Question

a survey found that womens heights are normally distributed with mean 63.2 in and standard deviation 2.2 in. the survey also found that mens heights are normally distributed with mean 68.7 in and standard deviation 3.5 in. most of the live - characters employed at an amusement park have height requirements of a minimum of 55 in. and a maximum of 64 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 9.01%.
(round to two decimal places as needed.)
since most men blank the height requirement, it is likely that most of the characters are blank

Explanation:

Step1: Calculate z - scores

For the lower - bound $x_1 = 55$ in, the z - score formula is $z_1=\frac{x_1-\mu}{\sigma}$, where $\mu = 68.7$ in and $\sigma=3.5$ in. So $z_1=\frac{55 - 68.7}{3.5}=\frac{- 13.7}{3.5}\approx - 3.91$. For the upper - bound $x_2 = 64$ in, $z_2=\frac{x_2-\mu}{\sigma}=\frac{64 - 68.7}{3.5}=\frac{-4.7}{3.5}\approx - 1.34$.

Step2: Find probabilities from z - table

We want to find $P(-3.91Since most men do not meet the height requirement (because only 9.01% meet it), it is likely that most of the characters are women.

Answer:

Since most men do not meet the height requirement, it is likely that most of the characters are women.