Question

A survey found that womens heights are normally distributed with mean 62.1 in and standard deviation 2.1 in the survey also found that mens heights are normally distributed with a mean 69.7 and SD 3.8 a) most of the live characters at an amusement park have height requirements with a minimum of 4ft 9in and a maximum of 6ft 4in 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 (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

For normal distribution, P(X < A) = P(Z < (A - mean)/standard deviation)

a) For women's heights. mean = 62.1 in

Standard deviation = 2.1 in

Percentage of women meeting the height requirement = P(height of a woman is between 4ft 9in and 6ft 4in)

= P(height of a woman is between 57 and 72 in)

= P(57 < X < 72)

= P(X < 72) - P(X < 57)

= P(Z < (72 - 62.1)/2.1) - P(Z < (57 - 62.1)/2.1)

= P(Z < 4.71) - P(Z < -2.43)

= 1 - 0.0075

= 0.9925

= 99.25%

b) For men's heights. mean = 69.7 in

Standard deviation = 3.8 in

Percentage of women meeting the height requirement = P(height of a woman is between 4ft 9in and 6ft 4in)

= P(height of a woman is between 57 and 72 in)

= P(57 < X < 72)

= P(X < 72) - P(X < 57)

= P(Z < (72 - 69.7)/3.8) - P(Z < (57 - 69.7)/3.8)

= P(Z < 0.61) - P(Z < -3.34)

= 0.7291 - 0.0004

= 0.7287

= 72.87%

c) Let the height requirement to exclude only the shortest 5% of women be W

P(X < W) = 0.05

P(Z < (W - 62.1)/2.1) = 0.05

Take Z score corresponding to probability of 0.0500 from standard normal distribution table

(W - 62.1)/2.1 = -1.645

W = 58.6 in

Let the height requirement to exclude only the tallest 5% of men be M

P(X > M) = 0.05

P(X < M) = 1 - 0.05 = 0.95

P(Z < (M - 69.7)/3.8) = 0.95

Take Z score corresponding to probability of 0.9500 from standard normal distribution table

(M - 69.7)/3.8 = 1.645

M = 76.0 in

he new height requirements are at least 58.6 in. and at most 76.0 in