An engineer is going to redesign an ejection seat for an airplane. The seat was
ID: 3152511 • Letter: A
Question
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 171 lb. The new population of pilots has normally distributed weights with a mean of 140 lb and a standard deviation of 26.2 lb.
a. If a pilot is randomly selected, find the probability that his weight is between 130 lb and 171 lb. The probability is approximately . (Round to four decimal places as needed.)
b. If 34 different pilots are randomly selected, find the probability that their mean weight is between 130 lb and 171 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 (a) because the seat performance for a sample of pilots is more important.
C. Part (b) because the seat performance for a sample of pilots is more important.
D. Part (a) because the seat performance for a single pilot is more important.
Explanation / Answer
as the distribution is given to be normal
the mean = 140
standard deviation = 26.2
the formula to be used = z =(x-mean)/standard deviation
a) p(130<x<171) =
For x = 130 , z = (130 - 140) /26.2 = -0.38 and for x = 171, z = (171 -140) / 26.2 = 1.18
Hence P(130 < x < 171) = P(-0.38 < z < 1.18) = [area to the left of z = 1.18] - [area to the left of -0.38]
= 0.881 - 0.3519 = 0.5291
b) the formula to be used = z = (x-mean)/(standarddeviation/sqrt(n))
For x = 130 , z = (130 - 140) /(26.2/sqrt(34)) = -2.22 and for x = 171, z = (171 -140) / (26.2/sqrt(34) = 6.90
Hence P(130 < x < 171) = P(-2.22 < z < 6.90) = [area to the left of z = 6.90] - [area to the left of -2.22]
= 1 - 0.0139 = 0.9861
c) probability of part b is more relevent because the performance of a single seat is more important as option A is correct.