INTRODUCTION TO BUSINESS STATISTICS SOLUTIONS – NOVEMBER 2011
SECTION A
Question One
(a)
Either Binomial distribution, or Poisson A1
Conditions – there is a fixed number of trials, A1
-
(b)
there are two possible outcomes called success and failure. A1
Let X be the number of vehicles serviceable per day. Then X ~ Poiss (  3) . M1
Then (i)
(ii)
P x  0  
 0e  
x!
0
P(at most 3)= Px  3  P( x  0 or 1or 2 or 3) M1
=
(c)
3
 3 0e!  0.04979 , M1, A1
3 0 e 3
0!
3
 3 1e!  3
1

  101  0.1 and ample proportion is p 
^
Then the 98% CI is p 1.64 
0.13  1.64 
 (1 )
n
0.1(10.1)
200
26
200
e 3
2!
2
3
 3 3e!  0.64723 , M2, A1
3
 0.13 , M1
^
 p  p 1.64 
 (1 )
 p  0.13  1.64 
n
, M1 (finding 1.64)
0.1(10.1)
200
, M1
i.e. 0.095  p  0.165 , M1, A1
Question Two
(a)
(i)
Independents events are events in which the occurrence of one event
does not affect the occurrence of the other event. A1
Example: Occurrence of a head of the first and occurrence of a tail on
the second toss. A2
(ii)
Mutually exclusive events are events that have no common elements. A1
Example: Drawing a red card and drawing a black heart when one card
is randomly selected from a pack of 52 cards. A2
(b)
(i)
Let A, B and B be events that entries were made by A, B and C
respectively.
Let E be the event an entry has errors.
Since total number of entries made = 3000 + 2500+4500 = 10000, M1
then
P(A)=0.3, P(B)=0.25 and P(C)=0.45. M1
Tree diagram (or contingency table): M2
E
0.01
A
0.99
E'
0.3
E
0.012
0.25
B
0.998
E'
0.45
C
0.02
E
0.98
E'
P(entry has errors) = P(E) = P( A  E or B  E or C  E ) , M1
= 0.3  0.01  0.25  0.012  0.45  0.02  0.015 , M1, A1
(i)
If found to have errors, probability it was entered by staff B
= P( B | E )  P (PB(E )E ) , M2
0.012
 0.2 , M1, A1
= 0.250.015
Question Three
(a)
(i)
Importance of index numbers:
- measuring relative changes prices or quantities of items
- may be used in salary negotiations
- used in calculating the real values of commodities, A3
(b)
(i)
Price indices:
11
100  110 , A1
= 10
Product A:
(ii)
Product B:
25
23
100  108.7 , A1
Product C:
17
17
100  100 , A1
Product D:
20
19
100  105.3 , A1
Laspeyres quantity index =

p0 q1
p0 q0
 100
Sum
p 0 q1
p0 q0
p1 q1
240
1173
1428
200
1265
1071
264
1275
1428
646
3487
532
3068
680
3647
M1
M1
Hence Laspeyres quantity index =
(iii)
Paasche price index =

