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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 $X$ $X$ and $Y$ $Y$ follow Poisson distributions.

Given that ${\text{E}}\left( X \right) = 3$ ${\text{E}}\left( X \right) = 3$ and ${\text{E}}\left( Y \right) = 4$ ${\text{E}}\left( Y \right) = 4$ , calculate

${\text{E}}\left( {2X + 7Y} \right)$ ${\text{E}}\left( {2X + 7Y} \right)$ .

[2]

Var $\left( {4X - 3Y} \right)$ $\left( {4X - 3Y} \right)$ .

[3]

${\text{E}}\left( {{X^2} - {Y^2}} \right)$ ${\text{E}}\left( {{X^2} - {Y^2}} \right)$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\text{E}}\left( {2X + 7Y} \right) = 2{\text{E}}\left( X \right) + 7{\text{E}}\left( Y \right) = 6 + 28 = 34$ ${\text{E}}\left( {2X + 7Y} \right) = 2{\text{E}}\left( X \right) + 7{\text{E}}\left( Y \right) = 6 + 28 = 34$ (M1)A1

[2 marks]

Var $\left( X \right) = {\text{E}}\left( X \right) = 3$ $\left( X \right) = {\text{E}}\left( X \right) = 3$ and Var $\left( Y \right) = {\text{E}}\left( Y \right) = 4$ $\left( Y \right) = {\text{E}}\left( Y \right) = 4$ (R1)

${\text{Var}}\left( {4X - 3Y} \right) = 16{\text{Var}}\left( X \right) + 9{\text{Var}}\left( Y \right) = 48 + 36$ ${\text{Var}}\left( {4X - 3Y} \right) = 16{\text{Var}}\left( X \right) + 9{\text{Var}}\left( Y \right) = 48 + 36$ (M1)

= 84 A1

[3 marks]

use of ${\text{E}}\left( {{U^2}} \right) = {\text{Var}}\left( U \right) + {\left( {{\text{E}}\left( U \right)} \right)^2}$ ${\text{E}}\left( {{U^2}} \right) = {\text{Var}}\left( U \right) + {\left( {{\text{E}}\left( U \right)} \right)^2}$ (M1)

${\text{E}}\left( {{X^2}} \right) = 3 + {3^2}$ ${\text{E}}\left( {{X^2}} \right) = 3 + {3^2}$ ; ${\text{E}}\left( {{Y^2}} \right) = 4 + {4^2}$ ${\text{E}}\left( {{Y^2}} \right) = 4 + {4^2}$ A1

${\text{E}}\left( {{X^2} - {Y^2}} \right) = {\text{E}}\left( {{X^2}} \right) - {\text{E}}\left( {{Y^2}} \right)$ ${\text{E}}\left( {{X^2} - {Y^2}} \right) = {\text{E}}\left( {{X^2}} \right) - {\text{E}}\left( {{Y^2}} \right)$ (M1)

= −8 A1

[4 marks]

Examiners report

[N/A]

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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