A continuous random variable X has probability density function f(x) given by $$ f(x) = \begin{cases} 12x^2(1-x), & 0 \leq x \leq 1 \\ 0, & \text{for all other values of } x \end{cases}$$ (a) Find the mode of X - HSC - SSCE Mathematics Advanced - Question 29 - 2023 - Paper 1

To find the cumulative distribution function F(x), we integrate the probability density function f(x) from 0 to x:

1. Setting up the integral:
   $F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} 12t^2(1-t) dt$

2. Carrying out the integration:

   = \left[ 4t^3 - 3t^4 \right]_{0}^{x} \\ = 4x^3 - 3x^4 $$ Thus, for 0 \leq x \leq 1, the cumulative distribution function is given by $$ F(x) = 4x^3 - 3x^4 $$.

To show that the mode is greater than the median, we look at the mode found in part (a) and use the properties of the cumulative distribution function:

1. Substituting the mode into the cumulative distribution function:
   We have found that the mode is \frac{2}{3}. We can express this in terms of the cumulative distribution function. We calculate F(\frac{2}{3}): F\left(\frac{2}{3}\right) = 4\left(\frac{2}{3}\right)^3 - 3\left(\frac{2}{3}\right)^4 \\ = 4\cdot \frac{8}{27} - 3\cdot \frac{16}{81} \\ = \frac{32}{27} - \frac{48}{81} \\ = \frac{32}{27} - \frac{16}{27} \\ = \frac{16}{27} \\ = 0.59 \ $$
2. Interpreting the result:
   Since the probability of the variable being less than \frac{2}{3} is greater than 0.5, we conclude that the mode (\frac{2}{3}) is greater than the median.