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Date November 2016 Marks available 3 Reference code 16N.2.hl.TZ0.1
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 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 E(X2).

[2]
a.

Find the value of Var(X).

[3]
b.

Markscheme

E(X2)=Σx2P(X=x)=10.37 (=10.4 3 sf)    (M1)A1

[2 marks]

a.

METHOD 1

sd(X)=1.44069    (M1)(A1)

Var(X)=2.08 (=2.0756)    A1

METHOD 2

E(X)=2.88 (=0.06+0.27+0.5+0.98+0.63+0.44)    (A1)

use of Var(X)=E(X2)(E(X))2     (M1)

 

Note: Award (M1) only if (E(X))2 is used correctly.

 

(Var(X)=10.378.29)

Var(X)=2.08 (=2.0756)    A1

 

Note: Accept 2.11.

 

METHOD 3

E(X)=2.88 (=0.06+0.27+0.5+0.98+0.63+0.44)    (A1)

use of Var(X)=E((XE(X))2)     (M1)

(0.679728++0.549152)

Var(E)=2.08 (=2.0756)    A1

[3 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.5 » Concept of discrete and continuous random variables and their probability distributions.
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