Question

In: Statistics and Probability

An engineer is going to redesign an ejection seat for an airplane. The seat was designed...

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 191 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 191 lb. The probability is approximately nothing. ​(Round to four decimal places as​ needed.)

b. if 40 different pilots are randomly selected, find the probability that their mean weight is between 130 lb and 191 lb.

Solutions

Expert Solution

Answer:

Given Data,

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 191 lb.

The new population of pilots has normally distributed weights with a mean of 140 lb and a standard deviation of 26.2 lb.

Given that,

Mean = 140 lb

Standard deviation = 26.3 lb

(a). To find the probability that his weight is between 130 lb and 191 lb:

P(X < A) = P(Z < (A - mean)/standard deviation)

P(130 < X < 191) = P(X < 191) - P(X < 130)

= P(Z < (191 - 140)/26.3) - P(Z < (130 - 140)/26.3)

= P(Z < 1.94) - P(Z < -0.38)

= 0.974- 0.349

= 0.565

(b). To find the probability that their mean weight is between 130 lb and 191 lb:

Given n=40

Mean = 140 lb

P(X < A) = P(Z < (A - mean)/standard error)

Standard error = 26.3/v40=4.158

Standard error =4.158

P(130 < X < 191) = P(X < 191) - P(X < 130)

= P(Z < (191 - 140)/4.158) - P(Z < (130 - 140)/4.158)

= P(Z < 12.025) - P(Z < -2.405)

= 1 -0.0081

= 0.99


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