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Stats - Normal Distribution-Modulus

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    X~N(15,4)

    Find w to 2 d.p.

    P(mod(X-15) < w ) = 0.9

    Im confused. Any help pleeeaase
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    Expand the modulus bracket inside the probability.
    • Thread Starter
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    I have, i get stuck @:

    P(-(x-15)< w < (x-15) )

    Next I modelled probability onto Z~N(0,1)

    P(((-x-15)-15)/2) < Z < (((x-15)-15)/2) =0.9

    P((x-30)/2)-P(-x/2) = 0.9
    P((x-30)/2) + P( x/2) - 1 = 0.9

    P( 1.2815 > w ) = 0.9

    (x-30)/2 + x/2 - 1 = 1.2815

    rearrage etc

    x = 11.521

    Plugged ths into P(mod(x-15) < w ) = 0.9

    w = 3.479

    The answer is 3.29

    And im not entirely sure you can do half the steps i just used
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    Look at it this way, Let Y be a normal random variable such that, Y = X - 15, now Y~N(0,4)

    And you want P(|y|<w); This means that the only area you are not looking for is 1-0.9 = 0.1; but since the normal is symmetric you have 0.05 of that on each side.

    So now look up the Z value for a probability of 0.05 and plug it in to the standardization equation: Z = (w - 0)/2. This obtains the answer.
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    I still do not understand.

    Can anyone please go through this question:

    If the random variable X is distributed as N(5,4), calculate:

    P(modulus(X -5) > 3 ).


    Thanks
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    (Original post by TheNightmare)
    I still do not understand.

    Can anyone please go through this question:

    If the random variable X is distributed as N(5,4), calculate:

    P(modulus(X -5) > 3 ).


    Thanks
    l,

    ok so this is the same as p (x-5>3)+p(x-5<-3)

    so its basically p(x>8)+p(x<2)

    Still want help with finding w?
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    (Original post by TheNightmare)
    I still do not understand.

    Can anyone please go through this question:

    If the random variable X is distributed as N(5,4), calculate:

    P(modulus(X -5) > 3 ).


    Thanks
    This is basically:
    P((x - 5) > 3) + P((x - 5) < -3)

    = P(x > 8) + P(x < 2)
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    (Original post by falcon pluse)
    l,

    ok so this is the same as p (x-5>3)+p(x-5<-3)

    so its basically p(x>8)+p(x<2)

    Still want help with finding w?
    Thanks a lot. I think I got the hang of it now!

    So the general rule is:

    P(modulus(X-y) > q) = P(X-y > q ) + P(X-y < -q) right?

    What if it was like this:

    P(modulus(X-y) < q) would this be = P(X-y < q ) + P(X-y > -q) ?
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    (Original post by MathematicsKiller)
    This is basically:
    P((x - 5) > 3) + P((x - 5) < -3)

    = P(x > 8) + P(x < 2)
    What is the answer?

    According to my text book it is: 0.0668

    but i'm somehow getting : 0.1336

    What am I doing WRONG???
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    (Original post by TheNightmare)
    What is the answer?

    According to my text book it is: 0.0668

    but i'm somehow getting : 0.1336

    What am I doing WRONG???
    Are you sure you posted the question correctly?
    I get the same answer as you.

    I believe the answer in the text book is the answer you would get if there were no modulus.
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    (Original post by MathematicsKiller)
    Are you sure you posted the question correctly?
    I get the same answer as you.

    I believe the answer in the text book is the answer you would get if there were no modulus.
    (Modulus is Bold and Underlined )

    Yes the question is P( X-5 > 3).

    So is the text book answer wrong?
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    (Original post by falcon pluse)
    l,

    ok so this is the same as p (x-5>3)+p(x-5<-3)

    so its basically p(x>8)+p(x<2)

    Still want help with finding w?
    What did you get as an answer?

    According to my text book the answer is: 0.0668

    but I am getting : 0.1336

    Is the text book answer wrong?
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    I miss statistics
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    (Original post by littleone271)
    I miss statistics
    Will you be able this question for me:

    If the random variable X is distributed as N(5,4), calculate:

    P((X -5) > 3 ).

    Modulus is in BOLD and is Underlined

    Thanks
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    (Original post by TheNightmare)
    Will you be able this question for me:

    If the random variable X is distributed as N(5,4), calculate:

    P((X -5) > 3 ).

    Modulus is in BOLD and is Underlined

    Thanks
    I used to be good at it and I did an AS in pure statistics but that was a couple of years ago and I havn't really done it properly since so I can't remember how to do it ... Sorry...

    I remember this book being pretty amazing though because it's got worked examples and everything in it. I had this one for s1b and the s2 and s3 ones and they were all really good.

    http://www.amazon.co.uk/Advancing-Ma...1519123&sr=1-1
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    (Original post by littleone271)
    I used to be good at it and I did an AS in pure statistics but that was a couple of years ago and I havn't really done it properly since so I can't remember how to do it ... Sorry...

    I remember this book being pretty amazing though because it's got worked examples and everything in it. I had this one for s1b and the s2 and s3 ones and they were all really good.

    http://www.amazon.co.uk/Advancing-Ma...1519123&sr=1-1
    Ahh, I'm doing Statistics 1 with Edexcel not AQA. But never mind, thanks anyway.
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    Can anyone answer this?

    X~N(5,4) -------------------------------- Calculate: P( |X-5| > 3 ).

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