# Random variable and Probability [closed]

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?

## closed as off-topic by zhoraster, Scientifica, pjs36, Namaste, user223391 Feb 21 '17 at 17:58

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• Have you attempted the problem yet? – John Smith Feb 19 '17 at 17:27
• yes ..i am stuck in these 2 – ernie Feb 19 '17 at 17:28
• For a) can you find the probability that none of them is over six feet tall? (You may want to invoke independence here.) If so, you just need the complement of that probability. For b), you want the distribution of the difference between two normally distributed random variables. That's normal too, as it turns out. – Theoretical Economist Feb 19 '17 at 17:34

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 $$P(Z<0) = \int_{-\infty}^{0} \frac{1}{\sqrt{2\pi}\cdot 1.72047}e^{-\frac{(x-13.8)^2}{2(2.96)}} \approx \boxed{5.25238\times 10^{-16}}.$$