Applied Statistics and Mathematics in Economics & Business

BS1501

Tutorial 2Pairach Piboonrungroj

(Champ)

me@Pairach.com

1. Mutually Exclusive?(a) Being an economist and being a professor:

Not mutually exclusive

(b) Throwing a 5 or 6 with one die:Mutually exclusive

(c) Being an M.P. and being 20 years of age. (Minimum age of an M.P. is 21):Mutually exclusive

(d) Drawing a red card or an ace out of a pack of cards:Not mutually exclusive

2.Accountancy employees

(a) Prob. of employees from Wales: P(W) = from Wales / All employees = 6/15

(b) Prob. of employees Scotland: P(S) = from Scotland / All employees = 4/15

NOT Wales Wales

2.Accountancy employees

(c) Prob. of employees NOT from Wales:

=All prob. – P(W) = 1 – P(W)

= 1-(6/15) = 9/15

15

915

6

2.Accountancy employees

(d) Prob. of employees From Wales and Scotland

= P(W) + P(S)

= (6/15) + (4/15)

= 10/15

15

4

15

6

3. Primary school in CardiffI = the event that the boy likes ice-creamC = the event that the boy likes cake

P(I) = 0.75, P(C) = 0.55, P (I and C) = 0.40

(a) P(I or C or both) = P(I) + P(C) – P(I and C) = 0.75 + 0.55 – 0.40 = 0.90

P(I) P(C)P ( I and C)

3. Primary school in Cardiff

I = the event that the boy likes ice-cream

C = the event that the boy likes cake

P(I) = 0.75, P(C) = 0.55, P (I and C) = 0.40

(b) Neither ice-cream nor cake

= 1 – P(I or C or Both)

= 1- 0.90

= 0.1

4. Accidents

Accident No accident

Under 45 years

25 35

45 years or older

5 35

4. Accidents

Accident No accident

Total

Under 45 years

25 35 60 (25 + 35)

45 years or older

5 35 40(5 + 35)

Total 30 (25 + 5)

70(35 + 35)

100

4. Accidents

(a)

Accident No accident

Total

Under 45 years

0.25

(25/100)

0.35

(35/100)

0.60

(60/100)

45 years or older

0.05

(5/100)

0.35

(35/100)

0.40

(40/100)

Total 0.30

(30/100)

0.70

(70/100)

1.0 (100/100)

4. Accidents(b) P(Accident | under 45)

= P(Accident and under 45) / P(Under 45) = 0.25 / 0.60 = 0.416

Accident No accident Total

Under 45 years

0.25

(25/100)

0.35

(35/100)

0.60

([25+35]/100)

45 years or older

0.05

(5/100)

0.35

(35/100)

0.40

(40/100)

Total 0.30

(30/100)

0.70

(70/100)

1.0

(100/100)

4. Accidents(c) P( Accident | over 45)

= P(Accident and over 45) / P(over 45) = 0.05 / 0.40 = 0.125

Accident No accident Total

Under 45 years

0.25

(25/100)

0.35

(35/100)

0.60

([25+35]/100)

45 years or older

0.05

(5/100)

0.35

(35/100)

0.40

(40/100)

Total 0.30

(30/100)

0.70

(70/100)

1.0

(100/100)

All students

5. Economics course

P(M) = P(F) =1/2

P(E | M) = 1/6

P(E | F) = 1/30

(e) Male and Economics

P(M and E) = P(E | M) * P(M)

= (1/6)*(1/2) = 1/12

P(M) P(E I M)

5. Economics course

P(M) = P(F) =1/2P(E | M) = 1/6P(E | F) = 1/30

P(E) = 10% = 1/10

(b) P(M or E) = P(M) + P(E) – P(M and E)= (1/2) + (1/10) – (1/12)= 31/60

P(M) P(E)P ( M and E)

6. Financial advisor

)!(!

