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**geton** The amount of vegetables eaten by a family in a week is a random variable W kg. The probability density function is given by:

$\displaystyle f(w) = \begin{cases} \frac {20}{5^5}w^3(5-w), &\text {$0 \leq w \leq 5$,} \\ 0, & \text {otherwise.} \end{cases} $

(a) Find the cumulative distribution function of W.

(b) Find the probability that the family eats between 2kg & 4kg of vegetables in one week.

(c) Verify that the amount, m, of vegetables such that the family is equally likely to eat more of less than m in any week is about 3.431 kg.

I have done part (a) & (b), and the answer of (b) is 0.650. But how could I verify (c)?