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Thread: Normal distribution, finding a value question!

  1. #1
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    Normal distribution, finding a value question!

    For a normal distribution, with a mean of $\displaystyle 10.5$ and a variance of $\displaystyle 16.7$, determine the value of $\displaystyle X_{0}$ so that $\displaystyle P(10.5 < x < X_{0}) = 0.4987$.

    I am very lost on this question, any and all help is greatly appreciated!

    Answer is supposedly $\displaystyle X_{0}=22.76$
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  2. #2
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    Re: Normal distribution, finding a value question!

    Apply the fact that $\displaystyle P(10.5 < x < X_{0}) = P(x < X_{0}) - P(x<10.5 ) $
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    Re: Normal distribution, finding a value question!

    Quote Originally Posted by pickslides View Post
    Apply the fact that $\displaystyle P(10.5 < x < X_{0}) = P(x < X_{0}) - P(x<10.5 ) $
    $\displaystyle 0.4987 = P(x < X_{0}) - 0.5$

    $\displaystyle 0.9987 = P(x < X_{0})$

    Is that correct so far? Not sure what I do next
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    Re: Normal distribution, finding a value question!

    Looks great

    $\displaystyle 0.9987 = P(x < X_{0})$

    $\displaystyle \displaystyle 0.9987 = P\left(z < \frac{X_{0}-10.5}{16.7}\right)$

    Using the Normal Distribution Table

    $\displaystyle \displaystyle 3.1 = \frac{X_{0}-10.5}{16.7}$
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