Let $X$ be a normal random variable with mean 1 and variance 4. Find $P(X^2 − 2X ≤ 8)$. (Answer key .86)
My attempt $$P(X^2-2X\le 8)=P((X+2)(X-4)\le 0)$$ and this is where I am lost. I did the following$$P(X\le -2)+P(X\le 4)=1$$ and noticed that answer is just $1$ by looking at the mean. So I am doing something wrong.
The condition $(X+2)(X−4) \leq 0$ is equivalent to $-2 \leq X \leq 4$. Therefore,
$$ P[(X+2)(X−4) \leq 0] = P[X \leq 4] - P[X < -2]. $$