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Math Help - continous random variable

  1. #1
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    continous random variable

    Y has a distribution function:

    F(y) = \left\{ \begin{array}{rcl} 0 & \mbox{for} & y \leq 0 \\ <br />
\\<br />
\frac{y}{8} & \mbox{for} & 0 < y < 2 \\<br />
\\<br />
\frac{y^2}{16} & \mbox{for} & 2 \leq y < 4 \\<br />
\\<br />
1 & \mbox{for} & y \geq y 4<br />
\end{array}\right.

    find \mu and \sigma^2

    since \mu = \int yf(y) dy \therefore \int^{2}_{0} y \left( \frac{y}{8} \right) dy + \int^{4}_{2} y \left( \frac{y^2}{16} \right) dy + \int^{\infty}_{4} y(1) dy= \frac{y^3}{24} \bigg{|}^{2}_{0} + \ \ \frac{y^4}{64} \bigg{|}^{4}_{2} + \ \ \frac{y^2}{2} \bigg{|}_{4} = \frac{1}{3} +\frac{256-16}{16} + 8 = \frac{145}{12}

    I know that my final solution is incorect since it's suppossed to be \frac{31}{12}, where did I go wrong?
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  2. #2
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    differentiate  F(y) to get  f(y) (not integrate). Then take  \int yf(y) \ dy to get  \mu .
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