Answer:
(a) 0.50928
(b) 0.857685.
Step-by-step explanation:
We are given that an engineer is going to redesign an ejection seat for an airplane. The new population of pilots has normally distributed weights with a mean of 155 lb and a standard deviation of 29.2 lb i.e.; = 160 lb and
= 27.5 lb
(A) We know that Z = ~ N(0,1)
Let X = randomly selected pilot
If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)
P(150 < X < 201) = P(X < 201) - P(X <= 150)
P(X < 201) = P( <
) = P(Z < 1.57) = 0.94179
P(X <= 150) = P( <
) = P(Z < -0.17) = 1 - P(Z < 0.17) = 1 - 0.56749
= 0.43251
Therefore, P(150 < X < 201) = 0.94179 - 0.43251 = 0.50928 .
(B) We know that for sampling mean distribution;
Z = ~ N(0,1)
If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);
P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)
P(X bar < 201) = P( <
) = P(Z < 9.84) = 1 - P(Z >= 9.84)
= 0.999995
P(X bar <= 150) = P( <
) = P(Z < -1.07) = 1 - P(Z < 1.07)
= 1 - 0.85769 = 0.14231
Therefore, P(150 < X bar < 210) = 0.999995 - 0.14231 = 0.857685.
C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.