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

Modeling
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 distributed, with a mean of 63.6 inches and a standard deviation of 2.5 inches. Modeling academy standards require women to be models taller than 66 inches (or 5 feet 6 inches). What percentage of women meet this requirement?

2021-02-12

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 $$=P(X\ >\ 66)\times 100\%$$
$$=P\left(\frac{X\ -\ \mu}{\sigma}\ >\ (66\ -\ \mu)\sigma\right)\times 100\%$$
$$=P\left(z\ >\ \frac{66\ -\ 63.6}{2.5}\right)\times 100\%$$
$$=P(z\ >\ 0.96)\times 100\%$$
$$=[1\ -\ P(z\ \leq\ 0.96)]\times 100\%$$
$$=0.1685\times 100\%$$ from the excel function, $$=1\ −\ NORM.DIST(0.96,\ 0,\ 1,\ TRUE)$$
$$=16.85\%$$