Business Mathematics and Statistics by Asim Kumar Manna (z-lib.org)

Solution of C.U. Question Paper–2017 (New Syllabus) A.7

or x – 6 = 0.65 (y – 7) or x – 6 = 0.65y – 4.55
or x = 0.65y – 4.55 + 6
or x = 0.65y + 1.45
Regression equation of y on x:

y – y = bxy (x – x)
or y – 7 = 0.45 (x – 6) or y – 7 = 0.45x – 2.70
or y = 0.45 × –2.70 + 7
or y = 0.45x + 4.3

or
If bxy = –0.4 and byx = –0.9 , find rxy
Solution: We know that, r = bxy × byx

= −0.4 × −0.9 = 0.36
= 0.6
Since, bxy and byx both are negative, therefore, r = –0.6.

Group – B

6. Answer the following questions:
(a) Find the Price Index Number by the method of arithmetic mean of
price relatives from the following.

Commodity Base Price Current Price
Wheat 5 7
Milk 8 10
Fish 25 32
Sugar 6 12

Solution: Calculation of Price Index Number by the method of arithmetic mean
of price relatives.

Commodity Base Price (p0) Current Price (pn) pn × 100
po

Wheat 5 7 140
Milk 8 10 125
Fish 25 32 128
Sugar 6 12 200
Total – – 593

A.8 Business Mathematics and Statistics

∑Price Index Number (Ion) =  pn × 100 
 po 
  593
=
N4

= 148.25

or

Find the general cost of living index of 2016 from the following table:

Class Food Clothing House Rent Fuel Miscellaneous
Group Index 620
Weight 30 575 325 255 280

20 25 15 10

Solution: Calculation of cost of living index

Class Group Index (I) Weight (W) I.W.
Food 620 30 18600
Clothing 575 20 11500
House Rent 325 25 8125
Fuel 255 15 3825
Miscellaneous 280 10 2800
Total 100 44,850

Cost of living index = ΣIW = 44850 = 448.5
ΣW 100

(b) From the following data, find an appropriate regression equation and
predict the value of y for x = 2.5:

x 1 2 3 4 5 7 10
y 2 2 5 4 6 9 12

Solution: In order to predict the value of y we have to find out the regression
equation of y on x.

Calculation of regression equation of y on x

x y x2 xy
1212
2244
3 5 9 15
4 4 16 16
5 6 25 30
7 9 49 63
10 12 100 120
32 40 204 250

Solution of C.U. Question Paper–2017 (New Syllabus) A.9

x = Σx = 32 = 4.57 (Here n = 7)
n7

y = Σy = 40 = 5.71
n 7

Σxy − Σx . Σy 250 − 32 . 40
n − n n 7 77
byx = =
Σx2  Σx 2  2
n n 204 − 32
7 7

= 35.71 − 4.57 × 5.71 35.71 − 26.095
29.14 − 20.885
29.14 − (4.57)2 =

= 9.615 = 1.165 .
8.255

Therefore, Regression equation of y on x

y − y = byx (x − x )

or y – 5.71 = 1.165 (x – 4.57)
or y – 5.71 = 1.165x – 5.32
or y = 1.165x – 5.32 + 5.71
or y = 1.165x + 0.39

When x = 2.5
Then, y =1.165 × 2.5 + 0.39

= 2.9125 + 0.39 = 3.3025

or

The coefficient of rank correlation of the marks obtained by 10 students in
Mathematics and Statistics was found to be 0.8. It was then detected that the
difference in ranks in the two subjects for one particular student was wrongly
taken to be in place of 7. What should be the correct rank correction coeffi-
cient?

Solution: We know that coefficient of rank correlation

R = 1 − 6Σd 2 [d = difference in ranks and n = no. of students]
n3 − n

Given that R = 0.8, n = 10

Putting these values we get

0.8 = 1 − 6Σd 2
103 −
10

A.10 Business Mathematics and Statistics

or 6Σd2 = 1 − 0.8
1000 − 10

or 6Σd2 = 0.2 or 6Σd2 = 198
990

or Σd2 = 33
Correct Σd2 = 33 − 32 + 72 = 33 − 9 + 49

= 33 + 40 = 73
Therefore, correct rank correlation coefficient

R = 1 − 6 × 73 = 1 − 438
1000 − 10 990

= 1 – 0.44 = 0.56.

