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Math Help - Normal distribution help..

  1. #1
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    Normal distribution help..

    I am not sure how to write it in latex.

    Let u = mean / mu - the greek letter m. (this is because I don't know how to write it in latex)

    The random variable X has a normal distribution. The mean is u (u > 0), and the variance is 1/4u^2.

    (a) Find P(X > 1.5u)

    So we have a distribution: X ~ N(u, 1/4u^2) .

    Using standardization form:
    Let Z = 1/0.25u (X - u)

    => P(Z > 1/0.25u(1.5u - u))
    => P(Z > 1/0.25u(0.5u))
    => P(Z > 0.5u/0.25u)
    P(Z > 2)  = 1 - \phi (2.0)

    But it won't give me the right answer as I am obviously doing something wrong...

    (b) Find the probability that X is negative.
    Here, it's P(Z > 0) or P(Z < 0) .. right?
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  2. #2
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    Quote Originally Posted by struck View Post
    I am not sure how to write it in latex.

    Let u = mean / mu - the greek letter m. (this is because I don't know how to write it in latex)

    The random variable X has a normal distribution. The mean is u (u > 0), and the variance is 1/4u^2.

    (a) Find P(X > 1.5u)

    So we have a distribution: X ~ N(u, 1/4u^2) .

    Using standardization form:
    Let Z = 1/0.25u (X - u)

    => P(Z > 1/0.25u(1.5u - u))
    => P(Z > 1/0.25u(0.5u))
    => P(Z > 0.5u/0.25u)
    P(Z > 2) = 1 - \phi (2.0)

    But it won't give me the right answer as I am obviously doing something wrong...

    (b) Find the probability that X is negative.
    Here, it's P(Z > 0) or P(Z < 0) .. right?
    (a) Z = \frac{X - \mu}{\sigma} = \frac{\frac{3 \mu}{2} - \mu}{\frac{\mu}{2}} = 1.

    (b) Z = \frac{X - \mu}{\sigma} = \frac{0 - \mu}{\frac{\mu}{2}} = -2. Therefore \Pr (X < 0) = \Pr(Z < -2).
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  3. #3
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    So I think I missed the sqrt(1/4) part in (a)

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