Question

A survey found that women's heights are normally distributed with mean 63.6 in. and standard deviation 3.7 in. The survey also found that men's heights are normally distributed with mean 68.7 in. and standard deviation 3.8 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. 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__%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.

Answer 1

Answer:

(a) The percentage of men meeting the height requirement is 3.79%

The percentage of women that are employed as characters at the amusement park is greater than the percentage of men.

(b) The new height requierements are a minimun of 62.45 in. and a maximun of 68.7 in.

Step-by-step explanation:

The normal probability density function for men's heights is given by

$m(x) = \frac{1}{\sqrt{2\pi}3.8}\exp{-\frac{(x-68.7)^{2}}{2(3.8)^{2}}}$ and

the normal probability density function for women's heights is given by

$w(x) = \frac{1}{\sqrt{2\pi}3.7}\exp{-\frac{(x-63.6)^{2}}{2(3.7)^{2}}}$.

You can use a table from a book or a computer to calculate the next probabilities.

We used the R statistical programming language.

(a) The percentage of men meeting the height requirement is given by

$\int\limits_{57}^{62}m(x) \, dx$ = 0.0379 (3.79 %)

(you should multiply by 100 this number)

(in R execute) pnorm(62, mean = 68.7, sd = 3.8) - pnorm(57, mean = 68.7, sd = 3.8)

The percentage of women that are employed as characters at the amusement park is greater than the percentage of men.

The percentage of women meeting the height requirement is given by

$\int\limits_{57}^{62}w(x) \, dx$ = 0.2955 (29.55 %)

(you should multiply by 100 this number)

(in R execute) pnorm(62, mean = 63.6, sd = 3.7) - pnorm(57, mean = 63.6, sd = 3.7)

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(b) To exclude the tallest 50% of men we should find the 50th percentile of the men's heights. This is the same as the median, and is equal to the mean for the normal distribution, i.e., 68.7 in.

maximun of 68.7 in.

To exclude the shortest 5% of men we should find the 5th percentile of the men's heights. We should find a height $x_{0}$

such that

$\int\limits_{-\infty}^{x_{0}}m(x)\, dx$ = 0.05

this number is $x_{0}$ = 62.45

minimun of 62.45

(in R execute) qnorm(0.05, mean = 68.7, sd = 3.8)

Answer 2

Using the normal distribution, we have that:

a) The percentage of men who meet the height requirement is 3.82%. It suggests that most of the characters employed at the park are women.

b) The new height requirements are a minimum of 62.45 in and a maximum of 68.7 in.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Men's heights have a mean of 68.7 inches, thus $\mu = 68.7$
• Standard deviation of 3.8 inches, thus $\sigma = 3.8$

Item a:

The proportion is the p-value of Z when X = 62 subtracted by the p-value of Z when X = 57, so:

X = 62:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{62 - 68.7}{3.8}$

$Z = -1.76$

$Z = -1.76$ has a p-value of 0.0392.

X = 57:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{57 - 68.7}{3.8}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.0010.

0.0392 - 0.0010 = 0.0382

0.0382 x 100% = 3.82%

The percentage of men who meet the height requirement is 3.82%. It suggests that most of the characters employed at the park are women.

Item b:

• The tallest 50% are the heights above the 50th percentile, which is the mean, so the upper limit is 68.7 in.
• The lower limit is the 5th percentile(shortest 5% of the mean), so X when Z has a p-value of 0.05, that is, X when Z = -1.645.

$Z = \frac{X - \mu}{\sigma}$

$-1.645 = \frac{X - 68.7}{3.8}$

$X - 68.7 = -1.645(3.8)$

$X = 62.45$

The new height requirements are a minimum of 62.45 in and a maximum of 68.7 in.

A similar problem is given at https://brainly.com/question/13490654