Normal Distribution Question

Quote:

Originally Posted by tommyhakinen

Hi, i have this question

The weight of a box of cereal has a normal distribution with mean = 340g and standard deviation = 5g.

(c) 30 boxes of this cereal are selected at random for weighing. Find the probability that the sample variance is more than 36.69.

how do i solve this? thank you.

Regards,

tommy

You should know that when samples of size n are taken from a normal population with variance $\sigma^2$ then

$\frac{n-1}{\sigma^2} S^2$ ~ $\chi^2_{n-1}$

where $S^2$ is the random variable variance of sample.

So for your problem $\frac{29}{25} S^2$ ~ $\chi^2_{29}$ .

$\Pr\left( S^2 > 36.69\right)$ $= \Pr\left( \frac{29}{25} \, S^2 > \left( \frac{29}{25} \right) \, 36.69 = 42.5604\right) = \Pr(\chi^2_{29} > 42.5604) = 0.05$ , correct to two decimal places.