# Normal distribution

• August 20th 2013, 01:38 AM
downthesun01
Normal distribution
$\\\text{If }X\text{has a normal distribution with mean }1 \text{ and variance }4\text{, then }\\\\P[X^2-2X\leq 8]=?$

My question whether there is another way of approaching this problem other than realizing the probability in question can be rewritten as:

$\\P[X^2-2X\leq 8]= & P[X^2-2X+1\leq8+1]\\\\ & =P[(X-1)^2\leq 9]\\\\ & =P[-3\leq X-1\leq3]\\\\ & =P[-1.5\leq \frac{X-1}{2}\leq1.5]$

The reason that I ask is that I had a hard time seeing it.
• August 20th 2013, 10:48 PM
chiro
Re: Normal distribution
Hey downthesun01.

The transformation you used is probably the best way to do it. Some alternatives include looking at (X-1)^2 [in terms of a chi-square 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.