Probability of two normal random variables when random samples are taken from a population This is sort of second section to my previous question, I should have included both together, but I forgot to.  Sorry for any inconvenience.
X= random height of a male
Y= random height of a female
X and Y are independent of each other
For x, μ=180 cm and σ^2=16 cm^2
For y, μ=170 cm and σ^2=25 cm^2
μ is expected value, σ^2 is variance
A random sample of 16 males and 16 females are selected from the population.
 i) What is the probability that the average height of the males is less than 178cm?
 ii) What is the probability that the average height of the males is more than 12cm
 greater than the average height of the females?
As for this question, I believe it is asking for the probability of (uX<178), but since I know the expected value, how would I compute for a probability for it.  Any sort of help would be appreciated, thanks.
 A: If $X_i$ is the height of the $i^{th}$ sampled male and $Y_j$ is the height of the $j^{th}$ sampled female, then - assuming these random variables are normally distributed - you know that $$ X_i \sim \mathcal{N}(180,4), ~~~ Y_j \sim \mathcal{N}(170,5) $$  where I'm writing $\mathcal{N}(\mu,\sigma)$ to denote a normal random variable and its parameters.  One new definition: lets write $S_X = {1\over 16} \sum_{i = 1}^{16} X_i$ and $S_Y = {1\over 16} \sum_{j = 1}^{16} Y_j$ for the averages of the male and female heights, respectively.
The question then asks for $P(S_X < 178)$ and $P(S_X > S_Y + 12)$.  
To get you started, note that each $X_i$ is independent of the others and has the same distribution.  This means the average of the $X_i$'s is also normally distributed.  You can calculate that $\mathbb{E} [ S_X ] = {1\over 16} \sum_{i=1}^{16} \mathbb{E}[X_i] = {1\over 16} \sum_{i=1}^{16} \mu_X = \mu_X$, and $Var(S_X) = {1\over 16^2} \sum_{i=1}^{16} Var(X_i) = {1\over 16^2} \sum_{i=1}^{16} 16 = 1$, so $$ S_X \sim \mathcal{N}(180,1).$$  Similar results can be calculated in the same way for $Y$.
The last tool you need is a way to calculate $P(S_X > S_Y + 12)$.  But here we can treat $S_X - S_Y$ as our random variable, and being the sum of independent normal variables, it too will be normal.  If you calculate its mean and variance you can find the probability that it exceeds 12 using a table of $z$-scores, as usual.  But
$$ \mathbb{E}[S_X - S_Y] = \mathbb{E}[S_X] - \mathbb{E}[ S_Y] = 180 - 170 $$
and 
$$ Var(S_X - S_Y) = Var(S_X) + Var(-S_Y) = Var(S_X) + Var(S_Y)$$
by independence.  The rest should be straightforward!
A: Let $\overline{X}$ be the random variable you get by taking a random sample of $16$ males and finding their average height. 
Then $X$ is normally distributed, with mean $\mu_X=180$, and variance $\frac{\sigma_X^2}{16}=1$.  
Now all you need to do is to compute the probability that a normal with mean $180$ and variance $1$ is less than $178$.
For the second problem, define $\overline{Y}$ analogously. Then $\overline{Y}$ is normally distributed, mean $170$, variance $\frac{25}{16}$.
Let $W=\overline{X}-\overline{Y}$. We want the probability that $W\gt 12$. The calculation now goes much like in the first part of your previous question.
A: The thing to know is that if $$X_1, X_2, \ldots, X_n \overset{\mathrm{iid}}{\sim} \operatorname{Normal}(\mu, \sigma^2)$$ is a sample of independent and identically distributed normal random variables with mean $\mu$ and variance $\sigma^2$, then the sample mean $$\bar X = \frac{X_1 + X_2 + \cdots + X_n}{n} \sim \operatorname{Normal}\Bigl( \mu, \frac{\sigma^2}{n} \Bigr).$$  This is exact, not asymptotic:  for the sum of independent normal random variables is itself normal, thus their average is also normal.
Using this concept, we can readily answer the first two parts:  you can see that for part (i), the sample consists of $n = 16$ males, so $$\bar X \sim \operatorname{Normal}\Bigl( \mu_{\overline X} = 180, \sigma_{\overline X}^2 = \frac{\sigma^2}{n} = 1 \Bigr),$$ hence $$\Pr[\bar X < 178] = \Pr\left[ \frac{\bar X - 180}{1} < \frac{178-180}{1} \right] = \Pr[Z < -2],$$ where $Z \sim \operatorname{Normal}(0,1)$ is a standard normal random variable.  For part (ii), note that we wish to determine $$\Pr[\bar X > \bar Y + 12].$$  This is equivalent to $$\Pr[\bar X - \bar Y > 12],$$ and again, since $\bar X \sim \operatorname{Normal}(\mu_{\overline X} = 180, \sigma_{\overline X}^2 = 1)$ and $\bar Y \sim \operatorname{Normal}(\mu_{\overline Y} = \mu_Y, \sigma_{\overline Y}^2 = \frac{\sigma_Y^2}{16})$ are both normal, the difference $\bar X - \bar Y$ is also normal:  $$\bar X - \bar Y \sim \operatorname{Normal}\left(\mu_{\overline X} - \mu_{\overline Y}, \sigma_{\overline X}^2 + \sigma_{\overline Y}^2\right).$$  Note that the mean is the difference of individual means, but the variance is the SUM of the individual variances.
