Date | November 2018 | Marks available | 3 | Reference code | 18N.3.AHL.TZ0.Hsp_1 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Calculate | Question number | Hsp_1 | Adapted from | N/A |
Question
Two independent random variables XX and YY follow Poisson distributions.
Given that E(X)=3E(X)=3 and E(Y)=4E(Y)=4, calculate
E(2X+7Y)E(2X+7Y).
Var(4X−3Y)(4X−3Y).
E(X2−Y2)E(X2−Y2).
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
E(2X+7Y)=2E(X)+7E(Y)=6+28=34E(2X+7Y)=2E(X)+7E(Y)=6+28=34 (M1)A1
[2 marks]
Var(X)=E(X)=3(X)=E(X)=3 and Var(Y)=E(Y)=4(Y)=E(Y)=4 (R1)
Var(4X−3Y)=16Var(X)+9Var(Y)=48+36Var(4X−3Y)=16Var(X)+9Var(Y)=48+36 (M1)
= 84 A1
[3 marks]
use of E(U2)=Var(U)+(E(U))2E(U2)=Var(U)+(E(U))2 (M1)
E(X2)=3+32E(X2)=3+32; E(Y2)=4+42E(Y2)=4+42 A1
E(X2−Y2)=E(X2)−E(Y2)E(X2−Y2)=E(X2)−E(Y2) (M1)
= −8 A1
[4 marks]