The problem: A weight of a bag is normally distributed with a $E(x)= 1000$ $\sigma = 100$. What is the probability that the mean value of 5 random values is bigger than 950.
Solution: The expected value of this mean is $E(\mu)=1000$ since expected value of each is thousand. Which I do understand. The part that i do not understand is that according to the solution key the variance of the mean value is 5 times less than original variance. $E(\mu) = \frac{\sigma^2}{5}$. I do not understand mathematics behind this. If anyone could help me understand it it would be of great help.
Two rules are used to prove this. The first is that for $X,Y$ two random variables with variance, we have $\text{Var}(X+Y) = \text{Var} (X) + 2 \text{cov}(X,Y) + \text{Var}(Y). $ If $X$ and $Y$ are independent, their covariance is 0 and thus $\text{Var}(X+Y) = \text{Var} (X)+ \text{Var}(Y)$. If they are identically distributed (like taking balls at random with replacement within your bag), then their variances are equal. Thus in your case $\text{Var}(X_1 + X_2 + X_3 + X_4 + X_5) = 5 \text {Var}(X_1)$.
The second rule is that if $c>0$ is a number and $X$ is a random variable with variance, then $\text{Var} (cX) = c^2 \text{Var}(X)$.
Putting these two rules together, we get that $$\text{Var}(\mu) = \text{Var}(\frac{1}{5}\cdot (X_1+ X_2 + X_3 + X_4 + X_5 ))= (\frac{1}{5})^2\cdot 5 \cdot \text{Var}(X_1) = \frac {1}{5}\cdot \text{Var}(X_1). $$