I have the following question:
Suppose $X_1,X_2, \ldots, X_{12}$ are identical independent uniform random variables on $[0,1]$. Let the sample mean(lets call it $m$) = $\frac{1}{12}(X_1 + X_2 + \ldots + X_{12})$. Use the central limit theorem to estimate the following: $P\left ( m \leq \frac{1}{3}\right )$.
I am taken to assume that the population mean ($E[X_i] = \mu = \frac{b+a }{ 2}$ for a uniform RV) is $\frac{1}{2}$ but as we have $12$ random variables, using the additive property we have $\frac{1}{2}\cdot 12 = 6$. Also $Var[X_i]= \frac{(b+a)^2}{12} = \frac{1}{12} \cdot 12$(12 random variables) $= 1$ and standard deviation(std) $= 1$
Using the central limit theorem or normal approximation: $\left [\frac{m-µ}{\frac{std }{\sqrt{n}} }\right ] \leq \frac{1}{3}$. However I ran into a problem after this as the $P(Z\leq z)$ returns a incredibly big $z$ value of $P(Z\leq-19.63)$ which does not seem correct. I assume it has something to do with $m = \frac{1}{12} \cdot (X_1+X_2+\ldots +X_{12})$ as I am not sure why that was included in the question.
Any help would be much appreciated!