A survey found that women\'s heights are normally distributed with mean 63.7 inc

Question

A survey found that women's heights are normally distributed with mean 63.7 inches and standard deviation 2.5 inches.The survey also found that men's heights are normally distributed with a mean 67.5 inches and standard deviation 2.6

A- Most live characters at an amusement parks have a height requirement with a minimum of 4ft 9 inches and maximum of 6ft 2 inches.
What is the % of women who meet the requirements?__

B-What's the % for men?__

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?_
?__

Explanation / Answer

Women: Mean height = 63.7 inches; stdev = 2.5 inches

Men: Mean height = 67.5 inches; stdev = 2.6 inches

a) Minimum height = 4ft 9 in = 4*12 + 9 = 57 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 = 57 and 74

Z = (xbar - mu)/stdev

= P[Z = (74 - 63.7)/2.5] - P[Z = (57 - 63.7)/2.5]

= P[Z = 4.12] - p[-2.6]

= 1 - 0.0046

= 0.9954

=99.54%

b) The percentage of men who meet the height requirement is given by area under the Z curve between points Xbar = 57 and 74

Z = (xbar - mu)/stdev

= P[Z = (74 - 67.5)/2.6] - P[Z = (57 - 67.5)/2.6]

= P[Z = 2.5] - p[-4.03]

= 0.9937 - 0.0000

=99.37%

c) Tallest 5% men

p = 0.95

Z = 1.645

1.645 = (Xbar - 67.5)/2.6

X bar = 72.77 inches

Shortest 5% women

p = 0.05

Z = -1.645

-1.645 = (Xbar - 63.7)/2.5

X bar = 59.58 inches

the new height requirements are at least _59.58__ in. and at most 72.77 in.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.