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A survey found that​ women's heights are normally distributed with mean 63.6 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?

User Ganesh Babu
by
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1 Answer

15 votes
15 votes

Answer:

(A)

Explanation:

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}μ=63.5 --- mean

\mathbf{\sigma = 2.5}σ=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}}z=

σ

x−μ

This gives

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

1

=

2.5

58−63.5

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

1

=

2.5

−5.5

\mathbf{z_1= -2.2}z

1

=−2.2

When x = 80, the z-score is:

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

2

=

2.5

80−63.5

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

2

=

2.5

16.5

\mathbf{z_2= 6.6}z

2

=6.6

So, the percentage of women is:

\mathbf{p = P(z < z_2) - P(z < z_1)}p=P(z<z

2

)−P(z<z

1

)

Substitute known values

\mathbf{p = P(z < 6.6) - P(z < -2.2)}p=P(z<6.6)−P(z<−2.2)

Using the p-value table, we have:

\mathbf{p = 0.9999982 - 0.0139034}p=0.9999982−0.0139034

\mathbf{p = 0.9860948}p=0.9860948

Express as percentage

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

\mathbf{p = 98.60948\%}p=98.60948%

Approximate

\mathbf{p = 98.61\%}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}}z=

σ

x−μ

Substitute \mathbf{z = -2.32635}z=−2.32635

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

2.5

x−63.5

Multiply through by 2.5

\mathbf{-2.32635 \times 2.5= x - 63.5}−2.32635×2.5=x−63.5

Make x the subject

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

\mathbf{x = 57.684125}x=57.684125

Approximate

\mathbf{x = 57.7}x=57.7

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

σ

x−μ

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

2.5

x−63.5

Multiply through by 2.5

\mathbf{2.05375\times 2.5= x - 63.5}2.05375×2.5=x−63.5

Make x the subject

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

\mathbf{x= 68.634375}x=68.634375

Approximate

\mathbf{x= 68.6}x=68.6

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

User Dpren
by
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