Date | November 2018 | Marks available | 2 | 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 X and Y follow Poisson distributions.
Given that E(X)=3 and E(Y)=4, calculate
E(2X+7Y).
[2]
a.
Var(4X−3Y).
[3]
b.
E(X2−Y2).
[4]
c.
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=34 (M1)A1
[2 marks]
a.
Var(X)=E(X)=3 and Var(Y)=E(Y)=4 (R1)
Var(4X−3Y)=16Var(X)+9Var(Y)=48+36 (M1)
= 84 A1
[3 marks]
b.
use of E(U2)=Var(U)+(E(U))2 (M1)
E(X2)=3+32; E(Y2)=4+42 A1
E(X2−Y2)=E(X2)−E(Y2) (M1)
= −8 A1
[4 marks]
c.
Examiners report
[N/A]
a.
[N/A]
b.
[N/A]
c.
Syllabus sections
Topic 4—Statistics and probability » AHL 4.14—Linear transformation of a single RV, E(X) and VAR(X), unbiased estimators
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