Miscellaneous Examples (with Solutions)
1.
(a)
Class marks, x
215
225
235
245
255
265
275
285
Mean,
x
f
2
5
4
8
10
5
4
2
40
fx
430
1125
940
1960
2550
1325
1100
570
10000
fx2
92450
253125
220900
480200
650250
351125
302500
162450
2513000
 fx  10000  $250
 f 40
Standard deviation,
S

n( fx 2 )  ( fx) 2
n( n  1)
40(251300)  (10, 000) 2
40(39)
 $18.26
(b)
Median  250 
20  19
10
10
 $251
(c)
Mean = $250
Standard deviation = $18.26
Median = $251
Pearson’s 2nd coefficient of skewness

3(mean  median) 3(250  251)

 0.164
Std. deviation
18.26
 the distribution is nearly symmetrical.
Miscellaneous Examples (with Solutions)
(d)
Percentage of workers whose daily income is at least $242 but less than $280
8
 10  5  4
 10
100%
40
25.4

100%  63.5%
40
8
(e)
S
 100%
x
18.26
CV (QMB) 
 100%  7.3%
250
400
CV (rival firm) 
 100%  5.7%
7000
Coefficient of Variation, CV 
CV (QMB)  CV (rival firm)
Therefore, QMB workers’ income is more variable.
2.
(a)
Frequency Table:
-5 - under 0
0 - under 5
5 - under 10
10 - under 15
15 - under 20
20 - under 25
25 - under 30
30 - under 35
Total
f
x
2
2
4
8
11
13
6
4
–2.5
2.5
7.5
12.5
17.5
22.5
27.5
32.5
50
Miscellaneous Examples (with Solutions)
Cumulative Frequency Table:
Under
-5
Cumulative frequency
0
0
5
10
15
20
25
30
35
2
4
8
16
27
40
46
50
 fx  910  fx 2  20, 212.5
mean 
910
 18.20 ($ million)
50
(50 / 2)  16
5
11
 19.09 ($ million)
median  15 
 20.212.5 9102 
standard deviation  


50  49 
 49
 8.63 ($ million)
Miscellaneous Examples (with Solutions)
(b) Cumulative frequency curve
Cumulative Frequency Curve
Cumulative total number of companies
60
50
40
30
20
10
0
-5
0
5
10
15
20
25
30
35
Annual profit ($ million)
(c)
Profit exceeded by 30% of the companies is equivalent to profit that is less than by
70% of the companies
= 23.1 ($ million)
(a)
Histogram
450
400
350
Number of customers
3.
300
250
200
150
100
50
0
0
5
10
15
Times per annum
20
25
30
Miscellaneous Examples (with Solutions)
(b)
 fx  26026,  fx2  394058,  f  2000
X
 fx 26026

 13.013
f
2000
Mode  11.5 
403  337
(3)  11.5  2.6053  14.1053
(403  337)  (403  393)
Median  11.5 
Q1  8.5 
1000  774
(3)  11.5  1.6824  13.1824
403
500  1177
(3)  8.5  .5608  9.0608
337
Q3  15.5 
1500  1177
(3)  15.5  2.4656  17.9656
393
1
 394058
26026 2  2
S

2000(1999) 
 1999
1
 [27.7047] 2  5.2635
Mean  Mode 13.013  14.1053

 0.2075
S
5.2635
3(Mean  Median) 3(13.013  13.1824)
or SK 2 

 0.0966
S
5.2635
SK1 
C.V . 
(c)
S
5.2635
100% 
100%  40.4480
13.013
x
The distribution of customer purchasing behaviour is slightly skewed to the left.
Miscellaneous Examples (with Solutions)
4.
Let A be the event of the award for design
Let B be the event of the award for material
Pr( A)  0.28,
Pr( B )  0.13,
Pr( A  B )  0.36
Pr( A  B )  Pr( A)  Pr( B )  Pr( A  B )
 0.28  0.13  0.36
 0.05
5.
Let G be the event of good loan
Let B be the event of bad loan
Let L be the event of long-term loan
Let S be the event of short-term loan
Pr(G )  0.8,
Pr( B)  0.2
Pr( L | G )  0.7, Pr( S | G)  0.3
Pr( L | B)  0.2, Pr( S | B)  0.8
Pr(G )  Pr( L | G )
Pr( L)
Pr(G )  Pr( L | G )

