Question

(1 point) The heights of women aged 20 to 29 follow approximately the N(64.2.6 distribution. Men the same age have height distributed as N60.3, 2.95). What percent of young men are taller than the mean height of young women?

Answer 1

This is a normal distribution question with
$\\Mean (\mu)= 69.3 \\Standard\;Deviation (\sigma)= 2.95 \\Since\; we\; know\; that \\z_{ score } = \frac{x-\mu}{\sigma}$
P(x > 64.0)=? (64 is the mean height of young women)
The z-score at x = 64.0 is,
$z = \frac{64.0-69.3}{2.95}$
z = -1.7966
This implies that
P(x > 64.0) = P(z > -1.7966) = 1 - 0.036199572207258536
$P(x > 64.0) = \textbf{0.9638}$


PS: you have to refer z score table to find the final probabilities.
Please hit thumbs up if the answer helped you