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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?

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    $\begingroup$ $\sigma$ is usually used for the standard deviation, not the variance, which is usually written $\sigma^2$. Variance adds for independent random variables. $\endgroup$ – Eric Auld Dec 14 '13 at 5:19
  • $\begingroup$ en.wikipedia.org/wiki/Variance#Properties $\endgroup$ – hhsaffar Dec 14 '13 at 5:20
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According to the Bienayme formula, the variance of the sum of uncorrelated random variables is the sum of their variances. If they are independent, then they are uncorrelated . http://en.wikipedia.org/wiki/Variance#Sum_of_uncorrelated_variables_.28Bienaym.C3.A9_formula.29

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 $\sigma^2$.

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Variance is additive for independent random variables. The way to prove this:

  1. Prove that the mean is additive
  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.
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