Pr(G )  Pr( L | G )  Pr( B)  Pr( L | B)
0.8  0.7

0.8  0.7  0.2  0.2
56

60
14

15
 0.933
Pr(G | L) 
Miscellaneous Examples (with Solutions)
6.
There are only 3 cases for at least one of each course,
i.e.
Management
2
1
1
Accountancy
1
2
1
Marketing
1
1
2
Therefore, the required probability
 6  7  3   6  7  3   6  7  3 
             
2 1 1
1 2 1
1 1 2
            
16 
 
4 
 0.45
7.
(a)
Let X be random variable of the sales volume in units,
X ~ N (10000, 20002 )
Pr(7000  X  13000)
13000  10000 
 7000  10000
 Pr 
Z

2000
2000


 Pr(1.5  Z  1.5)
 1  2  0.0668
 0.8664
(b)
Let y be the sales volume for break even,
y (20  16)  30000
30000
4
 7500
y
Miscellaneous Examples (with Solutions)
Pr(at least break even)
 Pr( y  7500)
7500  10000 

 Pr  Z 

2000


 Pr( Z  1.25)
 1  0.1056
 0.8944
8.
P defective ball pen is among the 3 chosen for testing

9.
19
C2  1 C1 171

 0.15
1140
20 C3
P 1 defective ball pen is found among the 3 chosen for testing

15 8 C2  2 C1 5 9 C2  1 C1

 
20
20
10 C3
10 C3

15 56 5 36

 
 0.425.
20 120 20 120
10. Let X be the life of the randomly selected motor.
Then X ~ N(12 , 32),
 10  12 X   15  12 
P(10  X  15)  P 




3 
 3
 P(0.67  Z  1)
 1  0.2514  0.1587
 0.5899
11. Let A be the event of the defective part detected by 1st inspector
B be the event of the defective part detected by 2nd inspector
P(A) = 0.85, P(B) = 0.85, P(A  B) = 0.8
(a)
P ( B | A)  P ( B  A) / P ( A)
P( B)  P( A  B)
1  P ( A)
0.85  0.80

 0.333
1  0.85

Miscellaneous Examples (with Solutions)
(b)
P( A  B)  P( A)  P( B)  P( A  B)
 0.85  0.85  0.80
 0.90
(c)
1   P( A  B)  1  0.903
3
 0.271
12.
Let X be the random variable of the number of customer who turn up in the 55
X ~ b (55, 0.85)
Mean = np = 55(0.85)
= 46.75
  npq  55(0.85)(0.15)
 2.6481
Required probability:
Pr( X  50)
By Normal approximation, the required probability
50.5  46.75 

Pr( X  50.5)  Pr  Z 

2.6481 

 Pr( Z  1.42)
 1  0.0778
 0.9222
Miscellaneous Examples (with Solutions)
13.
Let the mean of the distribution of the number of errors per page be  and X be the
Poisson random variable. Then
Pr( X  0) 
e  
 0.14
0!
Hence  = 1.966 and
e1.966 (1.966)1
1!
 0.275
Pr( X  1) 
The required percentage is 27.5%.
14. (a) Let X be the random variable of the number of error per 2 pages.
  4 error per 2 pages, X ~ Po( )
e 4 4 x
x!
x 0
2
P( X  3)  
 40 41 42 
 e 4    
 0! 1! 2! 
 0.2381
(b)
P ( X  5)  1  P ( X  5)
e 4 4 x
 1 
x!
x 0
5

43 4 4 4 5 
 1  e 1  4  8  
 
3! 4! 5! 

 0.2149
4
15.
(a) Let H be the event of an employee who was in favour of the modified health care
plan
W be the event of an employee who was in favour of the proposal to change
the work schedule.
P( H )  0.42, P(W )  0.22, P(W | H )  0.34
Miscellaneous Examples (with Solutions)
P( H  W )  P( H )  P(W )  P( H  W )
 0.42  0.34  0.1428
(b) P( H  W )  P( H )  P(W )  P( H  W )
 0.42  0.22  0.1428  0.4972
(c) P( H | W ) 
P( H  W ) 0.1428

 0.6491
0.22
P(W )
16. (a) Let X be the rate of return (%)
X ~ N (12.2, 7.22 )
20  12.2 

P( X  20)  P  Z 

7.2 

 P( Z  1.08)  0.1401
(b)
17.
0  12.2 

P( X  0)  P  Z 

7.2 

 P( Z  1.69)  0.0455.
n1  n2  2.20,
pˆ1  0.36,
pˆ 2  0.30
Let P1 , P2 be the True proportion of response to single-sheet questionnaire and two-sheet
questionnaire respectively.
Then a 95% C.I. for P1  P2 is
( pˆ1  pˆ 2 )  1.96
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )

n1
n2
 (0.36  0.30)  1.96
0.36(0.64) 0.30(0.70)

