Answer

We know that for a normal population,

\[ \begin{aligned} \frac{(n-1)S^{2}}{\sigma^{2}} \sim \chi_{n-1}^{2} \end{aligned} \]

In the given problem, \(n = 15\) and \(\sigma^{2} = 9\). Hence,

\[ \begin{aligned} P(S^{2} > 12) &= P(\frac{14 S^{2}}{9} > \frac{14}{9} 12)\newline &= P(\chi_{14}^{2} > \frac{56}{3})\newline &= 0.178 \quad\text{from standard tables or chi-square calculators} \end{aligned} \]