Probability - continuous uniform distribution?

The random variable X has uniform continuous distribution on the interval [0,10]. Find the distribution of Y = X^2 and P(Y>50).

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  • ?
    Lv 7
    8 years ago
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    The cdf of Y is F_Y(y) = P(Y <= y).

    Note that 0 <= Y <= 100, so clearly F_Y(y) = 0 if y < 0 and F_Y(y) = 1 if y > 100.

    If 0 <= y <= 100, then we have

    F_Y(y) = P(X^2 < y)

    = P(|X| < sqrt(y))

    = P(-sqrt(y) < X < sqrt(y))

    = (sqrt(y) - 0)/(10 - 0), since we are given X is uniform on [0,10]

    = sqrt(y)/10.

    In summary F_Y(y) = sqrt(y)/10 if 0 <= y <= 100, 0 if y < 0, and 1 if y > 100.

    The pdf of y is

    f_Y(y) = (d/dy) of F_Y(y) = 1/[20sqrt(y)] if 0 <= y <= 10, and 0 otherwise.

    P(Y>50) = 1 - P(Y<=50)

    = 1 - F_Y(50)

    = 1 - [sqrt(50)/10]

    = 1 - [sqrt(2)/2].

    Lord bless you today!

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