The random variable Y is such that ${\text{E}}(2Y + 3) = 6{\text{ and Var}}(2 - 3Y) = 11$.

Calculate

(i)     E(Y) ;

(ii)     ${\text{Var}}(Y)$ ;

(iii)     ${\text{E}}({Y^2})$ .

Independent random variables R and S are such that

$$R \sim {\text{N}}(5,{\text{ 1}}){\text{ and }}S \sim {\text{N(8, 2).}}$$

The random variable V is defined by V = 3S – 4R.

Calculate P(V > 5).

(i)     ${\text{E}}(2Y + 3) = 6$

$2{\text{E}}(Y) + 3 = 6$     M1

${\text{E}}(Y) = \frac{3}{2}$     A1

(ii)     ${\text{Var}}(2 - 3Y) = 11$

${\text{Var}}( - 3Y) = 11$     (M1)

$9{\text{Var}}(Y) = 11$

${\text{Var}}(Y) = \frac{{11}}{9}$     A1

(iii)     ${\text{E}}({Y^2}) = {\text{Var}}(Y) + {\left[ {{\text{E}}(Y)} \right]^2}$     M1

$= \frac{{11}}{9} + \frac{9}{4}$

$= \frac{{125}}{{36}}$     A1     N0

[6 marks]

 E(V) = E(3S – 4R)

 = 3E(S) – 4E(R)     M1

 = 24 – 20 = 4     A1

 Var(3S – 4R) = 9Var(S) + 16Var(R) , since R and S are independent random variables     M1

 =18 + 16 = 34     A1

 $V \sim {\text{N}}(4,{\text{ 34}})$

 ${\text{P}}(V > 5) = 0.432$     A2     N0

 [6 marks]

E(Y) was calculated correctly but many could not go further to find $Var(Y){\text{ and }}E({Y^2})Var(2)$ was often taken to be 2. V was often taken to be discrete leading to calculations such as $P(V > 5) = 1 - P(V \leqslant 5)$.

E(Y) was calculated correctly but many could not go further to find $Var(Y){\text{ and }}E({Y^2})Var(2)$ was often taken to be 2. V was often taken to be discrete leading to calculations such as $P(V > 5) = 1 - P(V \leqslant 5)$.