Simple problem. We role a die 100 times and we add the results. What is the probability of getting sum between 330 and 380 ?
I got this: $P(330 \le X \le 380) = P\left( \frac{330 - n * EX}{\sigma = \sqrt{100 * \frac{1}{6} * \frac{5}{6}} } \le Y \le \frac{380 - n * EX}{\sigma = \sqrt{100 * \frac{1}{6} * \frac{5}{6}} } \right) = P\left( \frac{-20}{3,72} \le Y \le \frac{30}{3,72} \right)$
But obviusly something is wrong. What and why ? $\sigma^2$? or what?
Let $Y_i$ be the number obtained on the $i$-th roll. Then $E(Y_i)=\frac{1+2+\cdots+6}{6}=3.5$.
To find the variance of $Y_i$, use the fact that it is equal to $E(Y_i^2)-(E(Y_i))^2$. And $E((Y_i)^2)=\frac{1^2+2^2+\cdots+6^2}{6}$.
So now we know the mean and the variance of $Y_i$. Our random variable $X$ is the sum of the $X_i$. So it has mean $100$ times $E(X_i)$, and by independence it has variance $100$ times the variance of $X_i$.
Now we can use the normal approximation, since $X$ is the sum of $100$ independent identically distributed "nice" $Y_i$.
We want $\Pr(330\le X\le 380)$. Maybe we should use the continuity correction (if that has been covered in your course). If $W$ is normal mean and variance the same as that of $X$, we have $$\Pr(330\le X\le 380)\approx \Pr(329.5\le W\le 380.5).$$