Step 1
From the given information, the mean of women is 63.6 inches and standard deviation is 2.5 inches.
Let X be the height of the women follows normal distribution with \(\mu=63.6\ and\ \sigma=2.5.\)
Modeling academy standards require women to be models taller than 66 inches
Step 2
The percentage of women meet this requirement is,
Percentage \(=P(X\ >\ 66)^{*}\ 100\%\)

\(=P\left(\frac{X\ -\ \mu}{\sigma}\ >\ (66\ -\ \mu)\sigma\right)^{*}\ 100\%\)

\(=P\left(z\ >\ \frac{66\ -\ 63.6}{2.5}\right)^{*}\ 100\%\)

\(=P(z\ >\ 0.96)^{*}\ 100\%\)

\(=[1\ -\ P(z\ \leq\ 0.96)]^{*}\ 100\%\)

\(=0.1685^{*}\ 100\%\) from the excel function, \(=1\ −\ NORM.DIST(0.96,\ 0,\ 1,\ TRUE)\)

\(=16.85\%\)

\(=P\left(\frac{X\ -\ \mu}{\sigma}\ >\ (66\ -\ \mu)\sigma\right)^{*}\ 100\%\)

\(=P\left(z\ >\ \frac{66\ -\ 63.6}{2.5}\right)^{*}\ 100\%\)

\(=P(z\ >\ 0.96)^{*}\ 100\%\)

\(=[1\ -\ P(z\ \leq\ 0.96)]^{*}\ 100\%\)

\(=0.1685^{*}\ 100\%\) from the excel function, \(=1\ −\ NORM.DIST(0.96,\ 0,\ 1,\ TRUE)\)

\(=16.85\%\)