Probability clarification $\chi^2$ distribution table I'm having some trouble understanding the solution of this probability question.

Ammeters produced by a manufacturer are marketed under the specification that the standard deviation of gauge readings is no larger than $.2$ amp. One of these ammeters was used to make ten independent readings on a test circuit with constant current. If the sample variance of these ten measurements is $.065$ and it is reasonable to assume that the readings are normally distributed, do the results suggest that the ammeter used does not meet the marketing specifications? [Hint: Find the approximate probability that the sample variance will exceed $.065$ if the true population variance is $.04$.]

The solution is as follows:

At the step $P(9S^2 / .04 >= 14.925)$, where does the $.10$ come from. I look at the $\chi^2$ distribution and I'm looking at $9$ degrees of freedom, but I still don't understand.

$\Bbb P\left(\frac{(n-1)S^2}{\sigma^2}\geq \frac{(n-1)s^2}{\sigma^2}\right) = \Bbb P({\raise{0.5ex}{\chi}}^2_{9}\geq 14.925) = ~$$0.0930...$ $~\approx 0.1{\small 0}$
The value comes from lookup tables or an online calculator.