p1q1
p0 q1
M1 (for products), M1 (for summing)
3487
3068  100
 100 
3647
3068
 113.7 ,
M1, A1
 100  104.6 t, M1, A1
Question Four
(a)
Linear constraints represent limited resources under which optimality is to be
achieved. They give a limit to how much resources may be used. A2
(b)
Let x11 , x12 , x13 be number of radios shipped from Kajiso to Jenda, Lizulu and
Songani respectively.
Similarly, x 21 , x 22 , x 23 be number of radios shipped from Oleriwa to Jenda, Lizulu
and Songani respectively.
And x31 , x32 , x33 be number of radios shipped from to Jenda, Lizulu and
Songani respectively. M1
Hence the LP model is:
Minimise Z  6 x11  4 x12  x13  x21  2 x22  5 x23  3x31  8 x32  12 x33 , A1
subject t: x11  x12  x13  300
x21  x22  x213  200
x31  x32  x33  200
x11  x21  x31  250
x12  x22  x32  200
x13  x23  x33  250 , A5
where xij  0 , for all 1  i, j  3 .
(c)
Accounting Rate of Return (ARR) =
(i)
Averageannualreturn
Initialcos t of investment
Investment A: Average annual return = K
Hence ARR=
280, 000
1, 000, 000
(ii)
220, 000
1, 000, 000
( 200 300 400 400100) 000
5
 K 280,000 , M1
100  28% , A1
Investment B: Average annual return = K
Hence ARR=
100 :
(125 275 0 500 200) 000
5
 K 220,000 , M1
100  22% , A1
Recommend Investment A on account of ARR being higher than in the
second investment. A2
SECTION B
Question Five
(a)
(i)
Let X be the number of girls. Then X: x = 0, 1, 2, 3, M1
 3  1   1 
 0  2   2 
0
3
Now Px  0       
 3  1 
1
, Px  1    
8
 1  2 
1
2
3
1
  
8
2
2
1
3
0
 3  1   1  3
 3  1   1 
1
and
Px  2       
Px  3        , M2
8
 2  2   2  8
 3  2   2 
Hence probability distribution: M1, A1
X: x
P(X=x)
(ii) (I)
(II)
0
1
2
3
1
8
3
8
3
8
1
8
EX    xP( X  x)  0  18  1 83  2  83  3  18  1.5 , M1, A1
 
 
Now EX    x P( X  x)  0   1   2 
Then VarX   Ex  1.5  3  2.25  0.75 , A1
VarX   E x 2  EX   E x 2  1.5 2 , M1
2
2
2
2
(b)
2
1
8
2
2
Calculating expected values for each alternative:
Low tender: 0.3 10  0.3 15  0.4 16  13.9 , M1
Medium tender: 0.3  5  0.3  20  0.4 10  11.5 , M1
High tender: 0.3  18  0.3  10  0.4  (5)  6.4 , M1
3
8
2
3
8
 32  18  3 , M2
The company should submit the low tender because it gives the highest
expected profit. M1, A1
(c)
Let X be the numbers of defects. Then X ~ Bin (n  12, p  0.1) , M1
Then P(at least 4 defects)
= P( X  4)  1  P( X  3) , M1
= 1  P( x  0)  P( x  1)  P( x  2)  P( x  3)
= 1  12 0.10 1  0.112  12 0.11 1  0.111  12 0.12 1  0.110  12 0.13 1  0.19   0.02564 ,M1, A1
 
 0 
1
 
2
 
3
 

Question Six
(a)
A, B are two events. Let number of events be labeled a, b and c as shown.
A
B
b
a
c
Then a+ b = 25, b+ c =15. M1
But n(A and B) =10, then b = 10. Hence a = 15, c = 5 and so a+b+c=15+10+5=30,
M1
(b)
(i)
P(A|B)=
P ( A B )
P(B)
2
 1 533 00  10
15  3 , M1, A1
(ii)
P(B|A)=
P ( B  A)
P ( A)

A 95% CI is x  1.96 
10

n
10
30
25
30
2
 10
25  5 , M1, A1
   x  1.96 
So the width is x  1.96 

n
 ( x  1.96 
Since width is 20,000, then 2 1.96 
21.96
20000
(c)

n

n
, M1

)  2  1.96 
n

n
, M1
 20000 , M1
1.96
9660000
   21.20000
  138 , M1, A1
 n or n   220000
2
2
Let x c and x v be mean time required for form completing and verification
processes respectively. Then

 

P x c  x v  P x c  x v  0 , M1
(i)


But x c  x v ~ N c  v , ncc  nvv , M1
2
2
where    c   v  20  15  5 and
 

Thus P x c  x v  0  P

xc  xv 

 
 0  P z 
 c2
nc
 nvv  1230.5  1040.5  7.96458 , M1
05
7.96458


2
2
2
  P( z  1.77)  0.0384 , M1, A1

P x c  x v  43 . But x c  x v ~ N  c   v , ncc  nvv , M1,
(ii)
where    c   v  20  15  35 and
2
 c2
nc
2
 nvv  1230.5  1040.5  7.96458 , M1
2
2
2
Thus

