Question

A survey found that​ women's heights are normally distributed with mean 62.6 in. and standard deviation 2.2 in. The survey also found that​ men's heights are normally distributed with a mean 67.3 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 8 in. and a maximum of 6 ft 2 in. Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is ​%. ​(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 ​%. ​(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 in. and at most in. ​(Round to one decimal place as​ needed.)

Answer 1

sol:

Women: Mean height = 62.6 inches; stdev = 2.2 inches

Men: Mean height = 67.3 inches; stdev = 2.8 inches

a) Minimum height = 4ft 8 in = 4*12 + 8 = 56 inches

Maximum height = 6ft 2in = 6*12 + 2 = 74 inches

The percentage of woment who meet the height requirement is given by area under the Z curve between points Xbar = 56 and 74

Z = (xbar - mu)/stdev

= P[Z = (74 - 62.6)/2.2] - P[Z = (56 - 62.6)/2.2]

= P[Z = 5.18] - p[-3]

= 1 - 0.0013

= 0.9987

=99.87%

b) The percentage of men who meet the height requirement is given by area under the Z curve between points Xbar = 56 and 74

Z = (xbar - mu)/stdev

= P[Z = (74 - 67.3)/2.8] - P[Z = (54 - 67.3)/2.8]

= P[Z = 2.39] - p[-4.75]

= 0.9916 - 0.00000

=99.16%

c) Tallest 5% men

p = 0.95

Z = 1.645

1.645 = (Xbar - 67.3)/2.8

X bar = 71.90 inches

Shortest 5% women

p = 0.05

Z = -1.645

-1.645 = (Xbar - 62.6)/2.2

X bar = 58.98 inches

the new height requirements are at least 58.98_ in. and at most 71.90 in.