HONG KONG STATISTICAL SOCIETY 2016

EXAMINATIONS − SOLUTIONS

HIGHER CERTIFICATE – MODULE 2

The Society is providing these solutions to assist candidates preparing for the

examinations in 2017.

The solutions are intended as learning aids and should not be seen as "model

answers".

Users of the solutions should always be aware that in many cases there are valid

alternative methods. Also, in the many cases where discussion is called for, there

may be other valid points that could be made.

While every care has been taken with the preparation of these solutions, the Society

will not be responsible for any errors or omissions.

The Society will not enter into any correspondence in respect of these solutions.

RSS

HC

Module 2 2016 Solutions

Each question has marks out of 20.

Question 1X ∼ Poisson(λ) x = 0, 1, 2, . . .

PX(x) =λxe−λ

x!

(i)

PX(x+ 1) =λx+1e−λ

(x+ 1)![1]

=λ

(x+ 1)

λxe−λ

x![1]

=λ

(x+ 1)Px(x) [1]

(ii)

E(X) =∞∑x=0

xP (X = x)

=∞∑x=0

xλxe−λ

x![1]

= λ∞∑x=1

λx−1e−λ

(x− 1)!

= λ∞∑y=0

λye−λ

y!y = x− 1 [1]

= λ× 1 as sum of density

= λ [1]

E(X(X − 1)) =∞∑x=0

x(x− 1)P (X = x)

=∞∑x=2

x(x− 1)λxe−λ

x!

= λ2∞∑x=2

λx−2e−λ

(x− 2)![1]

= λ2∞∑y=0

λye−λ

y!y = x− 2

= λ2 × 1 as sum of density

= λ2 [1]

Var(X) = E(X(X − 1)) + E(X)− (E(X))2

= λ2 + λ− λ2 [1]

= λ

(iii) Y ∼ Poisson(µ) independent of X.

PW (w) =w∑x=0

P (X = x)P (Y = w − x).

=w∑x=0

λxe−λ

x!

µw−xe−µ

(w − x)![1]

=w∑x=0

(λ

λ+ µ

)x e−λx!

(µ

λ+ µ

)w−x×e−µ(λ+ µ)w

(w − x)![1]

=e−(λ+µ)(λ+ µ)w

w!

w∑x=0

w!

x!(w − x)!

×(

λ

λ+ µ

)x ( µ

λ+ µ

)w−x[1]

=(λ+ µ)we−(µ+λ)

w![1]

⇒W ∼ Poisson(λ+ µ).

(iv) (a) Two conveyor belts - type C and D.X =number of breakdowns per day on C.Y =number of breakdowns per day on D.

X ∼ Poisson(λ = 1.5)

Y ∼ Poisson(µ = 0.5)

P (exactly one breakdown) = P (X = 1, Y = 0) + P (X = 0, Y = 1)

= P (X = 1)× P (Y = 0) + P (X = 0)× P (Y = 1) [1]

= 1.5e−1.5 × e−0.5 + e−1.5 × 0.5e−0.5

= e−2(1.5 + 0.5)

= 2e−2 [1]

= 0.2707

P (1 breakdown on C|exactly one breakdown) =P (X = 1, Y = 0)

P (exactly one breakdown)

=1.5e−1.5 × e−0.5

2e−2[1]

=1.5

2

=3

4[1]

(b) In 5 days, the rate of occurrence of failures is 5×2 = 10. Let T=number of failures in 5 days. Then

T ∼ Poisson(10)

[1,1]

P (T ≤ 1) = 0.0005 [1]

Question 2 (i) Let X = number of failures before the first success.

PX(x) = (1− p)xp [1]

for x = 0, 1, 2, 3, 4, . . . [1]

∞∑x=0

(1− p)xp = p∞∑x=0

(1− p)x [1]

= p1

1− (1− p)sum of GP [1]

= p× 1

p= 1 [1]

(ii)

E(X) =∞∑x=0

xPX(x)

=∞∑x=0

x(1− p)xp [1]

Using the fact that:

d

dp

∞∑x=0

(1− p)x = −∞∑x=0

x(1− p)x−1

=d

dp

1

p[1]

= − 1

p2[1]

Hence

E(X) = p(1− p)∞∑x=0

x(1− p)x−1

= p(1− p)× 1

p2

=(1− p)

[1]p

NB Other correct solutions would be given credit.