 
P x c  x v  43  P
xc  xv 


43 

  Pz 
4335
7.96458
  P( z  2.83)  0.0023 , M1, A1
Question Seven
(a)
Seasonal variation are short term fluctuations due to seasons or period of the
year while cyclic variation are longer term fluctuations lasting between four and
seven years usually caused by economic factors. A2
(b)
(i)
Graphing the data:
Labeled axes: M2, Plotted and joining points: M1, A1
F
r
a
u
d
90
80
70
60
50
c
a
s
e
s
40
30
20
10
0
0
5
10
Time points
15
20
(ii)
Trend values:
Upper group: 84, 53, 60, 75, 81, 57, 51, 73: x u  84538.... 73  66.75 , M1, A1
Lower group: 69, 37, 40, 77, 73, 46, 39, 63: x l 
6937.... 63
8
 55.5 , M1, A1
Trend line plotted: M1
Trend values read off (approximate values): M1, A1
Year
2007
2008
2009
2010
(ii)
Quarter 1
67
62
57
53
Quarter 2
65
61
56
52
Quarter 3
64
60
55
51
Quarter 4
63
58
54
50
Seasonal factors:
Table of y/t: M1
2007
2008
2009
2010
Year
Quarter 1
1.25
1.31
1.21
1.38
Quarter 2
0.82
0.93
0.66
0.88
Quarter 3
0.94
0.85
0.73
0.76
Quarter 4
1.19
1.26
1.43
1.26
Total
Average
Adjusted s. factors
5.15
1.2875
1.222
3.29
0.8225
0.781
3.28
0.82
0.778
5.14, M1
1.285, M1
1.219, A1
Sum of average = 4.215
Correction factor =
(iii)
period
total average
4
 4.215
 0.94899 ,M1
Estimated trend for 2011 Qrtr 1 = 49, M1
Hence forecast 49 1.222  59.878 , approximately 60 cases. A1
Question eight
(a)
(b)
Four major steps:
(i)
Formulating the null and alternative hypotheses
(ii)
Choosing and applying an appropriate test statistic
(iii)
Using an appropriate significance level to come up with a decision rule
(iv)
Making a conclusion based on parts (i) and (iii). A4
Hypotheses: H 0 :   15,000 , H1 :   15,000 , M2
Test Statistic: Since n = 45 is large, use the normal distribution.
Hence z 
x

15000
 14300
 -2.3479 , M1
2000
n
45
Decision Rule: At 5% level of significance, reject H 0 if z > 1.96 or z < -1.96. M1
Conclusion: Since z = -2.3479 < -1.96, H 0 has to be rejected i.e. there is significant
evidence that to suggest that the average salary is different from K15,000. M1,
A1
(c)
(i)
Type I error is committed when H 0 is rejected when true while type II error
is an error that is made when H 0 is accepted when false. A2
(ii)(I)
Hypotheses: H 0 : Difference in opinion is independent of age
H 1 : Difference in opinion depends on age, M1
Test Statistic:
2  
O  E 2
E

(100121) 2
121
184)
10)
 ( 200184
 ....  (1510
 (1588)  24.6985 , M3
2
Expected frequencies (in brackets) calculated: M3
2
2
Age in years
Opinion
Very unfair
Fair
Don’t Care
No opinion
20 but less than 30
100(121)
200(184)
50(42)
30(33)
30 but less than 40
300(270)
400(412)
80(93)
70(74)
20(29)
40(44)
15(10)
15(8)
40 and older
Decision Rule: At 5%, reject H 0 if  2   02.[(0541)( 31)6 ]  12.6 , M1
Conclusion: Since  2  24.6985  12.6 , reject H 0 i.e. difference in opinion
is not independent of age. A1
(II)
Type I may have been made. A1