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Let $X_1$, $X_2$, ... be a sequence of i.i.d random variables such that $P(X_1 = 2) = .4$, $P(X_1 = 1) = .2$, $P(X_1 = 0) = .4$. Calculate $E[X_1]$, standard deviation of $X_1$. And calculate approximately: $P(15 \leq X_1 +\dots + X_{25} \le 30)$.

I got the $E(X_1) = 1$ and the standard deviation to be the square root of 1.8, but how can I get the last part? My thinking was let $Z = X_1 + X_2 +\dots + X_{25}$ so then we will have $E[Z] = E[n X_1] = n \cdot 1 = 25 \cdot 1 = 25$. Am I on the right track?

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    @NateEldredge Thanks Nate for editing and is it the Normal distribution theorem ?2012-09-26

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You are right about the mean of the $X_i$, and the mean of "$Z$." But call the sum by some other name, since $Z$ is kind of reserved for the standard normal. Call the sum $Y$. So $E(Y)=1$.

For the variance of the $X_i$, there was a slip. Either use $E(X_i-\mu)^2$, or $E(X_i^2)-(E(X_i))^2$. You will I think get $0.8$.

So the variance of $Y$ is $(25)(0.8)$.

Now for the probability, hold your nose and pretend that the sum of our random variables is normal. So we want the probability that a normal with mean $25$ and variance $20$ lies between $15$ and $30$. I do not know whether you are expected to use the continuity correction.

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    So you want $\Pr(Z\lt 5/\sqrt{20})-\Pr(Z\lt -11/\sqrt{20})$, where $Z$ is standard normal.2012-09-26