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?

\$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\$.

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\$\$sqrtfrac2k=fracsqrt24\$\$

Hence \$k = 16.\$

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

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

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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