\frac{191-136}{28.5}=\frac{55}{28.5}\approx 1.93[/tex]

Using a z table, we can see that the area under the curve to the left of z = -0.21 is 0.4168; the area under the curve to the left of z = 1.93 is 0.9732. This means the area between them is

0.9732-0.4168 = 0.5564.

For part B,

We will find the z score for each value again and subtract them; however, since we have a sample we will use the z score for the mean of a sample:

[tex]z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}[/tex]

Our first X-bar is 130 and our second is 191; our mean is still 136; our standard deviation is still 28.5; and our sample size, n, is 36:

[tex]z=\frac{130-136}{28.5\div \sqrt{36}}=\frac{-6}{28.5\div 6}\approx -1.26\\\\z=\frac{191-136}{28.5\div \sqrt{36}}=\frac{55}{28.5\div 6}\approx 11.58[/tex]

The area under the curve to the left of -1.26 is 0.1038; the area under the curve to the left of 11.58 is 1.00:

1.00-0.1038 = 0.8962

For part C,

We want the probability that each individual pilot will be safe in these seats, so the value in part A is more important.