A survey found that womens heights are normally distributed

A survey found that womens heights are normally distributed with mean 63.8 in and standard deviation 2.5 in the survey also found that mens heights are normally distributed with a mean 67.9 and SD 2.7

a) most of the live characters at an amusement park have height requirements with a minimum of 4ft 9 in and a maximum of 6ft 2 in. Find the percentage of women meeting the height requirement.

the percentage of woment who meet the height requirement?

(round to two decimal places as needed)

b) find the percentage of men meeting the height requirement

the percentage of men meeting 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)

Solution

Women: Mean height = 63.2 inches; stdev = 2.7 inches

Men: Mean height = 68.1 inches; stdev = 2.7 inches

a) Minimum height = 4ft 9 in = 4*12 + 9 = 57 inches

Maximum height = 6ft 4in = 6*12 + 4 = 76 inches

The percentage of woment who meet the height requirement is given by area under the Z curve between points Xbar = 57 and 76

Z = (xbar - mu)/stdev

= P[Z = (76 - 63.2)/2.7] - P[Z = (57 - 63.2)/2.7]

= P[Z = 4.74] - p[-2.30]

= 1 - 0.0107

= 0.9893

=98.93%

b) The percentage of men who meet the height requirement is given by area under the Z curve between points Xbar = 57 and 76

Z = (xbar - mu)/stdev

= P[Z = (76 - 68.1)/2.7] - P[Z = (57 - 68.1)/2.7]

= P[Z = 2.93] - p[-4.11]

= 0.9983 - 0.00000

=99.83%

c) Tallest 5% men

p = 0.95

Z = 1.645

1.645 = (Xbar - 68.1)/2.7

X bar = 72.54 inches

Shortest 5% women

p = 0.05

Z = -1.645

-1.645 = (Xbar - 63.2)/2.7

X bar = 58.76 inches

the new height requirements are at least _58.76__ in. and at most 72.54 in.