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 \(P(V > 5) = 1 - P(V \leqslant 5)\).