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

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?
the new height requirements are a minimum of in and a maximum of in.
(round to one decimal place as needed.)

Explanation:

Step1: Find z - score for the lower 5%

The lower 5% corresponds to a cumulative probability of 0.05. Looking up in the standard - normal distribution table (z - table), the z - score $z_1$ such that $P(Z

Step2: Use the z - score formula to find the lower height

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 68.7$ in (mean of men's heights) and $\sigma = 2.2$ in (standard deviation of men's heights). Rearranging for $x$ gives $x=\mu+z\sigma$. Substituting $z = z_1=-1.645$, $\mu = 68.7$, and $\sigma = 2.2$:
$x_1=68.7+( - 1.645)\times2.2=68.7-3.619 = 65.1$ in.

Step3: Find z - score for the upper 50%

The upper 50% corresponds to a cumulative probability of 0.5. The z - score $z_2$ such that $P(Z < z_2)=0.5$ is $z_2 = 0$.

Step4: Use the z - score formula to find the upper height

Substituting $z = z_2 = 0$, $\mu = 68.7$, and $\sigma = 2.2$ into $x=\mu+z\sigma$ gives $x_2=68.7+0\times2.2 = 68.7$ in.

Answer:

The new height requirements are a minimum of 65.1 in and a maximum of 68.7 in.