# Probability - Normal Distribution

• Mar 7th 2010, 01:00 PM
bhuang
Probability - Normal Distribution
Given that X~N(-3,5^2), calculate the probability.

P(|X|>5)

What I did was transform it and then put calculate the probability with my graphing calculator and got 0.0547992894.
Then, I multiplied it by 2 to get 0.1096.
However, the answer in the book is 0.3994.
Can someone else please do this question and tell me what you get?
• Mar 7th 2010, 01:52 PM
mr fantastic
Quote:

Originally Posted by bhuang
Given that X~N(-3,5^2), calculate the probability.

P(|X|>5)

What I did was transform it and then put calculate the probability with my graphing calculator and got 0.0547992894.
Then, I multiplied it by 2 to get 0.1096.
However, the answer in the book is 0.3994.
Can someone else please do this question and tell me what you get?

You require Pr(X > 5) + Pr(X < -5). If you draw a diagram showing the required area under the curve it will be clear that your use of symmetry is not valid.
• Mar 7th 2010, 03:33 PM
bhuang
Probability - Normal Distribution
I am finding Pr(X > 5) + Pr(X < -5). And I transformed it, so the normal distribution would have a mean of 0 and standard deviation of 1. But the answer is still different from that of the book's.
• Mar 7th 2010, 04:02 PM
mr fantastic
Quote:

Originally Posted by bhuang
I am finding Pr(X > 5) + Pr(X < -5). And I transformed it, so the normal distribution would have a mean of 0 and standard deviation of 1. But the answer is still different from that of the book's.

• Mar 7th 2010, 05:51 PM
bhuang
Probability - Normal Distribution
I know what I did wrong now.
I figured that because I transformed it, then |X|>5 would be symmetrical, but it isn't the case because when I transform it, the two random variables are different. Since I figured it was symmetrical, I multiplied the transformed random variables by 2. But the right way to do it is to add the probabilities of each distinct random variable.
• Mar 8th 2010, 12:41 AM
mr fantastic
Quote:

Originally Posted by bhuang
I know what I did wrong now.
I figured that because I transformed it, then |X|>5 would be symmetrical, but it isn't the case because when I transform it, the two random variables are different. Since I figured it was symmetrical, I multiplied the transformed random variables by 2. [snip]

Aha. So you re-read post #2 ....