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]