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Math Help - standard normal

  1. #1
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    standard normal

    Let X follow a normal distribution
    mean=80 and Var=100

    the probability is 0.08 that X is in the symmetric interval about the mean between which two numbers?


    I did:

    P( a <x< b)= .08
    a= mean - k
    b= mean + k

    P( -k/stand dev < Z< k/st dev) =.08
    F(k/st dev)- F(-k/st dev) =.08
    F(k/st dev) - [ 1- F(k/ st dev)]= 0.08
    2F(k/st dev)=1.08 from which k= 1 and a= 79

    but the result on the book is a= 62.5

    Help me, please. Thank you
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  2. #2
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    Quote Originally Posted by 0123 View Post
    Let X follow a normal distribution
    mean=80 and Var=100

    the probability is 0.08 that X is in the symmetric interval about the mean between which two numbers?


    I did:

    P( a <x< b)= .08
    a= mean - k
    b= mean + k

    P( -k/stand dev < Z< k/st dev) =.08
    F(k/st dev)- F(-k/st dev) =.08
    F(k/st dev) - [ 1- F(k/ st dev)]= 0.08
    2F(k/st dev)=1.08 from which k= 1 and a= 79

    but the result on the book is a= 62.5

    Help me, please. Thank you
    X is N(80 , 100) and Z = \frac{X - 80}{10}

    so if P(80-k < X < 80 + k) = .08, then
    P\left({\frac{-k}{10} < Z < \frac{k}{10}}\right) = P\left({|Z| < \frac{k}{10}}\right) = 0.08

    this implies that P\left({Z < \frac{k}{10}}\right) = 0.08 + \left({0.5 - \frac{0.08}{2}}\right) = 0.54
    which implies that \frac{k}{10} = .1004 \implies k=1.004

    indeed, you might be right..
    try checking this.. Z table - Normal Distribution
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