SOLUTION: An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 140 lb and 181 lb. The new population of pilots has normally di

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Question 1135425: An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 140 lb and 181 lb. The new population of pilots has normally distributed weights with a mean of 150 lb and a standard deviation of 28.6 lb.
a. If a pilot is randomly​ selected, find the probability that his weight is between 140 lb and 181 lb.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
seat designed for pilots weighing between 140 and 181 pounds.

new population of pilots has a mean weight of 150 pounds and a standard deviation of 28.6 pounds.

find probability that a randomly selected pilot will weigh between 140 and 181 pounds.

the following calculator does the job very nicely for you.

it can be found at http://davidmlane.com/hyperstat/z_table.html

this calculator tells you hat the probability that a randomly selected pilot will weight between 140 and 181 pounds is equal to .4975.

here's what the results of the use of this calculator looks like.

$$$

you can also do it the normal way by finding the z-score and looking up the area in the normal distribution table

doing it that way, you need to find the lower z-score and the upper z-score and then find the area in between.

the z-score formula is z = (x - m) / sd

z is the z-score.
x is the raw score.
m is the mean.
sd is the standard deviation

your mean is 150 and your standard deviation is 28.6.

your lower x-score is 140.
your upper x-score is 181.

the z-score formula is z = (x-m) / sd

the lower z-score would be z = (140 - 150) / 28.6 = -.3496503497.

the upper z-score would be z = (181 - 150) / 28.6 = 1.083916084.

to use the tables, you would round these to -0.35 and 1.08.

in the z-score table, the area to the left of -0.35 equals .36315 and the area for the left of 1.08 = .85993.

the area in between is .85993 minus ..36315 = .49676.

that's pretty close to .4975 using the online calculator, the difference being due to rounding.

here's another normal distributor z-score calculator by university of iowa.

https://homepage.stat.uiowa.edu/~mbognar/applets/normal.htmlhttps://homepage.stat.uiowa.edu/~mbognar/applets/normal.html

this calculator is used similar to how you would use the z-score table, except it allows more details in the z-score and also interpolates to give you a more accurate answer.

the calculator works off the raw score and the mean and the standard deviation, or it works off the z-score.

when in z-score mode, the mean is 0 and the standard deviation is 1.

the first calculator referenced also works off the z-score or off the raw score and the mean and the standard deviation.

here's the results of using the second referenced calculator.

$$$

$$$

the area getween the two z-scores is .8608 minus .3633 = .4975.

that agrees with the resulted from the use of the first referenced calculator.

the z-score table i used can be found at https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf