Random variable $X$ has a continuous distribution with the probability density function below
$$f_X(x) = \begin{cases} 60x^3(1-x)^2& 0 \leq x \leq 1\\ 0 & \text{otherwise} \\ \end{cases} $$
(a) Determine the prediction of for a future $X$, if we have decided to predict using the median of the distribution
Attempt
$f(x) = 60x^3(1-x)^2 = 60x^3 (1-2x+x^2) = 60(x^3 -2x^4+x^5)$
$F(x) = 60 \left(\frac{x^4}{4} - 2 \frac{x^5}{5} + \frac{x^6}{6} \right)$
we need to solve
$60 \left(\frac{x^4}{4} - 2 \frac{x^5}{5} + \frac{x^6}{6} \right) - 0.5 = 0$
which has roots
$m \approx -0.375$ and $m \approx 0.579$
now what?
Since $m$ represents a median, and the support of $X$ is on $[0,1]$, the negative root is extraneous, and the desired median is the unique solution satisfying $0 \le m \le 1$.
That said, this is a strange way of "predicting" a future observation. Usually, when we talk about prediction in statistics, we do so in the context of previously observed data; so for example, what would make this question more interesting is if a sample of some size were drawn from such a distribution, and based on this sample, you would try to predict the next observation.