# Normal distribution with modulus

• January 12th 2013, 11:47 AM
Mukilab
Normal distribution with modulus
P(|X-80|<b)=0.9 X~N(80,15)

I literally have absolutely no idea of the rules of modulus. Apparently it goes to P(80-b<X<80+b) but I have no idea why... I've been searching the net for information and I've found nothing.

I'm very comfortable doing all normal distribution equations, I just don't understand how you get past the modulus part, as in how do you expand |X-80| to get something normal like P(X<b+80) which is easy to solve [not saying that is the right thing]

• January 12th 2013, 12:14 PM
HallsofIvy
Re: Normal distribution with modulus
Are you saying that you do not know what "modulus" or "absolute value" means? |x| is equal to x if $x\ge 0$, -x is x< 0. In particular that means that |x- 80| is equal to x- 80 if [itex]x\ge 80[/b] and equal to -(x- 80)= 80- x if x< 80. Saying that |x- 80|< b means that either x- 80< b or -(x- 80)< b. That last is the same as x- 80> -b. Putting them together, if |x- 80|< b then -b< x- 80< b. That is, |x- 80|< b as long as x- 80 lies between -b and b. If you want you can add 80 to each part to say that if |x- 80|< b then x itself lies between 80- b and 80+ b.

Essentially, that says that x lies in an interval having 80 at its center and extending a distance "b" on either side. We can think of |x- 80| as a "distance". |x- 80|< b means that the distance form x to 80 is less than b.
• January 12th 2013, 12:26 PM
Mukilab
Re: Normal distribution with modulus
But it seems that on the one hand you said
-b<x-80<b

but then after that you said 'you can add 80 to each part...' surely giving -b+80<x<b+80 which completely contradicts what you said before?
• January 13th 2013, 09:05 AM
Mukilab
Re: Normal distribution with modulus
I hope this thread hasn't gone into section as if it were answered, I still don't understand, please can you or anyone else help further?
• January 13th 2013, 07:08 PM
Barioth
Re: Normal distribution with modulus
Quote:

Originally Posted by Mukilab
But it seems that on the one hand you said
-b<x-80<b

but then after that you said 'you can add 80 to each part...' surely giving -b+80<x<b+80 which completely contradicts what you said before?

-b < x-80 < b

it is the same as
-b < x-80 and x-80<b