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$ f(x) = \begin{cases} k\sqrt{x}, 0

I know that $E[X] = \frac{2k}{5}$ and $Var[X] = \frac {2k}{7}$

Then what can I do to find $k$ with these few information?

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    However, there's a typo mistake, it should be 0 instead of x>02012-10-23

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Updated to match the corrected version of the question:

You must have $\int_{-\infty}^\infty f(x)~dx=1$ in order for $f$ to be a probability density function. In this case

$\int_{-\infty}^\infty f(x)~dx=\int_0^1 k\sqrt x~dx=k\int_0^1 x^{1/2}~dx\;,$

so you need only solve the equation

$k\int_0^1 x^{1/2}~dx=1$

for $k$.

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    @Justin: maybe you should verify $Var[X] = \frac{2k}{7}$ to make sure the problem makes sense.2012-10-23