My book gives the following lemma: if $X_1, X_2, \dots, X_n$ are independent normally distributed random variables, such that each $X_i$ has mean $\mu$ and variance $\sigma^2$, then the mean of $\bar{x}$ is normally distributed with mean $\mu$ and variance $\sigma^2$ over $n$.
Following this, I attempted a problem where each $X_i$ had mean 25 and standard deviation 1. I had to find the probability that the mean of $X_1, X_2, X_3, X_4$ was greater than 26. I ended up with the probability being zero - I do $P(X < 26) = P(Z < (26-25)/0.25) = P(Z < 4)$. I think I'm doing something wrong somewhere. Can someone point out where I'm going wrong please?