Let $X_1,\ldots,X_n$ be a random sample of size $n = 10$ from a population which is Normally distributed with mean $= 48$ and variance $= 36$.

What is the probability that the sample variance of such a sample lies between $25$ and $60$?

  • $\begingroup$ Do you know what distribution the sample variance follows? $\endgroup$ – Stefan Hansen Jun 3 '13 at 18:21

For a normal population, the quantity $${{(n-1)s^2} \over{\sigma^2} }$$ is known to have a chi-squared distribution with $n-1$ degrees of freedom.

So then $$ P[25 < s^2 < 60]=P\left[{{9(25)} \over {36} } < {{9s^2} \over 36} < {{9(60)} \over {36} } \right]=P \left[ {25 \over 4}<\chi^2_{(9)}< 15 \right].$$

Using spreadsheet software, this evaluates to 0.6237


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