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Consider a continuous random variable X with probability density function given by:

$f(x)=4x(1-x^2)$ for $0 \le x \le 1$

Find the median.

So to calculate the median, I calculated the CDF and then set that equal to 0.5 and solve for x:

$F(x)=2x^2-x^4$

$0.5=2x^2-x^4\tag{1}$

So now we just have to solve equation (1) for x. We can do this by quadratic formula by setting $y=x^2$.

$y^2-2y+0.5=0\tag{2}$

$\implies y = \cfrac{2 \pm \sqrt{2}}{2}$

$\implies y= 1.71, y=0.293$

The answer in my book is $x_{0.5}=\cfrac{2 - \sqrt{2}}{2}$. Don't we have to solve for x by taking the sqrt of y to get the final answer? In other words, shouldn't the answer be $\sqrt{.293}$? We eliminate $\sqrt{1.71}$ because it's not in the domain...

Thanks in advance.

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  • $\begingroup$ I agree you need to take the root. The reason to choose $.293$ is probably because the variable takes values in $[0,1]$, so the median "has to" be in that interval, whereas $\sqrt{1.71}$ isn't, as it is greater than 1. $\endgroup$ – MickG Dec 9 '15 at 14:43
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I believe you are 100% correct.

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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ – Shobhit Oct 29 '13 at 4:46
  • $\begingroup$ I don't agree with your comment. The question is essentially "shouldn't the answer be $\sqrt{.293}$"?, to which I answered yes. Maybe OP can chime in here? $\endgroup$ – timidpueo Oct 29 '13 at 5:30
  • $\begingroup$ I actually didn't mind his answer. The only part that worried me a little was when he said "I believe". If someone provides an answer, I think it should be provided certainty. If not THEN it should be listed as a comment.... but that's just my 2 cents. Thanks guys!! $\endgroup$ – user1527227 Oct 29 '13 at 13:51
  • $\begingroup$ What if I said I was 99% sure you are 100% correct ;-) $\endgroup$ – timidpueo Oct 29 '13 at 14:02

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