(c) Compute the seasonal index quarterly average for the following data:

Year 1st quarter 2nd quarter 3rd quarter 4th quarter
2010 75 60 54 59
2011 86 65 63 80
2012 90 72 66 85
2013 100 78 72 93

Solution: Calculation of seasonal index

Year 1st 2nd 3rd 4th Total
quarter quarter quarter quarter
2010 –
75 60 54 59 –
2011 86 65 63 80 –
90 72 66 85 –
2012 100 78 72 93 1198
351 275 255 317 299.5
2013 87.75 68.75 63.75 79.25 74.875
– – – – 400.00
Total 117.2 91.82 85.14 105.84

Average

Grand Average

Seasonal Index

 Average × 
 Grand Average 100

Solution of C.U. Question Paper–2017 (New Syllabus) A.11

Group – C

7. (a) Construct 5-yearly moving averages of the number of students
studying in a college shown below:

Year 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015
No. of 332 317 357 392 402 405 410 427 405 431 467 483
students

Solution: Calculation of 5-yearly moving averages

Year No. of students 5-year moving total 5-year moving average
2004 332 – –
2005 317 – –
2006 357
2007 392 1800 360.0
2008 402 1873 374.6
2009 405 1966 393.2
2010 410 2036 407.2
2011 427 2049 409.8
2012 405 2078 415.6
2013 431 2140 428.0
2014 467 2213 442.6
2015 483
– –
– –

or
Fit a straight line trend equation by the method of least squares and estimate
the trend values:

Year 2008 2009 2010 2011 2012 2013 2014 2015 2016
Values 117 123 130 139 143 151 158 163 170

Hence estimate the production for the year 2017.

Solution: Let y = a + bt be the equation of the straight line trend. Here the
number of years is odd, so mid-year 2012 is taken as origin and one year as unit.

Fitting a straight line trend

Year Values (y) t = year – 2012 t2 y.t
2008 117 –4 16 –468
2009 123 –3 9 –369
2010 130 –2 4 –260
2011 139 –1 1 –139

A.12 Business Mathematics and Statistics

Year Values (y) t = year – 2012 t2 y.t
2012 143 0 0 0
2013 151 1 1 151
2014 158 2 4 316
2015 163 3 9 489
2016 170 4 16 680
Total Σt2 = 60 Σyt = 400
Σy = 1294 Σt = 0

Normal equations are Σy = na + bΣt … (i)
and Σyt = aΣt + bΣt2 … (ii)

From (i), 1294 = 9a + b.0 or 9a = 1294 or a = 143.78
From (ii), 400 = a.0 + b.60 or 60b = 400 or b = 6.67
Therefore, the trend equation is

y = 143.78 + 6.67.t with origin at 2012 and t unit = 1 year.
The value of t for the year 2017 will be 5. Hence the estimate of production
for the year 2017 is

y = 143.78 + 6.67 × 5 = 143.78 + 33.35 = 177.13
8. (b) Using the following data, verify that Paasche’s formula does not

satisfy Factor Reversal Test:

Commodity Base year Current year

X Price Quantity Price Quantity
Y
Z 4 10 6 15

6 15 4 20

85 10 4

Solution: Let p0, pn denote prices in the base year and current year and q0, qn the
quantities in the base year and current year respectively.

Commodity p0 q0 pn qn p0 q0 pn q0 p0 qn pn qn
X 4 10 6 15 40 60 60 90
Y 6 15 4 20 90 60 120 80
Z 8 5 10 4 40 50 32 40
170 170 212 210

Σp0q0 = 170,Σpnq0 = 170,Σp0qn = 212,Σpnqn = 210

Omitting the factor 100 from each index,

P0n = Σpnqn = 210 = 0.99
Σp0qn 212

Qon = Σpnqn = 210 = 1.235
Σpoqn 170

Solution of C.U. Question Paper–2017 (New Syllabus) A.13

∑∑Value ratio = pnqn = 210 = 1.235
p0q0 170

∑∑Clearly, P0n × Q0n ≠ pnqn
p0q0

That is the product of the price index and quantity index is not equal to the
value ratio of the current period and base period.

Therefore, Paasche’s index formula does not satisfy Factor Reversal Test.

PART–III (HONS.) – 2017

Group – C

3. Answer the following equations:
(a) Four men in a company of 10 employees are engineers. If 2 men are
selected at random, then find the probability that exactly one of them
will be an engineer.

Solution: Out of 10 employees four are engineers.

Therefore, other than engineers = 10 – 4 = 6

2 men are selected at random

So, total number of cases = 10C = 10 × 9 = 45
2 2×1

Out of 2 men selected at random, one will be an engineer, so, second man will

be other than engineer.

Therefore, the number of favourable cases = 6C1 × 4C1

= 6 × 4 = 24

Therefore, the probability that exactly one will be an engineer = 24 8
=
45 15

or

A problem of Mathematics is given to three students A, B and C whose chances

of solving it are 1 , 1 , 1 respectively. What is the chance that the problem will be
234

solved?

Solution: Let X, Y and Z denote the respective event that A, B and C can solve

the given problem.

