# Sequence of Random Variables.

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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I fixed the $\LaTeX$. You can get multiple characters in subscripts with braces: X_{25}. Also it is best to put an entire equation inside $ signs, not just the variables. – Nate Eldredge Sep 26 '12 at 5:07 Hint: What famous theorem tells you about the distribution of a sum of iid random variables? – Nate Eldredge Sep 26 '12 at 5:08 @NateEldredge Thanks Nate for editing and is it the Normal distribution theorem ? – Q.matin Sep 26 '12 at 5:23 add comment ## 1 Answer 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. - So since the variance is 20 here we will have the standard deviation to be the square root of 20 so that will be our sigma in this case? and our mu will be 25 ? and it lies btwn 15 and 30 so it the probability will be .85552835? – Q.matin Sep 26 '12 at 5:55 I did a hurried look at a normal table, got about$0.865$, without continuity correction. With continuity correction, it would be larger, for at the top we would be looking at$\Pr(Z\lt 5.5/\sqrt{20}$. And not subtracting a lot at the bottom. – André Nicolas Sep 26 '12 at 6:00 We never learned continuity correction so I guess your first answer of 0.865 is correct. Thank you very much !!! – Q.matin Sep 26 '12 at 6:08 So you want$\Pr(Z\lt 5/\sqrt{20})-\Pr(Z\lt -11/\sqrt{20})$, where$Z\$ is standard normal. –  André Nicolas Sep 26 '12 at 6:11
Thank you very much! –  Q.matin Sep 26 '12 at 6:14