A discrete random variable X has the distribution $U(11)$. The mean of $50$ observations of $X$ is denoted by $\bar{X}$ . Use an approximate method, which should be justified, to find $P(\bar{X} \le 6 .10)$.
As $n$ is sufficiently large for CLT to apply it can be said that $X$ can be approximated by a normal distribution with $\mu = \frac12(n+1)$ and $\sigma^2 = \frac{1}{12}(n^2-1)$ so $X \sim N(6,10)$. This means that $\bar{X} \sim N(6,0.2)$.
We want to calculate $P(\bar{X} \le 6 .10)$, however the answer then goes on to calculate $P(\bar{X} \le 6 .11)$ due to a continuity correction. Could someone explain this to me, I don't see why we would want to add 0.1 to the region.
Am quite puzzled, but here is an idea. Your $X_i$ are discrete, and so $\bar{X}$ is inherently discrete as well. Step size of each $X_i$ is $1$, so really $\sum X_i$ is a discrete r.v. with support on $[50, 550]$, and hence $\bar{X}$ is discrete on $[1,11]$ with step size of $1/50 = 0.02$.
Hence, in reality, $\mathbb{P}[\bar{X} \le x]$ is the same for all real $x \in [6.1,6.12)$. To approximate that with the continuous normal from the CLT, you compute the cdf at the average of the interval, i.e. $(6.12+6.1)/2 = 6.11$.