Then we have P(X) = 1 , P(Y) = 1 and P(Z) = 1
23 4

Therefore, P(Xc ) = 1 − P (X ) = 1 − 1 = 1

22

P(Y c ) = 1 − P (Y ) = 1 − 1 = 2

33

A.14 Business Mathematics and Statistics

P(Zc ) = 1 − P(Z) = 1 − 1 = 3

44

The probability that the problem will not be solved

( ) ( ) ( ) ( )= P Xc ∩ Y c ∩ Zc = P Xc .P Y c .P Zc (As the events are independent)

= 1 × 2 3 = 1
×

234 4

Therefore, the probability that the problem will be solved =

( )P (X ∪ Y ∪ Z ) = 1 − P Xc ∩ Y c ∩ Zc

= 1− 1 = 3 .
44

FOR ALL OTHER CATEGORIES OF CANDIDATES

MODULE – I

Group – A

1. Answer the following questions:
(a) If nP5 : nP3 = 2 : 1, find the value of n.

Solution: nP5 : nP3 = 2 : 1

or (n n! : (n n! = 2 : 1

− 5)! − 3)!

or n! (n − 3)(n − 4)(n − 5)! 2
(n − 5)! × =
n! 1

or (n − 3)(n − 4) = 2.1 = (5 − 3)(5 − 4)

or n = 5
or

If nCx = 56 and nPx = 336, find n and x.

Solution: nCx = 56

or n! x)! = 56 … (i)

x!(n −

nPx = 336

or n! = 336 … (ii)

(n − x)!

Dividing equation (ii) by equation (i), we get

Solution of C.U. Question Paper–2017 (New Syllabus) A.15

(n n! ÷ n! x)! = 336 ÷ 56

− x)! x!(n −

or n! × x!(n − x)! = 336
56
(n − x)! n!

or x! = 6 = 3.2.1 = 3!

Therefore, x = 3

Putting the value of ‘x’ in equation (ii) we get

n! = 336 or n(n − 1)(n − 2)(n − 3)! = 336
−x
(n )! (n − 3)!

or n(n − 1)(n − 2) = 8.7.6

= 8(8 − 1)(8 − 2)

Therefore, n = 8

(b) Find the number of different odd numbers of 5 digits that can be formed
with the digits 1, 2, 3, 4, 5, 6 without repetition.

Solution: A number is odd if its unit place is occupied by an odd number. In this
case there are 3 odd numbers namely 1, 3, 5. Thus, the unit place can be filled in
3P1 ways = 3 ways.

After filling the unit place, the remaining 4 places can be filled by the remaining
5 numbers in 5P4 ways, i.e. 5 × 4 × 3 × 2 × 1 = 120 ways.

Therefore, required odd numbers = 3 × 120 = 360.

(c) Find the power set of {2, 3, 5}

Solution: Let A = {2, 3, 5}
Subsets of set A = {2}, {3}, {5}, {2, 3}, {3, 5}, {2, 5}, {2, 3, 5} F
Therefore, power set of set A
P(A) = [{2}, {3}, {5}, {2, 3}, {3, 5}, {2, 5}, {2, 3, 5}, F]
or
Given A = {1, 2, 3} and B = {4, 6, 8}. Find A × B

Solution: A = {1, 2, 3} and B = {4, 6, 8}
Therefore, A × B = {(1, 4), (1, 6), (1, 8), (2, 4), (2, 6), (2, 8), (3, 4), (3, 6), (3, 8)}
(e) If A = {x : x is a natural number and x ≤ 6},
B = {x : x is the natural number and 3 ≤ x ≤ 8},

Find A – B and A ∩ B.

Solution: A = {x : x is a natural number and x ≤ 6}
= {1, 2, 3, 4, 5, 6}

A.16 Business Mathematics and Statistics

B = {x : x is the natural number and 3 ≤ x ≤ 8}
= {3, 4, 5, 6, 7, 8}

Therefore, A – B = {1, 2}
A ∩ B = {3, 4, 5, 6}.

Group – B

2. Answer the following questions:
(a) From 6 bowlers, 2 wicket keepers and 8 batsmen; in how many ways
can a team of 11 players consisting of at least 4 bowlers, 1 wicket
keeper and at least 5 batsmen be formed?

Solution: The problem can be arranged as under.

Situation Bowlers (6) Wicket keepers (2) Batsmen (8) Total
(i) 4 1 6 11
(ii) 5 1 5 11

Therefore, total number of ways of selection

= 6C4 × 2C1 × 8C6 + 6C5 × 2C1 × 8C5

= 6×5 × 2 × 8×7 + 6 × 2 × 8×7×6
2 ×1 2 ×1 3 × 2 ×1

= 840 + 672 = 1512 ways.

or

A committee of 7 is to be chosen from 13 students of whom 6 are science
students and 7 are commerce students. In how many ways can the selection be
made so as to retain a majority in the committee for commerce students?

