The stature of men is normally distributed, with a mean of 69.0 inches and a standard deviation of 2.8 inches. The height of women is normally distrib

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 $\mu =63.6and\sigma =2.5.$ Modeling academy standards require women to be models taller than 66 inches Step 2 The percentage of women meet this requirement is, Percentage $=P(X>66)\times 100\mathrm{\%}$
$=P(\frac{X-\mu }{\sigma }>(66-\mu )\sigma )\times 100\mathrm{\%}$
$=P(z>\frac{66-63.6}{2.5})\times 100\mathrm{\%}$
$=P(z>0.96)\times 100\mathrm{\%}$
$=[1-P(z\le 0.96)]\times 100\mathrm{\%}$
$=0.1685\times 100\mathrm{\%}$ from the excel function, $=1-NORM.DIST(0.96,0,1,TRUE)$
$=16.85\mathrm{\%}$