A random variable $X$ has probability density function

$$f(x) = \left\{ \begin{array}{r}ax + b,\\0,\end{array} \right.\begin{array}{*{20}{c}}{2 \le x \le 3}\\{{\rm{ otherwise}}}\end{array},a,b \in \mathbb{R}$$

(a)     Show that $5a + 2b = 2$.

Let ${\text{E}}(X) = \mu$.

(b)     (i)     Show that $a = 12\mu  - 30$.

          (ii)     Find a similar expression for b in terms of $\mu$.

Let the median of the distribution be 2.3.

(c)     (i)     Find the value of $\mu$.

          (ii)     Find the value of the standard deviation of X.

(a)   $\int_2^3 {(ax + b){\text{d}}x{\text{ }}( = 1)}$     M1A1

$\left[ {\frac{1}{2}a{x^2} + bx} \right]_2^3{\text{ }}( = 1)$     A1

$\frac{5}{2}a + b = 1$     M1

$5a + 2b = 2$     AG

[4 marks]

 

(b)     (i)     $\int_2^3 {\left( {a{x^2} + bx} \right){\text{d}}x{\text{ }}( = \mu )}$     M1A1

          $\left[ {\frac{1}{3}a{x^3} + \frac{1}{2}b{x^2}} \right]_2^3{\text{ }}( = \mu )$     A1

          $\frac{{19}}{3}a + \frac{5}{2}b = \mu$     A1

          eliminating b     M1

          eg

          $\frac{{19}}{3}a + \frac{5}{2}\left( {1 - \frac{5}{2}a} \right) = \mu$     A1

          $\frac{1}{{12}}a + \frac{5}{2} = \mu$

          $a = 12\mu  - 30$     AG

 

Note:     Elimination of b could be at different stages.

 

          (ii)     $b = 1 - \frac{5}{2}(12\mu  - 30)$

          $= 76 - 30\mu$     A1

 

Note:     This solution may be seen in part (i).

 

[7 marks]

 

(c)     (i)     $\int_2^{2.3} {(ax + b){\text{d}}x{\text{ }}( = 0.5)}$     (M1)(A1)

          $\left[ {\frac{1}{2}a{x^2} + bx} \right]_2^{2.3}{\text{ }}( = 0.5)$

          $0.645a + 0.3b{\text{ }}( = 0.5)$     (A1)

          $0.645(12\mu  - 30) + 0.3(76 - 30\mu ) = 0.5$     M1

          $\mu  = 2.34{\text{ }}\left( { = \frac{{295}}{{126}}} \right)$     A1

          (ii)     ${\text{E}}\left( {{X^2}} \right) = \int_2^3 {{x^2}(ax + b){\text{d}}x}$     (M1)

          $a = 12\mu  - 30 =  - \frac{{40}}{{21}},{\text{ }}b = 76 - 30\mu  = \frac{{121}}{{21}}$     (A1)

          ${\text{E}}\left( {{X^2}} \right) = \int_2^3 {{x^2}\left( { - \frac{{40}}{{21}}x + \frac{{121}}{{21}}} \right){\text{d}}x = 5.539 \ldots {\text{ }}\left( { = \frac{{349}}{{63}}} \right)}$     (A1)

          ${\text{Var}}(X) = 5.539{\text{K}} - {(2.341{\text{K}})^2} = 0.05813 \ldots$     (M1)

          $\sigma  = 0.241$     A1

[10 marks]

 

Total [21 marks]

[N/A]