Free Essay

# Qm Ama2101

Submitted By teamboy
Words 557
Pages 3
AMA2101 Homework 2010/2011 Semester 2

Department of Applied Mathematics AMA2101 Quantitative Methods for Business Homework 2010/2011 Semester 2 Suggested outline solution Q1. (a) Class Mark (x) Frequency (f) 25.5 75.5 125.5 175.5 225.5 300.5 400.5 9 15 22 27 16 8 3

Amount less than (\$) 0.5 50.5 100.5 150.5 200.5 250.5 350.5 450.5

Cumulative frequency 0 9 24 46 73 89 97 100

∑f x= = 100

∑ fx = 16075 ∑ fx

2

= 3286675

16075 = \$160.75 100 50 − 46 × ( 200.5 − 150.5 ) = \$157.9074 27

median = 150.5 +

100 ( 3286675 ) − 160752 s= = \$84.2446 100 (100 − 1) CV = 82.446 × 100% = 52.41% 160.75 20 − 9 (100.5 − 50.5 ) = \$87.1667 15

(b) D2 = 50.5 +

(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 ( A ) = 0.8

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

(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