1
$\begingroup$

Ok, I can't figure out how to exactly calculate the median. I have a pdf for a continuous random variable f(x) = x/2, for 0 < x < 2, and 0 otherwise. I THINK the median is found using something like $$1/2 = \int_0^m x/2 \,dx$$, but I don't know what the lower bound is supposed to be in this context.

$\endgroup$
2
  • 1
    $\begingroup$ Never mind, I figured it out. I had it right. According to a website Let X be a continuous random variable. The median of X is the number M such that P(X ≤ M) = 1/2. $\endgroup$
    – PCRevolt
    Mar 5 '17 at 18:41
  • $\begingroup$ @johny The median is not the mean. $\endgroup$
    – mlc
    Mar 5 '17 at 18:44
0
$\begingroup$

An answer just for completeness:

In general for a continuous random variable the median is the value $n$ which satisfies $$ \int_{-\infty}^m f(x) \,dx = \frac12$$ which, with the particular density you mention corresponding to a triangular distribution with parameters $a=0, b=2, c=2$, indeed gives the expression in your question, leading you to solve $\frac{m^2}{4}=\frac12$, i.e. $m= \sqrt{2}$, slightly more than the mean of $\frac43$ and rather less than the mode of $2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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