$ f(x) = \begin{cases} k\sqrt{x}, 0<x<1 \\ 0, \text{elsewhere}\\ \end{cases}$
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. – Justin H. Oct 23 '12 at 07:56
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However, there's a typo mistake, it should be $0<x<1$ instead of $x>0$ – Justin H. Oct 23 '12 at 07:56
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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 $k$.
Brian M. Scott
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Sorry sir, there's a typo mistake, it should be $0<x<1$ instead of $x>0$ in the first line. – Justin H. Oct 23 '12 at 07:57
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@Justin: Yes, in that case $k=\frac32$; I’ll update the answer. – Brian M. Scott Oct 23 '12 at 08:00
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@Justin: maybe you should verify $Var[X] = \frac{2k}{7}$ to make sure the problem makes sense. – Stefan Smith Oct 23 '12 at 21:03