# Sample mean of independent, normally-distributed variables

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?

• If you want $P(\bar X > 26)$ (as your text says), why write $P( X < 26)$ (as your mathematics has)? That's wrong in two different ways. Then you got the standard error of the mean wrong when you standardized. – Glen_b -Reinstate Monica May 18 '14 at 1:06

Yes, the sample mean is normally distributed with mean $E(\overline{X})=E(X)=\mu$ and variance $\text{Var}(\overline{X})=\text{Var}(X)/n=\sigma^2/n$. But when you standardize you must divide by the standard deviation $\sqrt{\text{Var}(\overline{X})}=\sqrt{\frac{\sigma^2}{n}}=\frac{\sigma}{\sqrt{n}}$, not by $\frac{\sqrt{\sigma^2}}{n}=\frac{\sigma}{n}$:
$$P(\overline{X}<26)=P\left(Z<\frac{26-25}{\sqrt{\sigma^2/4}}\right) =P\left(Z<\frac{26-25}{1/2}\right)=P(Z<2)$$