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$X$ is a continuous random variable with probability density function $f(x)= \dfrac{2x}{15}$ where $1≤x≤4$.

What is the median of X?

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Welcome to MSE! It really helps readability to format questions using MathJax (see FAQ). Do you have any thoughts or have tried anything that you can share? Regards – Amzoti May 20 '13 at 20:24
A median (in some strange cases there may be more than one) is a number $m$ such that $\int_{-\infty}^m f(x)\,dx=\frac{1}{2}$, where $f(x)$ is the density function. In your case you want $\int_0^m \frac{2x}{15}\,dx=\frac{1}{2}$. – André Nicolas May 20 '13 at 20:33

Hint: To find the median, you want to find $c$ such that $\mathbb{P}(1 \le X \le c) = \mathbb{P}(c \le X \le 4)$. That's just the definition of the median: it's the number $c$ for which the probabilities on both of its sides are the same.

The integral you get here shouldn't be hard to carry out.

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Hint: You have to find $u$ such that $P(-\infty < x < u) = 0.5$.

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it's equal to 2.915 = SQRT(8.5) !

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