(iii) (a) When p = 0.01,

P (X = 80) = 0.99790.01 [1]

= 0.00452 [1]

(b) EITHER

P (X > 80) = 1− P (X ≤ 80) [1]

= 1−80∑x=0

x(1− p)xp [1]

= 1− p(

1− (1− p)80

1− (1− p)

)[1]

= 1− 0.01((1− 0.9980))

0.01[1]

= 1− 0.5525

= 0.4475 [1]

OR As it will take more than 80 attempts if and only if the first 80are all failures [2](Either get 2 marks or 0)

P (X > 80) = 0.9980 [2]

(either get 2 marks or 0)

= 0.4475. [1]

(iv) Let Y= total number of failures before the second success. The prob-ability mass function is

PY (y) =

(y + 1

1

)(1− p)yp2 = (y + 1)(1− p)yp2 y = 0, 1, 2, . . . [1]

The experiment to obtain Y can be considered as running the firstexperiment until the first success and then repeating an independentexperiment to obtain a further success. [1]Hence

E(Y ) = E(X1) + E(X2) [1]

=(1− p)p

+(1− p)p

[1]

= 2(1− p)p

Question 3 (a) Let L be the volume of the large bottle in litres.Let S be the volume of the small bottle in litres.

L ∼ N(1.5, 0.012)

S ∼ N(0.5, 0.0082)

(i) Take a random sample of 1 large bottle and 3 small bottles.

P (L > 3S) = P (L− 3S > 0) [1]

Let Y = L− 3S.

E(L− 3S) = E(L)− 3 E(S) = 1.5− 3× 0.5 = 0

[1]

Var(L−3S) = Var(L)+9∗Var(S) = 0.012+9∗0.0082 = 0.000676

6

sd(L− 3S) = 0.026

NB The variance is not needed here, but some may not realisethat. It is not necessary to calculate, or even mention, the vari-ance.

P (Y > 0) = P

(Z >

0− 0

0.026

)= P (Z > 0)

= 1− 0.5

= 0.5 [1]

(ii) Let T = 10L+ 3S.

E(10L+ 3S) = 10 E(L) + 3 E(S) [1]

= 10× 1.5 + 3× 0.5

= 16.5 [1]

Var(10L+ 3S) = 102 Var(L) + 32 Var(S) [1]

= 100× 0.012 + 9× 0.0082

= 0.01 + 0.000576

= 0.010576 [1]

T is a linear combination of normal r.v.’s so T is also normal [1]

T ∼ N(16.5, 0.010576) [1]

(b) n = 50, µ = 30g and σ = 5g.

(i) As n is large the distribution of the sample mean tends to nor-mality by the Central Limit Theorem. [1]

E(X̄) = µ = 30g

[1]

Var(X̄) =σ2

n=

25

50=

1

2

[1]

X̄ ∼ N(30, 0.5)

(ii)

P (X̄ < 29) = P

(Z <

29− 30√0.5

)[1]

= P (Z < −√

2)

= P (Z < −1.41) [1]

= P (Z > 1.41)

= 1− P (Z < 1.41)

= 1− 0.9207

= 0.0793 [1]

(iii) Sample of size n.

P

Z <29− 30

5√n

= 0.05 [1]

⇒

29− 305√n

< −1.645 [1]

−1 < −1.645× 5√n

‘ [1]

−√n < −1.634× 5√n > 8.225 [1]

The minimum value of n is 68. [1]

Question 4 (a) The continuous random variable X is distributed withprobability density function

f(x) = α(1− x)α−1 0 < x < 1;α > 0

(i)

F (x) =

∫ x

0α(1− t)α−1dt [1]

= α

[−(1− x)α

α

]x0

= −(1− x)α + 1

= 1− (1− x)α [1]

F (x) = 0 x ≤ 0 [0.5]

= 1− (1− x)α

= 1 x ≥ 1 [0.5]

(ii)

P (0.25 < X < 0.75) = F (0.75)− F (0.25) [1]

= 1− (1− 0.75)α − [1− (1− 0.25)α] [1]

= 0.75α − 0.25α [1]

= 0.25α(3α − 1)

=3α − 1

4α[1]

(iii) Let M be the median.

F (M) =1

2[1]

1− (1−M)α =1

2

(1−M)α =1

2[1]

1−M = 2−1α

M = 1− 2−1α [1]

(b)

Y ∼ Ga(2, 3) f(y) = 9ye−3y

(i)

E(Y ) =

∫ ∞0

yf(y)dy

= 9

∫ ∞0

y2e−3ydy [1]

= 9

[[y2e−3y

(−3)

]∞0

+2

3

∫ ∞0

ye−3ydy

][1]

= 6

∫ ∞0

ye−3ydy [1]

= 6

[[ye−3y

(−3)

]∞0

+1

3

∫ ∞0

e−3ydy

][1]

= 2

∫ ∞0

e−3ydy

= 6

[e−3y

(−3)

]∞0

[1]

=2

3[1]

NB in line 3 the last integral is a multiple of the integral of thepdf and hence integrates to 1, hence the last term is just 2

3 × 1.Full marks are to be given for this solution.

(ii)

P (X < 3) = 9

∫ 3

0xe−3xdx [1]

= 9

[xe−3x(−3)

]30

−∫ 3

0

e−3x

(−3)dx

[1]

= −9e−9 + 3

∫ 3

0e−3xdx

= −9e−9 + 3

[e−3x

(−3)

]30

[1]

= 1− 10e−9 [1]

= 0.9988

NB The final evaluation is not required for the final mark.