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 $X^2 - 2X \leq 9$.

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    $\begingroup$ Hint: The quadratic inequation will give you an interval. Then you have to compute the probability of that interval. $\endgroup$ – 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?


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