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Date	May 2019	Marks available	2	Reference code	19M.3.AHL.TZ0.Hsp_2
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	Find	Question number	Hsp_2	Adapted from	N/A

Question

Employees answer the telephone in a customer relations department. The time taken for an employee to deal with a customer is a random variable which can be modelled by a normal distribution with mean 150 seconds and standard deviation 45 seconds.

Find the probability that the time taken for a randomly chosen customer to be dealt with by an employee is greater than 180 seconds.

[2]

Find the probability that the time taken by an employee to deal with a queue of three customers is less than nine minutes.

[4]

At the start of the day, one employee, Amanda, has a queue of four customers. A second employee, Brian, has a queue of three customers. You may assume they work independently.

Find the probability that Amanda’s queue will be dealt with before Brian’s queue.

[6]

Markscheme

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

Note: In question 2, accept answers that round correctly to 2 significant figures.

$X \sim N\left( {150{\text{,}}\,\,{{45}^2}} \right)$

${\text{P}}\left( {X > 80} \right) = 0.252$ (M1)A1

[2 marks]

Note: In question 2, accept answers that round correctly to 2 significant figures.

required to find ${\text{P}}\left( {{X_1} + {X_2} + {X_3} < 540} \right)$

let $S = {X_1} + {X_2} + {X_3}$

${\text{E}}\left( S \right) = 450$ (A1)

${\text{Var}}\left( S \right) = 3\,{\text{Var}}\left( X \right)$ (M1)

$= 3 \times {45^2}\,\left( { \Rightarrow \sigma = 45\sqrt 3 } \right)\,\left( { = 6075} \right)$ (A1)

${\text{P}}\left( {S < 540} \right) = 0.876$ A1

Note: In (b) and (c) condone incorrect notation, eg, $3X$ for ${X_1} + {X_2} + {X_3}$ .

[4 marks]

Note: In question 2, accept answers that round correctly to 2 significant figures.

let $Y = \left( {{X_1} + {X_2} + {X_3} + {X_4}} \right) - \left( {{X_5} + {X_6} + {X_7}} \right)$ (M1)

${\text{E}}\left( Y \right) = {\text{E}}\left( X \right) = 150$ (A1)

${\text{Var}}\left( Y \right) = 4\,{\text{Var}}\left( X \right) + 3\,{\text{Var}}\left( X \right)\left( { = 7\,{\text{Var}}\left( X \right)} \right)$ (M1)

= 14175 (A1)

required to find ${\text{P}}\left( {Y < 0} \right)$ (M1)

= 0.104 A1

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 4—Statistics and probability » AHL 4.14—Linear transformation of a single RV, E(X) and VAR(X), unbiased estimators

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Topic 4—Statistics and probability

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