Solving Normal Random Variable Equations for P(X(X-1) > 2) and P(|X| > a)

twoski
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Homework Statement



X is a normal random variable with mean 1, variance 4.

1. Find P( X(X-1) > 2 )

2. Find a value 'a' for which P(|X| > a ) = .25


The Attempt at a Solution



I had no idea how to start 1.

For 2, i got this far then got stuck:

P(|X| > a) = 1 - P((X-1)/2 <= (a-1)/2) = 1 - Ф((a-1)/2)
 
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twoski said:

Homework Statement



X is a normal random variable with mean 1, variance 4.

1. Find P( X(X-1) > 2 )

2. Find a value 'a' for which P(|X| > a ) = .25


The Attempt at a Solution



I had no idea how to start 1.

##X^2-X>2## is the same as ##X^2-X-2>0## or ##(X-2)(X+1)>0##. What values of ##X## satisfy that?

For 2, i got this far then got stuck:

P(|X| > a) = 1 - P((X-1)/2 <= (a-1)/2) = 1 - Ф((a-1)/2)

##P(X>|a|)=P(X\le -a)+P(X\ge a)##

Is that enough to get you going?
 
For the 1st bit it's the complement of P(-1<X<2) I think.
 
twoski said:

Homework Statement



X is a normal random variable with mean 1, variance 4.

1. Find P( X(X-1) > 2 )

2. Find a value 'a' for which P(|X| > a ) = .25


The Attempt at a Solution



I had no idea how to start 1.

For 2, i got this far then got stuck:

P(|X| > a) = 1 - P((X-1)/2 <= (a-1)/2) = 1 - Ф((a-1)/2)

This is incorrect; start over, and be more careful. Draw a picture first, before trying to compute anything!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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