A continuous random variable X has probability density function

$$f(x) = \left\{ {\begin{array}{*{20}{c}}   {12{x^2}(1 - x),}&{{\text{for }}0 \leqslant x \leqslant 1,} \\   {0,}&{{\text{otherwise}}{\text{.}}} \end{array}} \right.$$

Find the probability that X lies between the mean and the mode.

Attempting to find the mode graphically or by using $f'(x) = 12x(2 - 3x)$     (M1)

${\text{Mode}} = \frac{2}{3}$     A1

Use of ${\text{E}}(X) = \int_0^1 {xf(x){\text{d}}x}$     (M1)

${\text{E}}(X) = \frac{3}{5}$     A1

$\int_{\frac{3}{5}}^{\frac{2}{3}} {f(x){\text{d}}x = 0.117\,\,\,\,\,\left( { = \frac{{1981}}{{16\,{\text{875}}}}} \right)}$     M1A1     N4

[6 marks]

A significant number of candidates attempted to find the mode and the mean using calculus when it could be argued that these quantities could be found more efficiently with a GDC. 

A significant proportion of candidates demonstrated a lack of understanding of the definitions governing the mean, mode and median of a continuous probability density function. A significant number of candidates attempted to calculate the median instead of either the mean or the mode. A number of candidates prematurely rounded their value for the mode i.e. subsequently using 0.7 for example rather than using the exact value of $\frac{2}{3}$. A few candidates offered negative probability values or probabilities greater than one.