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Math Help - [SOLVED] Normal Distrubtion Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Normal Distrubtion Question

    Question:
    X has a normal distribution, and P( X > 73.05 = 0.0289 ). Give that the variance of the distribution is 18, find the mean.

    Attempt:

    = 1 - 0.0289
    = 0.9711
    = -1.897

    \sigma^2 = 18 , \sigma = 3\sqrt2

    P\left( \frac{X - \mu}{\sigma} > 73.05 \right) = -1.897

    Don't know how to find the mean...
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by looi76 View Post
    Question:
    X has a normal distribution, and P( X > 73.05 = 0.0289 ). Give that the variance of the distribution is 18, find the mean.

    Attempt:

    = 1 - 0.0289
    = 0.9711
    = -1.897

    \sigma^2 = 18 , \sigma = 3\sqrt2

    P\left( \frac{X - \mu}{\sigma} > 73.05 \right) = -1.897

    Don't know how to find the mean...

    The z-score corresponding to a cumulative standard normal probability of 0.9711 is 1.897.

    So:

     <br />
\frac{73.05-\mu}{\sqrt{18}}=1.897<br />

    Now solve for \mu.

    RonL
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  3. #3
    Flow Master
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    Quote Originally Posted by looi76 View Post
    Question:
    X has a normal distribution, and P( X > 73.05 = 0.0289 ). Give that the variance of the distribution is 18, find the mean.

    Attempt:

    = 1 - 0.0289
    = 0.9711
    = -1.897

    \sigma^2 = 18 , \sigma = 3\sqrt2

    P\left( \frac{X - \mu}{\sigma} > 73.05 \right) = -1.897

    Don't know how to find the mean...
    A question of similar ilk that might be of interest: http://www.mathhelpforum.com/math-he...n-problem.html
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