How to combine standard deviation and mean from two samples? Question: 
We are given, (all normally distributed)
Male height: 
$\mu_1 = 178$cm
$\sigma_1 = 4$cm
Female height: 
$\mu_2 = 170$cm
$\sigma_2 = 3$cm
Also suppose that a random sample of 25 males and 25 females are selected from the population. 
Suppose that we take the average of all 50 heights. What is the probability that this average height is between 172 and 175. 
My Question: 
I know how to calculate probability for individual sample sizes (male or female), but when combining the two, how does one calculate the new mean and standard deviation of the combined samples? 
$\mu_1 + \mu_2 = \frac{178 + 170}{2}$? 
And what about standard deviation?
 A: Let $X_{i,1}$ and $X_{i,2}$ denote the males and females, respectively. Then the total average is given by 
$$
\overline{X}=\frac{\sum_{i=1}^{25}X_{i,1}+\sum_{i=1}^{25}X_{i,2}}{50}
$$
Obviously $\overline{X}$ is normal. An easy calculation shows the mean is given by $E[\overline{X}]=\mu_{\overline{x}}=\frac{25\mu_1+25\mu_2}{50}$, and 
$$
Var(\overline{X})=\sigma_{\overline{x}}^2=\frac{25\sigma_1^2+25\sigma_2^2}{50^2}
$$
The rest is straight forward by converting to standard normal with the formula $Z=\frac{\overline{X}-\mu_{\overline{x}}}{\sigma_{\overline{x}}}$.
A: I assume that the samples of the male heights are independent. Then $Var\left(\overline X_1\right)=\frac{\sigma_1^2}{n_1}=\frac{4^2}{25}$. Equivalently for the female heights $Var\left(\overline  X_2\right)=\frac{\sigma_2^2}{n_2}=\frac{3^2}{25}$
Then the sum of the (independent) means has a variance of $Var\left(\overline X_1+\overline X_2 \right)=\frac{4^2}{25}+\frac{3^2}{25}=1$
Consequently $Var(Z)=Var\left(\frac{\overline X_1+\overline X_2}{2} \right)=\frac14$
That means that the standard deviation of the average height is $\boxed{\sigma_z=\sqrt{\frac14}=\frac12}$
The expectation of the average is straightforward 
$\mathbb  E\left(\frac{\overline X_1+\overline X_2}{2} \right)=\frac{\mathbb  E\left(\ X_1\right)+\mathbb  E\left(\ X_2 \right)}{2}=\frac{178+170}2=174$. I agree with your result here. Thus $$Z=\frac{\overline X_1+\overline X_2}{2}\sim \mathcal N\left(174,\frac14\right)$$
