SOLUTION: I got the answer for part a but cannot figure out part b. Thanks.
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing be
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Probability-and-statistics
-> SOLUTION: I got the answer for part a but cannot figure out part b. Thanks.
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing be
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Question 1125237: I got the answer for part a but cannot figure out part b. Thanks.
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 140 lb and 191 lb. The new population of pilots has normally distributed weights with a mean of 150 lb and a standard deviation of 30.4 lb.
a. If a pilot is randomly selected, find the probability that his weight is between 140 lb and 191 lb.
The probability is approximately
0.5402
0.5402. (Round to four decimal places as needed.)
b. If 35 different pilots are randomly selected, find the probability that their mean weight is between 140 lb and 191 lb.
The probability is approximately
nothing. (Round to four decimal places as needed.) Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The SE is sd/sqrt(35)=5.14
Now the z-value (if this is a known sd, which it sounds like from the wording)is (140-150)/5.14 to (191-150)/5.14
z=(x bar-mean)/sigma/sqrt(n)
-1.95 to 7.98
probability is 0.9744
This says in essence that the likelihood of the MEAN of a sample being in this region is much higher than a single value.
