So the question is to discover the median of $X$ if the mode of the circulation is at $x = sqrt2/4$. And the random variable $X$ has the density function

$$f(x) = left{ eginarrayll kx & mboxfor $0 le x le sqrtfrac2k$ ;\ 0 & mboxotherwise.endarray ideal.$$

I understand that the median of a PDF is such that the integral is equated to half. If I proceed that way via this trouble, I am acquiring a response of $1$. But the answer is offered is $frac14$. Also, I am not sure why the mode is offered. If I equate $f"(x)$ to $sqrt2/4$ (which is the given mode), I end up obtaining random options.

Can someone help me out right here, please?


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$egingroup$ Siong Thye Goh has actually currently gave a nice explacountry for your question. The mode is offered simply to discover the worth of $k$. The p.d.f of $X$ is a raising attribute (in fact a straight line via slope $k>0$) in the interval $left<0, sqrtfrac2k ight>$. Since mode exists it need to be at the finish suggest of this interval. This gives you $x=fracsqrt24=sqrtfrac2k$ $endgroup$
You are told the mode so that we recognize the value of $k$.

You are watching: How to find the median of a pdf

$$sqrtfrac2k=fracsqrt24$$

Hence $k = 16.$

$$int_0^m 16x , dx = frac12$$

Try to deal with for $m$ from the above equation.


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The mode is in this case $sqrt2/k$ so the reality that it likewise equalizes $sqrt2/4$ tells us that $k=16$.

Now we have the right to discover the median $m$ on base of $$int^m_016xdx=frac12$$causing $$m=frac14$$


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