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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)

E(X)=(1×0.60)+(2×0.30)+(3×0.03)+(4×0.05)+(5×0.02)E(X)=(1×0.60)+(2×0.30)+(3×0.03)+(4×0.05)+(5×0.02)

E(X)=1.59E(X)=1.59($)     (A1)

use of E(1.20X+2.40)=1.20E(X)+2.40E(1.20X+2.40)=1.20E(X)+2.40      (M1)

E(T)=1.20(1.59)+2.40E(T)=1.20(1.59)+2.40

=4.31=4.31($)      A1

 

METHOD 2

attempting to find the probability distribution for T      (M1)

     (A1)

attempting to use the expected value formula       (M1)

E(T)=(3.60×0.60)+(4.80×0.30)+(6.00×0.03)+(7.20×0.05)+(8.40×0.02)E(T)=(3.60×0.60)+(4.80×0.30)+(6.00×0.03)+(7.20×0.05)+(8.40×0.02)

=4.31=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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Topic 4—Statistics and probability

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