Mathematics, 24.02.2020 22:33 dayjionneam

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and 191 lb. The new population of pilots has normally distributed weights with a mean of 155 lb and a standard deviation of 29.2 lb.
(a) If a pilot is randomly selected, find the probability that his weight is between 150 lb and 201 lb.
(b) If 39 different pilots are randomly selected, find the probability that their mean weight is between 150 lb and 201 lb.
(c) When redesigning the ejection seat which probability is more relevant
Part (a) or Part (b)?

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Answer from: slonekaitlyn01

(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.; $\mu$ = 160 lb and $\sigma$ = 27.5 lb

(A) We know that Z = $\frac{X - \mu}{\sigma}$ ~ 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( $\frac{X - \mu}{\sigma}$ < $\frac{201 - 155}{29.2}$ ) = P(Z < 1.57) = 0.94179

P(X <= 150) = P( $\frac{X - \mu}{\sigma}$ < $\frac{150 - 155}{29.2}$ ) = 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 = $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ 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( $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{201 - 155}{\frac{29.2}{\sqrt{39} } }$ ) = P(Z < 9.84) = 1 - P(Z >= 9.84)

= 0.999995

P(X bar <= 150) = P( $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{150 - 155}{\frac{29.2}{\sqrt{39} } }$ ) = 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.

Answer from: Quest

idk

step-by-step explanation:

Agroup of people were asked what from of exercise they participate in. they were asked whether they

Answer from: Quest

Understanding what? lol

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