A survey found that womens heights are normally distributed

A survey found that womens heights are normally distributed with mean 62.6 in. and standard deviation 2.5 in. The survey also found that mens heights are normally distributed with a mean 68.5. and standard deviation 2.5 in. The survey also found that mens heights are normally distributed with a mean 68.5 and standard deviation 2.7. Complete parts a through c. a)Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. Find the percentage of women meeting the height requirement. b) Find the percentage of men meeting the height requirement c)If the height requirement are changed to exclude only the tallest 5% of men and the shortest 5% of women, wgat are the new height requirements? The new height requirements are at least_____ in.And at most_________in. Round all to two decimal places as needed.

Solution

a) Mean = 62.6

sd = 2.5

P(X > 4 ft 9 in) = P(X > 57)

= P( Z > (57 - 62.6)/2.5)

= P(Z > -2.24)

= 0.9875

= 98.75%

--------------------------------------------------------------------

b) mean = 68.5

sd = 2.7

P(Y > 57) = P(Z > (57-68.5)/2.7)

= P(Z > -4.259)

= 1

= 100%

--------------------------------------------------------------------------------

Tallest 5 % men:

P(Y < y) = 0.95

=> (y - 68.5)/2.7 = 1.645

=> y = 72.94

Shortest 5% women:

P(X < x) = 0.05

=> (x - 62.6)/2.5 = -1.645

=> x = 58.49

The new height requirements are at least 58.49 in.And at most 72.94 in