$X \sim N(0;4) ,Y=2X^2+4X-5$. Find $P(Y<X)$

probability

$X \sim N(0;4) , Y=2X^2+4X-5$.

Find $P(Y<X)$.

Denote $Z=\frac{X-\mu}{\sigma}=\frac{X}{2} \sim N(0;1)$ (since $X \sim N(0;4)$).

I have to find the distribution of Y.

Firstly, I know that $4X-5 \sim N(4\cdot0-5,4^2\cdot4)$ but how I deal with $2X^2$?

$P(Y<X)=1-P(X<Y)=1-P(X<2X^2+4X-5)=1-P(0<2X^2+3X-5)$

I am really get stuck here and need some help.

Thanks !


\begin{align} P(Y<X) &= P(2X^2+4X-5<X) =\\ &= P(2X^2+3X-5<0)= \\ &= P(-1 < X<\frac52)\end{align} where the last equality is obtained by solving the quadratic inequality.

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