!

xnx

n

x

n

−=

)1*2*3*4(

7*8*9*10=

)3*8(

7*8*)3*3(*10=

n = 10

x = 4

(n-x) = 6!6!4

!6*7*8*9*10=

2101

7*3*10 ==

!6!4

!10

4

10=

Combination:Number of ways of selecting x object from a set of n, when the order in unimportant

7. Company’s Demand

Weekly unit Demand (x) Probability: f(x)

800 0.10

900 0.25

1000 0.40

1100 0.20

1200 0.05

Sum Expected value

E (x) =x*f(x)

80.00

225.00

400.00

220.00

60.00

985.00

x – E (x)

-185.00

-85.00

15.00

115.00

215.00

Variance

Std Deviation

[x-E(x)]^2*f(x)

3422.50

1806.25

90.00

2645.00

2311.25

10275.00

101.37

7. Company’s Demand

(a) E(x) = Sum [x*f(x)]= 800(0.10) + 900(0.25) + 1000(0.40) + 1100(0.20) + 1200(0.05) = 985

(b) Var(x) = Sum [(x-E(x)^2)*f(x)]= (800-985)^2(0.10) + (900 -985)^2(0.25) + (1000-985)^2(0.40)+(1100-985)^2(0.20) +(1200-985)^2(0.05) = 10,275

Standard deviation = sqrt(10,275) = 101.36

8. Insurance Salesperson

(a) 088.0)3/1()3/2(6

6)6( 06 =

==xP )!(!

!

xnx

n

x

n

−=

xnx xpxpx

nxP −−

= )](1[)]([)(

(b) 329.0)3/1()3/2(4

6)4( 24 =

==xP

8. Insurance Salesperson

(c)60 )3/1()3/2(

0

6)0(

==xP

)]1()0([1)1( =+=−=> xPxPxP

4)3/2(6 === npµ

51 )3/1()3/2(1

6)1(

==xP

)!(!

!

xnx

n

x

n

−=

xnx xpxp

x

nxP −−

= )](1[)]([)(

(d)

98.0

02.01

=−=

9. Fatal Rate

A) No death in a one year period x = 0

5.1)1( =yearλ

2231.0!0

5.1

!)0(

5.10

====−− e

x

exP

x λλ

Poisson Distribution: Estimating the number of occurrence of an event within specific interval of time

!)(

x

exP

x λλ −

=

9. Fatal Rate

b) No death in 6 months period x = 0

!)(

x

exP

x λλ −

=

75.02/5.1)6( ==monthsλ

4724.0!0

75.0)0(

75.00

===−e

xP

Poisson Distribution: Estimating the number of occurrence of an event within specific interval of time

9. Fatal Rate

c) P(>= 2 accidents in 6 months)

= 1 – P (x=0) – P (x=1)

= 1 – 0.4724 – 0.3543

= 0.1733

75.0)6( =monthsλ

10. Train (Cardiff – London)

−≥=>

10

120130)130( ZpxP

(a) )10,120(~ == σµNx

−>=>

σµ

0

0)(

xZpxxP

= P(Z>+1)

= 0.1586

10. Train

−<<−=<<

10

120125

10

120110)125110( ZPxP

)5.01( +<<−= ZP

(b) )10,120(~ == σµNx

−

≥=>σ

µ0

0)(

xZpxxP

10. Train

)1()1( >=−< ZPZP

)5.01( +<<− ZP(b)

= 1- [P(Z<-1) + P(Z>0.5)]

= 1 – 0.15866 – 0.30854 = 0.5328

Z = -1 Z = 0.5

= 1- [P(Z>1) + P(Z>0.5)]

10. Train

−≥=

10

12010.0 0

xZp

)10,120(~ == σµNx

−≥=>

10

120)( 0

0

xZpxxP

(c)

10.0%10)(0

==> xxP

−

≥=>σ

µ0

0)(

xZpxxP

10. Train

−≥=

10

12010.0 0

xZp

−=

10

1202816.1 0

x

X0 = 132.816

minutes

Thank You for Your Attention

• To download this slide (PPT & PDF)visitwww.pairach.com/teaching

• Any further question?Please email tome@pairach.com