Applications of integral calculus — Question 283

Original QCE Vault practice · 8 marks

Q283 · Practice questionTechnology-activeComplex unfamiliar8 marks

QUESTION 283 (8 marks)

A waiting time T has density

f(t)=kt e^{-2t}

for

t\ge0

and zero otherwise, with

k>0

. For this question you may use

E(T)=\int_0^\infty tf(t)\,dt,\qquad E(T^2)=\int_0^\infty t^2f(t)\,dt,\qquad\operatorname{Var}(T)=E(T^2)-[E(T)]^2.

All relevant integrals converge; polynomial factors times

e^{-2t}

tend to zero at infinity.

Determine k by normalising the density, showing the integration.

[2 marks]

Use integration by parts to determine the mean and variance.

[3 marks]

For 50 independent waits, approximate

P(\overline T>1.2)

using the central limit theorem. Give four decimal places and explain why the population need not be normal.

[3 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

8 marks

ANSWER

(a)

k=4

. (b)

E(T)=1

\operatorname{Var}(T)=1/2

. (c)

P(\overline T>1.2)\approx0.0228

Worked solution

(a) Integration by parts gives

\int_0^\infty te^{-2t}dt=[-te^{-2t}/2]_0^\infty+\tfrac12\int_0^\infty e^{-2t}dt=1/4

. Thus

k/4=1

, so

k=4

. The density is nonnegative.

(b) Let

I_n=\int_0^\infty t^n e^{-2t}dt

. Parts gives

I_n=(n/2)I_{n-1}

, with

I_0=1/2

. Hence

I_2=1/4

and

I_3=3/8

. Thus

E(T)=4I_2=1

and

E(T^2)=4I_3=3/2

, yielding variance

1/2

N(1,(1/2)/50)=N(1,0.01)

in variance notation. Standardising 1.2 gives z=2 and upper-tail probability 0.0228. The CLT concerns the distribution of averages of independent observations with finite variance; it does not assert that T itself is normally distributed.

Equivalent justified methods accepted; respect any requested proof method.

Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.

Part a: Evaluate the normalising integral using parts.

[1 mark]

Part a: Impose integral one and determine k.

[1 mark]

Part b: Apply parts to obtain the needed moments.

[1 mark]

Part b: Calculate mean 1 and second moment

3/2

[1 mark]

Part b: Subtract the squared mean to obtain variance

1/2

[1 mark]

Part c: Use the correct sample-mean standard deviation 0.1.

[1 mark]

Part c: Standardise and calculate the upper-tail probability.

[1 mark]

Part c: Explain the use and approximation of the CLT.

[1 mark]

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Applications of integral calculus — Question 283

QUESTION 283 (8 marks)

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