SOLUTION: The baggage weights for passengers using a particular airline are normally distributed with a mean of 22 lb and a standard deviation of 5 lb. If the limit on total luggage weight i
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Question 875039: The baggage weights for passengers using a particular airline are normally distributed with a mean of 22 lb and a standard deviation of 5 lb. If the limit on total luggage weight is 2161 lb., what is the probability that the limit will be exceeded for 90 passengers? (Give your answer correct to four decimal places.) Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! mean of 22 lb and a standard deviation of 5 lb
2161/90 = 24
P(x >24) = 2/(5/sqrt(90)) = 2/ .527 = 3.7951
P(z > 3.7951) = .0001 0r .01%
