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Math Help - Normal distribution with modulus

  1. #1
    Senior Member Mukilab's Avatar
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    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]


    Thanks for your help
    Last edited by Mukilab; January 12th 2013 at 11:51 AM.
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  2. #2
    MHF Contributor

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    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.
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  3. #3
    Senior Member Mukilab's Avatar
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    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?
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  4. #4
    Senior Member Mukilab's Avatar
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    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?
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  5. #5
    Junior Member Barioth's Avatar
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    Re: Normal distribution with modulus

    Quote Originally Posted by Mukilab View Post
    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?
    Hi, There is no contradictions. if we start with
    -b < x-80 < b

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

    if we add 80 everywhere

    -b+80 < x and x < b+80

    since x = x. -b + 80 < x = x <b +80
    -b + 80 < x < b + 80


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