Mathematics, 13.03.2020 03:14, smartboy2296

A survey found that women's heights are normally distributed with mean 63.5 in and standard deviation 2.5 in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

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answered: mantooth

The survey follows of women's height a normal distribution.

The height of 98.51% of women that meet the height requirement are between 58 inches and 80 inches.The new height requirements would be 57.7 to 68.6 inches

The given parameters are:

$\mathbf{\mu = 63.5}$ --- mean

$\mathbf{\sigma = 2.5}$ --- standard deviation

(a) Percentage of women between 58 and 80 inches

This means that: x = 58 and x = 80

When x = 58, the z-score is:

$\mathbf{z= \frac{x - \mu}{\sigma}}$

This gives

$\mathbf{z_1= \frac{58 - 63.5}{2.5}}$

$\mathbf{z_1= \frac{-5.5}{2.5}}$

$\mathbf{z_1= -2.2}$

When x = 80, the z-score is:

$\mathbf{z_2= \frac{80 - 63.5}{2.5}}$

$\mathbf{z_2= \frac{16.5}{2.5}}$

$\mathbf{z_2= 6.6}$

So, the percentage of women is:

$\mathbf{p = P(z$

Substitute known values

$\mathbf{p = P(z$

Using the p-value table, we have:

$\mathbf{p = 0.9999982 - 0.0139034}$

$\mathbf{p = 0.9860948}$

Express as percentage

$\mathbf{p = 0.9860948 \times 100\%}$

$\mathbf{p = 98.60948\%}$

Approximate

$\mathbf{p = 98.61\%}$

This means that:

The height of 98.51% of women that meet the height requirement are between 58 inches and 80 inches.

So, many women (outside this range) would be denied the opportunity, because they are either too short or too tall.

(b) Change of requirement

Shortest = 1%

Tallest = 2%

If the tallest is 2%, then the upper end of the shortest range is 98% (i.e. 100% - 2%).

So, we have:

Shortest = 1% to 98%

This means that:

The p values are: 1% to 98%

Using the z-score table

When p = 1%, z = -2.32635

When p = 98%, z = 2.05375

Next, we calculate the x values from $\mathbf{z= \frac{x - \mu}{\sigma}}$

Substitute $\mathbf{z = -2.32635}$

$\mathbf{-2.32635 = \frac{x - 63.5}{2.5}}$

Multiply through by 2.5

$\mathbf{-2.32635 \times 2.5= x - 63.5}$

Make x the subject

$\mathbf{x = -2.32635 \times 2.5 + 63.5}$

$\mathbf{x = 57.684125}$

Approximate

$\mathbf{x = 57.7}$

Similarly, substitute $\mathbf{z = 2.05375}$ in $\mathbf{z= \frac{x - \mu}{\sigma}}$

$\mathbf{2.05375= \frac{x - 63.5}{2.5}}$

Multiply through by 2.5

$\mathbf{2.05375\times 2.5= x - 63.5}$

Make x the subject

$\mathbf{x= 2.05375\times 2.5 + 63.5}$

$\mathbf{x= 68.634375}$

Approximate

$\mathbf{x= 68.6}$

Hence, the new height requirements would be 57.7 to 68.6 inches

Read more about probabilities of normal distributions at:

link

answered: calebabaltimore

a) 98.51% women meet the height requirement. This is a low percentage, so there are not many women being denied the opportunity to join this branch of the military because they are too short or too tall.

b) The new height requirements would be between 57.7in and 68.6 in.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 63.5, \sigma = 2.5$

a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall?

This is the pvalue of Z when X = 80 subtracted by the pvalue of Z when X = 58.

X = 80

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

$Z = \frac{80 - 63.5}{2.5}$

Z = 6.6

Z = 6.6 has a pvalue of 1.

X = 58

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

$Z = \frac{58 - 63.5}{2.5}$

Z = -2.2

Z = -2.2 has a pvalue of 0.0139.

1 - 0.0139 = 0.9861

98.51% women meet the height requirement. This is a low percentage, so there are not many women being denied the opportunity to join this branch of the military because they are too short or too tall.

b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

Between the 1st and the 100-2 = 98th percentile

1st percentile

value of X when Z has a pvalue of 0.01. So X when Z = -2.325.

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

$-2.325 = \frac{X - 63.5}{2.5}$