# QM AMA2101

Topics: Mathematics, Customer, Customer service Pages: 7 (263 words) Published: November 21, 2013
AMA2101 Homework 2010/2011 Semester 2

Department of Applied Mathematics
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