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Date	May Specimen paper	Marks available	4	Reference code	SPM.2.AHL.TZ0.6
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	6	Adapted from	N/A

Question

In a city, the number of passengers, X, who ride in a taxi has the following probability distribution.

After the opening of a new highway that charges a toll, a taxi company introduces a charge for passengers who use the highway. The charge is $ 2.40 per taxi plus $ 1.20 per passenger. Let T represent the amount, in dollars, that is charged by the taxi company per ride.

Find E(T).

[4]

a.

Given that Var(X) = 0.8419, find Var(T).

[2]

b.

Markscheme

METHOD 1

attempting to use the expected value formula (M1)

${\text{E}}\left( X \right) = \left( {1 \times 0.60} \right) + \left( {2 \times 0.30} \right) + \left( {3 \times 0.03} \right) + \left( {4 \times 0.05} \right) + \left( {5 \times 0.02} \right)$

${\text{E}}\left( X \right) = 1.59$ ($) (A1)

use of ${\text{E}}\left( {1.20X + 2.40} \right) = 1.20{\text{E}}\left( X \right) + 2.40$ (M1)

${\text{E}}\left( T \right) = 1.20\left( {1.59} \right) + 2.40$

$= 4.31$ ($) A1

METHOD 2

attempting to find the probability distribution for T (M1)

(A1)

attempting to use the expected value formula (M1)

${\text{E}}\left( T \right) = \left( {3.60 \times 0.60} \right) + \left( {4.80 \times 0.30} \right) + \left( {6.00 \times 0.03} \right) + \left( {7.20 \times 0.05} \right) + \left( {8.40 \times 0.02} \right)$

$= 4.31$ ($) A1

[4 marks]

a.

METHOD 1

using Var(1.20 X + 2.40) = (1.20)² Var(X) with Var(X) = 0.8419 (M1)

Var(T) = 1.21 A1

METHOD 2

finding the standard deviation for their probability distribution found in part (a) (M1)

Var(T) = (1.101…)²

= 1.21 A1

Note: Award M1A1 for Var(T) = (1.093…)² = 1.20.

[2 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Syllabus sections

Topic 4—Statistics and probability » AHL 4.14—Properties of discrete and continuous random variables

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