0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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
    May 18, 2014 at 1:06

1 Answer 1

0
$\begingroup$

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

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

$\endgroup$
3
  • $\begingroup$ Subtract your value from 1 and you got the answer to the original question. $\endgroup$
    – Michael M
    May 17, 2014 at 21:13
  • $\begingroup$ @Michael Mayer: of course. I just pointed out the wrong step. $\endgroup$
    – Sergio
    May 17, 2014 at 21:30
  • $\begingroup$ @gung, ok. I'm going to edit. $\endgroup$
    – Sergio
    May 17, 2014 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.