# QM AMA2101

Department of Applied Mathematics

AMA2101 Quantitative Methods for Business

Homework 2010/2011 Semester 2 Suggested outline solution

Q1. (a)

Class Mark (x) Frequency (f)

Amount less than ($)

Cumulative frequency

25.5

75.5

0

9

22

27

100.5

150.5

24

46

225.5

300.5

16

8

200.5

250.5

73

89

400.5

x=

0.5

50.5

125.5

175.5

∑f

9

15

3

350.5

450.5

97

100

= 100

∑ fx = 16075 ∑ fx

2

= 3286675

16075

= $160.75

100

median = 150.5 +

50 − 46

× ( 200.5 − 150.5 ) = $157.9074

27

100 ( 3286675 ) − 160752

s=

= $84.2446

100 (100 − 1)

CV =

82.446

× 100% = 52.41%

160.75

(b) D2 = 50.5 +

20 − 9

(100.5 − 50.5 ) = $87.1667

15

(c) Estimated proportion of customers spent more than $180

200.5 − 180

ˆ

× 27 + 16 + 8 + 3 100 = 0.3807

p =

200.5 − 150.5

(d)

A 95% confidence interval for the proportion of customers will be invited 0.3807 ± 1.96

0.3807 (1 − 0.3807 )

100

, i.e. 0.2855 < p < 0.4759

1

AMA2101 Homework 2010/2011 Semester 2

Q2. (a) (i)

Pr(selected 2 Mathematics subjects)

6 14

2 3

= = 0.3522

20

5

(ii) X – number of students out of 3 will select 2 Mathematics subjects

X ~ B ( 3, 0.3522 )

3

2

Pr ( X = 1) = 0.35221 (1 − 0.3522 ) = 0.4434

1

(b)

A – user satisfied with new features

B – user rated the price as reasonable

Pr ( B ) = 0.86 Pr ( A B ) = 0.9

Pr ( A ) = 0.8

(i) Pr ( A ∩ B ) = 1 − ( 0.8 + 0.86 − 0.86 × 0.9 ) = 0.114

(

)

(ii) Pr B A = 1 −

0.86 × 0.9

= 0.0325

0.8

(c)

A – category A customer

B – category B customer

C – category C customer

D – monthly expenditure of $1001 - $2000 on entertainment

Pr ( A ) = 0.25 Pr ( B ) = 0.45 Pr ( C ) = 0.3

Pr ( D A ) = 0.3

(

)

Pr C D = 1 −

Pr ( D B ) = 0.4

Pr ( D C ) = 0.48

0.3 × 0.48

= 0.6391

0.25 × 0.3 + 0.45 × 0.4 + 0.3 × 0.48

Q3. (a) X – service time of bank teller for individual customers in minutes X~

( 3.5, 0.6 )

2

(i) Pr ( X > 5 ) = Pr ( Z > 2.5 ) = 0.00621

2

AMA2101 Homework 2010/2011 Semester 2

(ii) Let k be the maximum service time required

Pr ( X < k ) = 0.9 ⇒

k − 3.5

= 1.282 ⇒ k = 4.27 minutes

0.6

(iii) Pr ( X < 2.3) = Pr ( Z < −2 ) = 0.0228

Y – number of customers have service time less than 2.3 minutes out of 10 Y ~ B(10, 0.0228)

10

0

10

Pr (Y ≥ 3) = 1 − ( 0.0228 ) (1 − 0.0228 )

0

10

10

1

9

2

8

− ( 0.0228 ) (1 − 0.0228 ) − ( 0.0228 ) (1 − 0.0228 ) 1

2

= 0.001261

(b) X – number of urgent requests out of 5

X ~ B(5, 0.8)

5

5

4

1

5

0

Pr ( X ≥ 4 ) = ( 0.8 ) (1 − 0.8 ) + ( 0.8 ) (1 − 0.8 ) = 0.73728 4

5

Y – number of days with at least 4 urgent requests out of 100 Y ~ B(100, 0.73728)

Since n>30, np>5, nq>5 and 0.1

Please join StudyMode to read the full document