Two lines and a point are guaranteed to be complainer if
1 answer:
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Answer:
A) 0.5737
B) 0.9884
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 160 lb and a standard deviation of 27.5 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 < 1.49) = 0.9319
P(X <= 150) = P( <
) = P(Z < -0.3636) = P(Z > 0.3636) = 0.3582
Therefore, P(150 < X < 201) = 0.9319 - 0.3582 = 0.5737 .
(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 < 9.311) = 1 - P(Z >= 9.311)
= 0.999995
P(X bar <= 150) = P( <
) = P(Z < -2.2709) = P(Z > 2.2709)
= 0.0116
Therefore, P(150 < X bar < 210) = 0.999995 - 0.0116 = 0.9884
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.