Date | May 2008 | Marks available | 6 | Reference code | 08M.3sp.hl.TZ2.1 |
Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ2 |
Command term | Calculate | Question number | 1 | Adapted from | N/A |
Question
The random variable Y is such that E(2Y+3)=6 and Var(2−3Y)=11.
Calculate
(i) E(Y) ;
(ii) Var(Y) ;
(iii) E(Y2) .
Independent random variables R and S are such that
R∼N(5, 1) and S∼N(8, 2).
The random variable V is defined by V = 3S – 4R.
Calculate P(V > 5).
Markscheme
(i) E(2Y+3)=6
2E(Y)+3=6 M1
E(Y)=32 A1
(ii) Var(2−3Y)=11
Var(−3Y)=11 (M1)
9Var(Y)=11
Var(Y)=119 A1
(iii) E(Y2)=Var(Y)+[E(Y)]2 M1
=119+94
=12536 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∼N(4, 34)
P(V>5)=0.432 A2 N0
[6 marks]
Examiners report
E(Y) was calculated correctly but many could not go further to find Var(Y) and E(Y2)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⩽5).
E(Y) was calculated correctly but many could not go further to find Var(Y) and E(Y2)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⩽5).