The random variable X is assumed to have probability density function f, where

\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\frac{x}{{18,}}}&{0 \leqslant x \leqslant 6} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]

Show that if the assumption is correct, then

\[{\text{P}}(a \leqslant X \leqslant b) = \frac{{{b^2} - {a^2}}}{{36}},{\text{ for }}0 \leqslant a \leqslant b \leqslant 6.\]

\({\text{P}}(a \leqslant X \leqslant b) = \int_a^b {\frac{x}{{18}}{\text{d}}x} \)     M1A1

\( = \left[ {\frac{{{x^2}}}{{36}}} \right]_a^b\)     A1

\( = \frac{{{b^2} - {a^2}}}{{36}}\)     AG

[3 marks]

This was the best answered question on the paper, helped probably by the fact that rounding errors in finding the expected frequencies were not an issue. In (a), some candidates thought, incorrectly, that all they had to do was to show that \(\int_0^6 {f(x){\text{d}}x = 1} \).