
Normal distribution
$\displaystyle \\\text{If }X\text{has a normal distribution with mean }1 \text{ and variance }4\text{, then }\\\\P[X^22X\leq 8]=?$
My question whether there is another way of approaching this problem other than realizing the probability in question can be rewritten as:
$\displaystyle \\P[X^22X\leq 8]= & P[X^22X+1\leq8+1]\\\\ & =P[(X1)^2\leq 9]\\\\ & =P[3\leq X1\leq3]\\\\ & =P[1.5\leq \frac{X1}{2}\leq1.5]$
The reason that I ask is that I had a hard time seeing it.

Re: Normal distribution
Hey downthesun01.
The transformation you used is probably the best way to do it. Some alternatives include looking at (X1)^2 [in terms of a chisquare distribution after centering) or using the transformation theorem to get the exact PDF of X^2 + 2X.
Again, the best thing in terms of simplicity is to do what you have done.