change of unit normally distributed random variable Assume that $X_{1}$,$X_{2}$,$X_{3}$ are independent continues random variables with $\mathcal{N}(30,12)$, what is the normal distribution of  $X_{average}$ (average of $X_{1}$,$X_{2}$,$X_{3}$)
solution given by teacher: 

$X_{average} = 1/3(X_{1}+X_{2}+X_{3})$ the normal distribution of $X_{average}$ is $\mathcal{N}(30,12/3)$ 

my approach :

according to formula given in the book normal random variable $X$ under change of unit $Y=rX+s$ where $r \neq 0$ has $\mathcal{N}(r\mu+s,r^2\sigma^2)$ .
  when $Y=X_{average} = 1/3(X_{1}+X_{2}+X_{3})$ then I thought $r=1/3$ and $s=0$  then  it would be true to say   $\mathcal{N}(30/3 ,12/9)$ .

please help, I donot know other way to solve it.
 A: You can just take the variance of it directly.
$$
V\left(\bar{X}\right)=V\left(\frac{X_1+X_2+X_3}{3}\right)=\frac{1}{9}V\left(X_1+X_2+X_3\right)=\frac{1}{9}\left(V(X_1)+V(X_2)+V(X_3)\right)\\
=\frac{1}{9}\times 3V(X)=\frac{12}{3}.
$$
What you did is that you created a new random variable:
$$
X^*_i=\frac{X_i}{3}
$$
whose moments are $E(X_i^*)=10$, $V(X_i^*)=12/9$. Equivalently, the sample mean is
$$
\bar{X}=\frac{X_1+X_2+X_3}{3}=X_1^*+X_2^*+X_3^*.
$$
So then,
$$
E(\bar{X})=3\times \frac{30}{3}=30\\
V(\bar{X})=3\times\frac{12}{9}=\frac{12}{3}.
$$
Edit:
What I used was the properties of the variance and expectations operators. In general, we have that
$$
E\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n E(X_i)
$$
where, if all $X_i$ have equal means $\mu$, is equal to $n\times \mu$. For the sample mean, we also divide by $n$. So if we have equal means:
$$
E\left(\frac{1}{n}\sum_{i=1}^n X_i\right)=\frac{1}{n}E\left(\sum_{i=1}^n X_i\right)=\frac{1}{n}n\times\mu=\mu.
$$
For the variance, if we have only two variables it is
$$
V(X_1+X_2)=V(X_1)+V(X_2)+2Cov(X_1, X_2)
$$
If the variables are independent (as in your case), the covariance is 0. Thus, for independent random variables it is true that
$$
V\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^nV(X_i)
$$
which if we have equal variances $\sigma^2$ is equal to $n\sigma^2$. For the sample mean, we also divide by $n$ inside the variance operator. If we have independence and equal variances, we get:
$$
V\left(\frac{1}{n}\sum_{i=1}^n X_i\right)=\frac{1}{n^2}V\left(\sum_{i=1}^n X_i\right)=\frac{1}{n^2}\sum_{i=1}^n V\left(X_i\right)=\frac{1}{n^2}n\sigma^2=\frac{\sigma^2}{n}.
$$
