# If $X\sim\mathcal{N}(\mu = 1,\sigma = 4)$ find $\textbf{P}(X^2 - 2X \leq 9)$

If $$X\sim\mathcal{N}(\mu = 1,\sigma = 4)$$ find $$\textbf{P}(X^2 - 2X \leq 9)$$.

I understand how to find the pdf of $$X$$, but I'm not sure how that would work for a function of $$X$$ like $$X^2 - 2X \leq 9$$.

• Hint: The quadratic inequation will give you an interval. Then you have to compute the probability of that interval. – Ertxiem Apr 29 at 2:18

To begin with, notice that \begin{align*} x^{2} - 2x \leq 9 \Longleftrightarrow (x^{2} - 2x + 1) = (x-1)^{2} \leq 10 \Longleftrightarrow \left(\frac{x-1}{4}\right)^{2} \leq \frac{5}{8} = 0.625 \end{align*}
Therefore we have \begin{align*} \textbf{P}(X^{2} - 2X \leq 9) = \textbf{P}\left(Z^{2} \leq \frac{5}{8}\right) = \textbf{P}\left(-\frac{\sqrt{10}}{4} \leq Z \leq \frac{\sqrt{10}}{4}\right) \end{align*}
where $$\textbf{Z}\sim\mathcal{N}(0,1)$$. Can you proceed from here?