220
220
 0.06  0.088  (0.028, 0.148)
Miscellaneous Examples (with Solutions)
18. Let P be the population proportion.
n  225, x  67
pˆ 
67
 0.298
225
 A 95% C.I. estimate for P is
 pˆ (1  pˆ ) 
pˆ  1.96 

n


 0.298(0.702) 
i.e. 0.298  1.96 

225


i.e. 0.298  0.060, i.e. (0.238,0.358)
19. H0 : Printing errors are independent of type size.
H1 : Printing errors are dependent on type size.
 = 0.05,   (2  1)(3  1)  2
2
 CR:  2   0.05
 5.991
Computation:
Pages with errors
Pages without errors
Total
A
23
(29.9)
241
(234.1)
264
Type Size
B
C
17
41
(22.7)
(28.4)
183
210
(177.3)
(222.6)
200
251
Total
81
634
715
6.92 5.72 12.62 6.92
5.72 12.62





29.9 22.7 28.4 234.1 177.3 222.6
 9.714
Test statistic,  2 
Conclusion: Reject H0.
Miscellaneous Examples (with Solutions)
20. (a)
Let 1 , 2 be the true mean absenteeism of recent ex-smokers and long-term exsmokers, respectively.
n1  34
x1  2.21 s1  2.21
n2  68 x2  1.47 s2  1.69
H 0 : 1   2 , i.e. 1   2  0
H1 : 1   2
  0.05
CR : Z  1.96 and
Test statistic, z 

Z  1.96
( x1  x2 )  ( 1   2 )
 s12 s22 
  
 n1 n2 
(2.21  1.47)  0
 2.212 1.692 



68 
 34
 1.717
Conclusion: Do not reject H0
(b) Level of significance () is the probability of committing type I error in a
significance test. In general,  is set equal to 0.05. However, if the consequence of
committing type I error in a significance test is serious in terms of monetary loss,
then a low value, for example, 0.01, should be used instead.
Miscellaneous Examples (with Solutions)
21. (a)
n  545 x  117
pˆ 
117
 0.2147
545
Let P be the population proportion of accountants finding cash flow the most
difficult estimates to derive.
H 0 : P  0.25; H1 : P  0.25
  0.05; CR : Z  1.645
Test statistic, z 
pˆ  p
0.2147  0.25

p (1  p )
0.25(0.75)
n
545
 1.903
Conclusion: Reject H0.
0.25(0.75)
545
 0.25  0.0305
(b) c  p  1.645
 0.2195
So probability of rejecting H0 when the true P = 0.28 is
0.2195  0.28 

P  pˆ  0.2195  P  Z 

0.28(0.72) 



545


 P{Z  3.145}
 0.00083.
22. (a)
n  8  x  192
 y  608
 x  4,902  xy  15, 032  y 2  47, 094

n  xy  ( x)( y ) 8(15, 032)  192(608)

n  x 2  ( x) 2
8(4,902)  192 2
3,520

 1.4966
2,352
b
Miscellaneous Examples (with Solutions)
a  y  bx  76  1.4966(24)  40.0816
 yˆ  40.0816  1.4966 x
(b)
yˆ  40.0816  1.4966(30)  84.9796
(c)
r

n  xy  ( x)( y )
[n  x  ( x) ][n  y  ( y ) ]
2
2
2
2

3520
2352(7088)
3520
 0.8621
4083.0107
This value of r shows a strong positive relationship between hours of study and
examination grade.
(d)
r 2  (0.8621)2  0.7432
This means hours of study can explain 74.32% of variation on examination grade in
the regression equation.
23. (a)
n  10 x  0.56 y  96.7
x 2  0.0367 xy  5.8
nxy  (x)(y )
n x 2  (  x ) 2
(10)(5.8)  (0.56)(96.7)

 72.060
(10)(0.0367)  0.56 2
b
a  y  bx 
96.7
0.56
 (72.060)(
)
10
10
 5.635
 yˆ  5.635  72.060 x
The slope of the linear regression line is 72.060 and this implies that the P/E ratio
of a company is expected to be increased by 0.72060 for each 0.01 increase in its
R/S ratio, i.e. for each 1 cent increase in research & development expenditure per
dollar of sales.
Miscellaneous Examples (with Solutions)
(b) If
x  0.060,
then
yˆ  5.635  (72.060)(0.060)
 9.96
Reliability of this estimate is low because n is equal to 10 only and r = 0.57
(i.e. only 0.32% of the variation in y is explained by the regression line).
24.
1  1.09 n 
5, 000, 000  1, 000, 000 

