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