Question

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and 201 lb. The new population of pilots has normally distributed weights with a mean of 155 lb and a standard deviation of 27.2 lb. a. If a pilot is randomly selected, find the probability that his weight is between 150 lb and 201 lb. The probability is approximately . (Round to four decimal places as needed.) b. If 38 different pilots are randomly selected, find the probability that their mean weight is between 150 lb and 201 lb. The probability is approximately |(Round to four decimal places as needed.) c. When redesigning the ejection seat, which probability is more relevant? A. Part (b) because the seat performance for a single pilot is more important. B. Part (b) because the seat performance for a sample of pilots is more important. C. Part (a) because the seat performance for a single pilot is more important. D. Part (a) because the seat performance for a sample of pilots is more important.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 155

standard deviation = $\sigma$ = 27.2

a) P( 150 < x < 201 ) = P[(150 - 155) / 27.2) < (x - $\mu$) /$\sigma$  < (201 - 155) / 27.2) ]

= P( - 0.18 < z < 1.69)

= P(z < 1.69) - P(z < - 0.18)

Using z table,

= 0.9545 -  0.4286

= 0.5259

b) n = 38

$\mu$_{$\bar x$} = 155

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 27.2 / $\sqrt$ 38 = 4.41

P(150 < < 201)  

T

= P[(150 - 155) / 4.41 < (
T
- $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (201 - 155) / 4.41)]

= P(- 1.13 < Z < 10.43)

= P(Z < 10.43) - P(Z < - 1.13)

Using z table,  

= 1 - 0.1292

= 0.8708

c) Correct option is C