Comparing Variability

Question: Constant velocity joints (CV joints) allow a rotating shaft to transmit power through a variable angle, at constant rotational speed, without an appreciable increase in friction or play. An after-market company produces CV joints. To optimize energy transfer, the drive shaft must be very precise. The company has two different branches that produce CV joints where the variability of the drive shaft is known to be $2$ mm. A sample of $n_1=10$ is drawn from the first branch and a sample of $n_2 = 15$ is drawn from the second branch. Suppose that the diameter follows a normal distribution. What is the probability that the drive shaft coming from the first branch will have greater variability than those of the second branch?

My attempt: I am having difficulty understanding the question itself. So from what I understand it's asking me to compare the variability by which I assume it means std dev. Now I think the $2$ mm is the population std dev. So if $s_1$ and $s_2$ are the sample std devs then I need to calculate $\Pr(s_1-s_2>0)$. I am not sure how to proceed further (assuming my understanding is correct). Can anyone help me out figuring out this question? Thanks.

My attempt number 2 after input from whuber After some reading I think the distribution of $s_1^2/\sigma^2$ and $s_2^2/\sigma^2$ is $\chi^2_9/9$ and $\chi^2_{14}/14$. And I could use the $F$ distribution to find the probability of the ratios as $s_1^2/\sigma^2/s_2^2/\sigma^2$ as $F_{9,14}$.

• (+1) You will make progress by restating your calculation in terms of $\Pr(X\gt 1)$ for $X=s_1^2/s_2^2$, because you can easily determine the distribution of $X$. An alternative formulation (often used) interprets "variability" in terms of the sample range: that version is answerable, too, but the calculations may be a little trickier.
– whuber
Oct 13, 2014 at 19:01

My attempt number 2 after input from whuber After some reading I think the distribution of $s_1^2/\sigma^2$ and $s_2^2/\sigma^2$ is $\chi^2_9/9$ and $\chi^2_{14}/14$. And I could use the $F$ distribution to find the probability of the ratios as $s_1^2/\sigma^2/s_2^2/\sigma^2$ as $F_{9,14}$.
$Pr(s_1^2/\sigma^2/s_2^2/\sigma^2 > 1) = 1-Pr(s_1^2/\sigma^2/s_2^2/\sigma^2 < 1) = 1-0.52=0.48$