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 \\(\square\\) %. (round to two decimal places as needed.)

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. For the lower limit $x_1 = 55$ inches, $\mu = 68.7$ inches and $\sigma=3.5$ inches. So, $z_1=\frac{55 - 68.7}{3.5}=\frac{- 13.7}{3.5}\approx - 3.91$. For the upper limit $x_2 = 64$ inches, $z_2=\frac{64 - 68.7}{3.5}=\frac{-4.7}{3.5}\approx - 1.34$.

Step2: Use the standard normal distribution table

We want to find $P(-3.91

Step3: Convert to percentage

To convert the probability to a percentage, we multiply by 100. So the percentage is $0.0901\times100 = 9.01\%$.

Answer:

$9.01$