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?

  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented May 18, 2014 at 1:06

1 Answer 1


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 $\frac{\sqrt{\sigma^2}}{n}=\frac{\sigma}{n}$:

$$P(\overline{X}<26)=P\left(Z<\frac{26-25}{\sqrt{\sigma^2/4}}\right) =P\left(Z<\frac{26-25}{1/2}\right)=P(Z<2)$$

  • $\begingroup$ Subtract your value from 1 and you got the answer to the original question. $\endgroup$
    – Michael M
    Commented May 17, 2014 at 21:13
  • $\begingroup$ @Michael Mayer: of course. I just pointed out the wrong step. $\endgroup$
    – Sergio
    Commented May 17, 2014 at 21:30
  • $\begingroup$ @gung, ok. I'm going to edit. $\endgroup$
    – Sergio
    Commented May 17, 2014 at 21:33

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