Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:



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


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

$\implies y= 1.71, y=0.293$

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.

share|cite|improve this question
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. – MickG Dec 9 '15 at 14:43
up vote 0 down vote accepted

I believe you are 100% correct.

share|cite|improve this answer
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – SHOBHIT GAUTAM Oct 29 '13 at 4:46
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? – timidpueo Oct 29 '13 at 5:30
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!! – user1527227 Oct 29 '13 at 13:51
What if I said I was 99% sure you are 100% correct ;-) – timidpueo Oct 29 '13 at 14:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.