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Math Help - Normal distribution

  1. #1
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    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.
    Last edited by downthesun01; August 20th 2013 at 01:59 AM.
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  2. #2
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    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.
    Thanks from downthesun01
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