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 3.1 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 0.01 %.
(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 68.7 in.
(round to one decimal place as needed.)

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. For men, $\mu = 68.7$ inches and $\sigma=3.5$ inches.

Step2: Find the z - score for the lower 5%

We look up the z - score in the standard normal distribution table (z - table). The area to the left of the z - score for the lower 5% is 0.05. The z - score corresponding to an area of 0.05 is approximately $z=-1.645$.

Step3: Solve for the lower - bound height $x_1$

Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we substitute $z = - 1.645$, $\mu = 68.7$, and $\sigma = 3.5$. Then $-1.645=\frac{x_1 - 68.7}{3.5}$. Cross - multiply: $x_1-68.7=-1.645\times3.5$. Calculate $-1.645\times3.5=-5.7575$. Then $x_1=68.7 - 5.7575=62.9425\approx62.9$ inches.

Step4: Find the z - score for the upper 50%

The area to the left of the z - score for the upper 50% is 0.5. The z - score corresponding to an area of 0.5 is $z = 0$.

Step5: Solve for the upper - bound height $x_2$

Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we substitute $z = 0$, $\mu = 68.7$, and $\sigma = 3.5$. Then $0=\frac{x_2 - 68.7}{3.5}$. Cross - multiply: $x_2-68.7=0\times3.5$. So $x_2 = 68.7$ inches.

Answer:

The new height requirements are a minimum of 62.9 in and a maximum of 68.7 in.