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Date	November 2016	Marks available	2	Reference code	16N.2.AHL.TZ0.H_1
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Determine	Question number	H_1	Adapted from	N/A

Question

A random variable $X$ has a probability distribution given in the following table.

N16/5/MATHL/HP2/ENG/TZ0/01

Determine the value of ${\text{E}}({X^2})$ .

[2]

Find the value of ${\text{Var}}(X)$ .

[3]

Markscheme

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

${\text{E}}({X^2}) = \Sigma {x^2} \bullet {\text{P}}(X = x) = 10.37{\text{ }}( = 10.4{\text{ 3 sf)}}$ (M1)A1

[2 marks]

METHOD 1

${\text{sd}}(X) = 1.44069 \ldots$ (M1)(A1)

${\text{Var}}(X) = 2.08{\text{ }}( = 2.0756)$ A1

METHOD 2

${\text{E}}(X) = 2.88{\text{ }}( = 0.06 + 0.27 + 0.5 + 0.98 + 0.63 + 0.44)$ (A1)

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

Note: Award (M1) only if ${\left( {{\text{E}}(X)} \right)^2}$ is used correctly.

$\left( {{\text{Var}}(X) = 10.37 - 8.29} \right)$

${\text{Var}}(X) = 2.08{\text{ }}( = 2.0756)$ A1

Note: Accept 2.11.

METHOD 3

${\text{E}}(X) = 2.88{\text{ }}( = 0.06 + 0.27 + 0.5 + 0.98 + 0.63 + 0.44)$ (A1)

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

$(0.679728 + \ldots + 0.549152)$

${\text{Var}}(E) = 2.08{\text{ }}( = 2.0756)$ A1

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 4—Statistics and probability » SL 4.7—Discrete random variables

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