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Date November 2016 Marks available 3 Reference code 16N.2.AHL.TZ0.H_1
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Find 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 E ( X 2 ) .

[2]
a.

Find the value of Var ( X ) .

[3]
b.

Markscheme

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

E ( X 2 ) = Σ x 2 P ( 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 ( X 2 ) ( E ( X ) ) 2      (M1)

 

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

 

( Var ( X ) = 10.37 8.29 )

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

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