The discrete random variable X has probability distribution:

(a)     Find the value of a.

(b)     Find ${\text{E}}(X)$.

(c)     Find ${\text{Var}}(X)$.

(a)     $\frac{1}{6} + \frac{1}{2} + \frac{3}{{10}} + a = 1 \Rightarrow a = \frac{1}{{30}}$     A1

(b)     ${\text{E}}(X) = \frac{1}{2} + 2 \times \frac{3}{{10}} + 3 \times \frac{1}{{30}}$     M1

$= \frac{6}{5}$     A1

Note:     Do not award FT marks if a is outside [0, 1].

[2 marks]

(c)     ${\text{E}}({X^2}) = \frac{1}{2} + {2^2} \times \frac{3}{{10}} + {3^2} \times \frac{1}{{30}} = 2$     (A1)

attempt to apply ${\text{Var}}(X) = {\text{E}}({X^2}) - {\left( {{\text{E}}(X)} \right)^2}$     M1

$\left( { = 2 - \frac{{36}}{{25}}} \right) = \frac{{14}}{{25}}$     A1

[3 marks]

Total [6 marks]

This was very well answered and many fully correct solutions were seen. A small number of candidates made arithmetic mistakes in part a) and thus lost one or two accuracy marks. A few also seemed unaware of the formula ${\text{Var}}(X) = {\text{E}}({X^2}) - {\text{E}}{(X)^2}$ and resorted to seeking an alternative, sometimes even attempting to apply a clearly incorrect ${\text{Var}}(X) = \sum {{{({x_i} - \mu )}^2}}$.