standard deviation of x= 121 divided by 121 $\{X_1, X2, \ldots, X_{121}\}$ are independent and identically distributed random variables such that $E(X_i)= 3$ and $\mathrm{Var}(X_i)= 25$.  What is the standard deviation of their average?  In other words, what is the standard deviation of $\bar X= {X_1+ X_2+ \cdots + X_{121} \over 121}$?
 A: We first compute the variance of $W$, where 
$$W=\frac{X_1+X_2+\cdots+X_{121}}{121}.$$
Here are the ingredients.
$1$) The variance of a sum of independent random variables is the sum of the variances. Each $X_i$ has variance $25$, so the variance of $X_1+X_2+\cdots+X_{121}$ is $(25)(121)$.
$2$) If $k$ is a constant, then the variance of $kY$ is $k^2$ times the variance of $Y$.  It follows, taking $k=\frac{1}{121}$, that the variance of $W$ is $\frac{(25)(121)}{121^2}$.
The variance of $W$ is therefore $\frac{25}{121}$ (we did some cancellation.) To find the standard deviation, take the square root of the variance. Thus $W$ has standard deviation $\frac{5}{11}$.
A: Let $\bar X=\frac{1}{n}\sum\limits_{i=1}^{n}X_i$.
We want $\rm{SD}(\bar X)$ (standard deviation of $\bar X$).
Hints:


*

*$\rm{Var}[\bar X]=\frac{\sigma^2}{n}$

*$\rm{SD}(\bar X)=\sqrt{\rm{Var}[\bar X]}$
where $n=$ number of random variables, $\sigma^2=$ population variance (25 in this case).
A: Since the $X_i$ are independent, their variance is additive: $\text{Var}(X_1 + \cdots + X_n) = \text{Var}(X_1) + \cdots + \text{Var}(X_n)$. Furthermore, $\text{Var}(\lambda X_i) = \lambda^2\text{Var}(X_i)$ for any constant $\lambda$. Thus the mean
$$\overline{X} = \frac{X_1 + \cdots + X_n}{n}$$
satisfies
$$\text{Var}(\overline{X}) = \text{Var}\left(\frac{X_1 + \cdots + X_n}{n}\right) = \frac{1}{n^2}\left(\text{Var}(X_1) + \cdots + \text{Var}(X_n)\right) = \frac{\sigma^2}{n},$$
where $\sigma^2$ is the common variance of the $X_i$.