 0.09 
5, 000, 000
 0.09  1  1.09 n
1, 000, 000
(a)
1.09 n  0.55
 n
n 0.55
 6.94.
n 1.09
So six payments of $1,000,000 are required.
(b) Let x be the final payment.
Then
x
1  1.096 
5, 000, 000  1, 000, 000 

7

 0.09  1.09
x
i.e. 5, 000, 000  4, 485,918.6 
1.828039
 x  939,760.8  $
Miscellaneous Examples (with Solutions)
25. Let i be the rate of interest being charged.
Then
 1  (1  i ) 12 
10, 000  1, 000 

i


 1  (1  i ) 12 


i


9954
10258
i
0.03
0.025
So i should be
10258  10000
 (0.03  0.025)
10258  9954
258
 0.025 
 0.005
304
 0.029 (i.e. 2.9% p.a.)
0.025 
26.
3,850, 000  R a240 0.007083
where i =
0.085
 0.007083
12
3,850, 000
3,850, 000

240
115.2308432
1  (1.007083)
0.007083
 $33, 411.19
R
Miscellaneous Examples (with Solutions)
27.
(a)
2
B3  200, 000a2 0.0875  200, 000a5 0.095 v0.0875
1  (1.0875) 2 1  (1.095) 5

 200, 000 

(1.0875) 2 
0.095
 0.0875

 200, 000 1.765094464  3.246682023
 200, 000 5.011776487 
 1, 002,355.297
I 4  0.0875 B3  0.0875(1, 002,355.297)
 $87, 706.09
P8  B7  B8  200, 000a3 .095  200, 000a2 .095
(b)
1  (1.095) 3 1  (1.095) 2 
 200, 000 


.095
.095


 200, 000  0.761653851  $152,330.77
1
1
L
L
L
2
2
 (0.05) 
28. 1000 
a10 0.05
2 S10 0.04
L
1000
1
2a10 0.05

 0.025 
1

100
0.06475  0.025  0.04165
2 s10 0.04
1000
 $7, 610.35
0.1314
29. Trading profit ( )  TR  TC
 (3500Q  8Q 2 )  (Q 3  6Q 2  120Q  600)
Q 3  2Q 2  3380Q  600
d
 0, i.e. 3Q 2  4Q  3380  0
dQ
i.e. 3Q 2  4Q  3380  0
Miscellaneous Examples (with Solutions)
4  (42  4  3  (3380))
 Q
23
4  201.4

6
 32.9 or  34.2 (unacceptable)
d 2
 6Q  4  0 when Q  32.9
dQ 2
 trading profit will be maximized when Q = 32.9
30. TR  PQ  (6e0.04Q )Q
d
TR  0
dQ
6e0.04Q  Q6e0.04Q (0.04)  0
6e0.04Q (1  0.04Q)  0
1
 25
0.04
P  6e0.04(25)  2.21
Q 
d2
TR  6e 0.04Q (0.04)  0.24e 0.04Q  0.24Qe 0.04 Q (0.04)
2
dQ
 0.48e 0.04Q  0.0096Qe 0.04Q
 e 0.04Q (0.0096Q  0.48)
 0.088 when Q  25
 when Q = 25 and then P = 2.21, total revenue will be maximized.
31.
Total interest = ($50,000)(0.12) = $6000
Monthly repayment =
$50, 000  $6, 000
 $4666.67
12
The flat rate tends to understate the effective rate since the flat rate is a percentage of
the amount originally borrowed, however, the balance owing from time to time
reduces as repayments are made.
Miscellaneous Examples (with Solutions)
32.
1  (1  i ) 240 
2,800, 000  R 

i


10%
i
 0.0083
12
 R  $27021
33. (a)
TR  PQ  200Q  Q 3
dTR
 200  3Q 2
dQ
dTC
MC 
 8  2Q
dQ
MR 
To maximize profit, MC = MR
 200  3Q 2  8  2Q
 3Q 2  2Q  208  0
(2)  (2) 2  4(3)(208)
2(3)
 Q  8.67(impossible) or Q  8.
Q
 The price should be P  200  (8)2  136
(b)
dQ
1

dP
2Q
P
E 2
2Q
136
 1.0625
2(8) 2
The demand is elastic at the point of maximum profit.
 when P  136 and Q  8, E  