The heights of men in the North Korea approximately follow a normal distribution with mean $175.9 cm$ and standard deviation $7.5 cm$. The heights of women in the N.Korea approximately follow a normal distribution with mean $162.1 cm$ and standard deviation $7.3 cm$.
a) Kim selects four Korean men at random. What is the probability at least one of them is over $6$ feet tall.
b) Kim select at random an man and, independently, an woman. What is the probability the woman is taller?
Let $X$ be the height of men in North Korea. Then $X\sim \text{Normal}(175.9,7.5^2)$, where $\text{Normal}(\mu_X,\sigma_X^2)$ is the normal distribution for the height of men with mean $\mu_X$ and variance $\sigma_X^2$.
Let $Y$ be the height of women in North Korea. Then $Y\sim \text{Normal}(162.1,7.3^2)$, where $\text{Normal}(\mu_Y,\sigma_Y^2)$ denotes the normal distribution for the height of women with mean $\mu_Y$ and variance $\sigma_Y^2$.
Since 1 feet converts to $30.48$ cm, $6$ feet would be equivalent to $182.88$ cm.
a) Since the $4$ men are chosen at random, we apply independence to obtain the probability: $$1-P(X_i<182.88)^4 = 1-\left(\int_{-\infty}^{182.88} \frac{1}{\sqrt{2\pi}\cdot 7.5}e^{-\frac{(x-175.9)^2}{2(7.5^2) }}\right)^4\approx \boxed{0.539021}.$$
b) Find the probability $P(Y>X)$, which is equivalent to $P(X-Y<0)$.
So let $Z=X-Y$ be the random variable with distribution $Z\sim\text{Normal}(\mu_X-\mu_Y,\sigma_X^2-\sigma_Y^2)$. Substituting in appropriate numbers, we have that $Z\sim \text{Normal}(13.8,2.96)$. So $Z$ is a normal distribution with mean $\mu_Z=13.8$ cm and variance $\sigma_Z^2 = 2.96$ cm.
So $$P(Z<0) = \int_{-\infty}^{0} \frac{1}{\sqrt{2\pi}\cdot 1.72047}e^{-\frac{(x-13.8)^2}{2(2.96)}} \approx \boxed{5.25238\times 10^{-16}}.$$