# Variance of sum of random variables 2

I need to rephrase the question I asked here: Variance of sum of random variables

Let $X_1,X_2,...,X_n$ be independent exam scores. Each $X_i$ is a random variable with $\mu=75$ and $\sigma^2=25.$ Let $S_n=\displaystyle\sum_{i=1}^n X_i.$ Then what does Var$(S_n)$ equal? Is it just $25$? If so, why?

• $\sigma$ is usually used for the standard deviation, not the variance, which is usually written $\sigma^2$. Variance adds for independent random variables. – Eric Auld Dec 14 '13 at 5:19
• en.wikipedia.org/wiki/Variance#Properties – hhsaffar Dec 14 '13 at 5:20

Hence, for your question the answer would be $\sigma^2 n = 25 n$.
Be careful with $\sigma$. $\sigma$ is the standard deviation, not the variance. The variance is $\sigma^2$.
2. Note that variance of the sum is $\mathbb{E}((X_1 - \mu_1 + X_2 - \mu_2 + \dotsb + X_n - \mu_n)^2)$
3. Expectation is linear, so it is $\mathbb{E}((X_1-\mu_1)^2) + \dotsb + \mathbb{E}((X_n-\mu_n)^2) + (\text{cross terms})$
4. Show that the cross terms are zero, using that $\mathbb{E}(X_1X_2)=\mathbb{E}(X_1)\mathbb{E}(X_2)$ by independence.