Use normal approximation to estimate uniform distribution Let $X_1,...,X_{10}\sim U(0,1)$. Use normal approximation to estimate
$$P\left(\sum^{10}_{k=1}X_k>5\right).$$
My solution so far:
$EX=\frac{1}{2}$ and $VarX=\frac{1}{12}$
$$P\left(\sum^{10}_{k=1}X_k>5\right)=1-P\left(\sum^{10}_{k=1}X_k\le5\right)=1-\left(\frac{\sum^{10}_{k=1}X_k-\frac{1}{2}}{\frac{1}{12}}\le\frac{5-\frac{1}{2}}{\frac{1}{12}}\right)=1-\Phi\left(\frac{5-\frac{1}{2}}{\frac{1}{12}}\right)=1-\Phi(54).$$
I am pretty new to normal approximation and something is going wrong here but I don't know what.
 A: Let say you have $X_1, X_2, \ldots , X_n$ iid random variables, where every $X_i$ has an expected value of $\mu$ and a variance of $\sigma^2$. Then you can probably use the central limit theorem. For a sufficiently large $n$  the sum of all $X_i$'s is approximately distributed as $\mathcal N(n\cdot \mu, n\cdot \sigma^2)$. With $\mu=0.5,\sigma^2=\frac1{12}$ and $n=10$ we get
$$\sum_{i=1}^{10} X_i \overset{\lower{0.5ex}{\cdot}}{\underset{\raise{1ex}{\cdot}}{\sim}} \mathcal N\left(5 , \frac{10}{12}\right)$$
So your line changes into
$$1-P\left(\sum^{10}_{k=1}X_k\le5\right)=1-P\left(\frac{\sum^{10}_{k=1}X_k-5}{\sqrt{\frac{10}{12}}}\le\frac{5-5}{\sqrt{\frac{10}{12}}}\right)$$
At the denominator we have to take the square root, since we need the standard deviation.
$$\approx 1-\Phi\left(\frac{5-5}{\sqrt{\frac{10}{12}}}\right)=1-\Phi(0)=1-0.5=0.5$$
I assumed indepdence of the random variables here. And one rule of thumb to apply the central limit theorem is that $n>30$, which is not fulfilled here. But probably $n=10$ is used to obtain $\Phi(0)$, which is easy to evaluate.
