A survey found that women\'s heights are normally distributed with mean 63.1 in.

Question

A survey found that women's heights are normally distributed with mean 63.1 in. and standard deviation 2.4 in. The survey also found that men's heights are normally distributed with a mean 67.4 in. and standard deviation 2.8 . Complete parts a through c below.

a. 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 3 in. Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is 99.45 %. (Round to two decimal places as needed.)

b. Find the percentage of men meeting the height requirement. The percentage of men who meet the height requirement is 99.66 %. (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 requirements are at least 59.2 in. and at most 72.0 in. (Round to one decimal place as needed.)

Having difficulty with c, just don't understand. Could please give a brief explanation and solution to this problem. Thanks.

Explanation / Answer

a) as 4 ft and 9 in =57 in and 6ft and 3 in =75 in

percentage of women who meet the height requirement :

b)

percentage of men meeting the height requirement:

c)

for top 5% and bottom 5% critiacal value of z score =-/+ 1.6449

therefore top 5% score =mean +z* std deviation =67.4 +1.6449*2.8 =72.0 inch

and bottom 5% score =mean +z* std deviation =63.1 -1.6449*2.4 =59.2 inch

( please revert for any clarification required)

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