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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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    oops...I approve the edit by Rick wrongly.2012-10-23
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    However, there's a typo mistake, it should be $0 instead of $x>0$2012-10-23

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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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    Sorry sir, there's a typo mistake, it should be $0 instead of $x>0$ in the first line.2012-10-23
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    If it's $0, is $k=\frac32 $?2012-10-23
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    @Justin: Yes, in that case $k=\frac32$; I’ll update the answer.2012-10-23
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    I got it. Thanks bro.2012-10-23
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    @Justin: You’re welcome.2012-10-23
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    @Justin: maybe you should verify $Var[X] = \frac{2k}{7}$ to make sure the problem makes sense.2012-10-23