SOLUTION: An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 120 lb and 171 lb. The new population of pilots has normally di
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Question 1121943: An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 120 lb and 171 lb. The new population of pilots has normally distributed weights with a mean of 127 lb and a standard deviation of 29.2 lb. a. If a pilot is randomly selected, find the probability that his weight is between 120 lb and 171 lb? Answer by Theo(13342) (Show Source):
you would set the mean at 127 and the standard deviation at 29.2 and then select between and enter 120 and 171.
the calculator tells you that the probability of getting a pilot who weighs between 120 and 171 pounds is equal to .5288.
that's the easy way.
without the use of this wonderful calculator, or with9out the use of any calculator, you would need to use the normal distribution tables.
they require a little more effort.
first you would need to get the z-score for a weight of 120 pounds and a weight of 171 pounds, given the mean is 127 and the standard deviation is 29.2
the formula for z-score is:
z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation.
in your problem:
on the low side:
x = 120
m = 127
s = 29.2
solve for z to get z = (120 - 127) / 29.2 = -.2397260274 which you would probably round off to - 0.24 since the normal distribution table shows the z-scores rounded to 2 decimal places.
on the high side:
x = 171
m = 127
s = 19.2
solve for z to get z = (171 - 127) / 29.2 = 1.506849315 which youo would probably round off to 1.51 since the normal distribution table shows the z-scores rounded to 2 decimal laces.
you would then look into the z-score table for the area to the left of a z-score of -0.24 and the area to the left of a z-score of 1.51.
you would then subtract the smaller area from the larger area to get the area in between.
that would be the probability that a randomly selected pilot would weigh between 120 and 171 pounds.
from the table, i got .......
area to the left of z-score of -0.24 = .4052
area to the left of a z-score of 1.51 = .9345
area in between = .9345 - .4052 = .5293
that's your probability.
compare to the use of the online calculator, which says that the probability is .5288.
not bad.
the difference is .0005 which is only off of the more detailed analysis by less than a tenth of a percent.
