An airliner carries 100 passengers and has doors with a height of 72 in. Heights of men are
normally distributed with a mean of 69.0 in and a standard deviation of 2.8 in.
If half of the 100 passengers are men, find the probability that the mean height of the 50 men is less than 72 in.
The probability is .
(Round to four decimal places as needed.)
This is a normal distribution question with
Sample size (n) = 50
Since we know that
P(x < 72.0)=?
The z-score at x = 72.0 is,
z = 7.5758
This implies that
PS: you have to refer z score table to find the final probabilities.
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