Question

A survey found that​ women's heights are normally distributed with mean 63.3 in. and standard deviation 2.3 in. The survey also found that​ men's heights are normally distributed with mean 69.1 in. and standard deviation 3.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57in. and a maximum of 62 in. Complete parts​ (a) and​ (b) below.

a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement​ park?

The percentage of men who meet the height requirement is ----- ​(Round to two decimal places as​ needed.)

b.If the height requirements are changed to exclude only the tallest​ 50% of men and the shortest​ 5% of​ men, what are the new height​ requirements?

Answer 1

Solution :

Given that ,

mean = = 69.1

standard deviation = = 3.1

a) P(57 < x < 62 ) = P[(57 - 69.1)/ 3.1) < (x - ) /  < (62 - 69.1) / 3.1) ]

= P(-3.90 < z < -2.29)

= P(z < -2.29) - P(z < -3.90)

Using z table,

= 0.0110 - 0

= 0.0110

The percentage of men who meet the height requirement is = 1.10%

b) Using standard normal table,

P(Z > z) = 50%

= 1 - P(Z < z) = 0.50  

= P(Z < z) = 1 - 0.50

= P(Z < z ) = 0.50

= P(Z < 0 ) = 0.50  

z = 0

Using z-score formula,

x = z * +

x = 0 * 3.1 + 69.1

x = 69.1

Using standard normal table,

P(Z < z) = 5%

= P(Z < z ) = 0.05

= P(Z < -1.645 ) = 0.05  

z = -1.645

Using z-score formula,

x = z * +

x = -1.645 * 3.1 + 69.1

x = 64.0

The new height​ requirements is 69.1 in. and 64.0 in.