SOLUTION: A survey found that women's heights are normally distributed with mean 62.7 in. and standard deviation 2.5 in. The survey also found that men's heights are normally distributed wit

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Question 836254: A survey found that women's heights are normally distributed with mean 62.7 in. and standard deviation 2.5 in. The survey also found that men's heights are normally distributed with a mean 68.6 in. and standard deviation 2.8.
Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 4 in. Find the percentage of women meeting the height requirement.
a)The percentage of women who meet the height requirement:
(Round to two decimal places as needed)
b)Find the percentage of men who meet the height requirement. The percentage of men who meet the height requirement is:
(round to two decimal places as needed)
c)If the height requirements are changed to exclude only the tallest 5% of men and the shortest 5% of women what are the new height requirements. The new height requirement is at least: in. and at most: in.
(round to one decimal place as needed)
c)
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
a)
Z*_Upper = (76 - 62.7)/2.5 = 5.32
Z*_Lower = (57 - 62.7)/2.5 = -2.28

The requirement is to get p(-2.28 < Z < 5.32) = p(Z<5.32) - p(Z<-2.28).
Use normal distribution table to get the answer for p and multiply with 100 to get the percentage.
For the rest, after this hint if you are still struggling to solve then contact me.
Cheers!