QCE Specialist Maths: Exponential random variables Q251

Q251 · Practice questionTechnology-activeVery complex unfamiliar12 marks

QUESTION 251 (12 marks)

Each round starts two independent unlocking processes. Their completion times

T_1

and

T_2

are exponential with means 2 and 3 minutes respectively. The round ends at

T=\min(T_1,T_2)

. All processes are reset independently between rounds. Let

E

be the event that process 1 finishes first. You may use

P(E\text{ and }T>t)=\int_t^\infty f_{T_1}(u)P(T_2>u)\,du

Determine the survival function and density of

T

[2 marks]

Find

P(E)

[2 marks]

Determine the distribution of

T

conditional on

E

, including its mean and standard deviation. Explain the surprising feature of your result.

[3 marks]

A four-hour event allows 240 minutes for rounds, with no changeover time. Using a normal approximation, determine the largest integer number of rounds

n

for which the chance of exceeding 240 minutes is at most

0.01

. Use

z_{0.99}=2.326

and check the integers on either side of your boundary.

[5 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

12 marks

ANSWER

(a)

P(T>t)=e^{-5t/6}

;

f_T(t)=\frac56e^{-5t/6}

for

t\ge0

. (b)

P(E)=3/5

. (c)

T\mid E

is exponential with rate

5/6

, mean and standard deviation

6/5

min; winner and round duration are independent. (d)

n=169

rounds (normal approximation).

Worked solution

(a) Independence gives

P(T>t)=P(T_1>t)P(T_2>t)=e^{-t/2}e^{-t/3}=e^{-5t/6}

. Its negative derivative gives the density.

(b)

P(E)=\int_0^\infty(1/2)e^{-u/2}e^{-u/3}\,du=(1/2)/(5/6)=3/5

. Simultaneous completion has probability zero.

P(E\text{ and }T>t)=(3/5)e^{-5t/6}

. Divide by

P(E)=3/5

to get

P(T>t\mid E)=e^{-5t/6}

. Thus conditioning on the winner does not change the duration law. The exponential mean and standard deviation both equal

6/5

(d) For total time

S_n

E(S_n)=1.2n

and

\operatorname{sd}(S_n)=1.2\sqrt n

. The CLT gives the requirement

(240-1.2n)/(1.2\sqrt n)\ge2.326

, equivalently

n+2.326\sqrt n\le200

. Setting

u=\sqrt n

gives

u^2+2.326u-200\le0

, so

n\le169.699\ldots

. The normal tail is about

0.00855

for

n=169

and

0.01070

for

n=170

. Since the boundary expression increases with

n

, 169 is the largest allowable integer under this approximation.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Obtain the survival function using independence.

[1 mark]

Part a: Obtain the density and its domain.

[1 mark]

Part b: Form the appropriate integral for the winning event.

[1 mark]

Part b: Evaluate it as

3/5

[1 mark]

Part c: Evaluate the joint survival-and-winner probability.

[1 mark]

Part c: Divide by

P(E)

and identify the unchanged exponential law.

[1 mark]

Part c: Give its mean and standard deviation and interpret the independence.

[1 mark]

Part d: Obtain the mean and standard deviation of total time.

[1 mark]

Part d: Form the correct one-sided normal probability requirement.

[1 mark]

Part d: Solve the quadratic boundary in

\sqrt n

[1 mark]

Part d: Check that 169 meets the probability requirement.

[1 mark]

Part d: Check that 170 fails and conclude the maximum integer.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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Exponential random variables — Question 251

QUESTION 251 (12 marks)

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