Solution: The problem can be arranged as under.

Situation Science students (6) Commerce students (7) Total
(i) 34 7
(ii) 25 7
(iii) 16 7

Therefore, total number of ways of selection

= 6C3 × 7C4 + 6C2 × 7C5 + 6C1 × 7C6

= 6×5×4 × 7×6×5 + 6×5 × 7×6 + 6 × 7
3 × 2 ×1 3 × 2 ×1 2 ×1 2 ×1

= 700 + 315 + 42 = 1057

(b) Prove that, log3  3 3 3..........∞  = 1

Solution of C.U. Question Paper–2017 (New Syllabus) A.17

Solution: Let x = 3 3 3..........¥ \ x2 = 3 3 3.....∞

[Squaring both sides]

or x2 = 3x or x2 – 3x = 0

or x (x – 3) = 0

or x – 3 = 0 (as x ≠ 0)

or x = 3

\ log3  3 3 3........∞  = log3 3 = 1

Or

If log  a + b  = 1 (log a + log b) , show that a + b = 7.
3 b a
2

Solution: log  a + b  = 1 .(log a + log b)
3 2

or log  a + b  = 1 log ab
3 2

or log  a + b  = 1 = log ab
3
log ab2

or a + b = ab
3

or a + b = 3 ab

or (a + b)2 = 9ab [Squaring both sides]

or a2 + b2 + 2ab = 9ab

or a2 + b2 = 7ab

or a2 b2 7ab [Dividing both sides by ab]
ab + ab = ab

or a + b = 7
b a

(c) If the coefficient of (r + 3)th term in the expansion of (1 + x)47 be the
same as the coefficient of (3r + 2)th term, find these two terms.

Solution:
(r + 3)th term in the expansion of (1 + x)47
tr+3 = t(r+2)+1

= 47Cr+2 ⋅ xr+2

A.18 Business Mathematics and Statistics

(3r + 2)th term in the expansion of (1 + x)47
t3r+2 = t(3r+1)+1 = 47C3r+1x3r+1

Therefore, the coefficients of (r + 3)th term and (3r + 2)th term are respectively
47Cr + 2 and 47C3r + 1

Now, 47Cr + 2 = 47C3r + 1 (according to the problem)
or (r + 2) + (3r + 1) = 47

or 4r + 3 = 47 or 4r = 44 or r = 11

(d) In a class of 100 students, 55 students read History, 41 students read
English and 25 students read both the subjects. Find the number of
students who study neither of the subjects.

Solution: Let H and E be the set of students who read History and English
respectively.

Then, n(H) = 55, n(E) = 41, n(H ∩ E) = 25 and N = 100
We are to calculate n(H′ ∩ E′)
We know, n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

= 55 + 41 – 25
= 96 – 25 = 71

Now, n(H′ ∩ E′) = n(H ∪ E)′ = 100 – n(H ∪ E)
= 100 – 71 = 29

or

For any three sets A, B, C prove that
A × (B ∩ C) = (A × B) ∩ (A × C)

Proof: Let (a, b) be an arbitrary element of A × (B ∩ C)
Therefore, (a, b) ∈ A × (A ∩ C)
⇒ a ∈ A and b ∈ (B ∩ C)
⇒ a ∈ A and {b ∈ B and b ∈ c}
⇒ {a ∈ A and b ∈ B} and {a ∈ A and b ∈ c}
⇒ (a, b) ∈ A × B and (a, b) ∈ A × C
⇒ (a, b) ∈ (A × B) ∩ (A × C)
Therefore, A × (B ∩ C) ⊆ (A × B) ∩ (A × C) … (i)
Again, let (c, d) be an arbitrary element of (A × B) ∩ (A × C)
Therefore, (c, d) ∈ (A × B) ∩ (A × C)
⇒ (c, d) ∈ (A × B) and (c, d) ∈ (A × C)
⇒ {C ∈ A and d ∈ B} and {C ∈ A and d ∈ c}
⇒ C ∈ A and {d ∈ B and d ∈ c}
⇒ C ∈ A and d ∈ (B ∩ C)
  ⇒ (c, d) ∈ A × (B ∩ C)

Solution of C.U. Question Paper–2017 (New Syllabus) A.19

Therefore, (A × B) ∩ (A × C) ⊆ A × (B ∩ C) … (ii)
From (i) and (ii) we can write

A × (B ∩ C) = (A × B) ∩ (A × C)

Group – C

3. (a) Find the present value of ` 10,000 due in 12 years at 6% p.a. compound
interest. [Given (1.06)12 = 2.012]

Solution: Here, A = ` 10,000, n = 12, i = 0.06
Present value (P) = A(1 + i)–n
= 10,000 . (1 + 0.06)–12
= 10,000 . (1.06)–12
= 10,000 = 10,000

(1.06)12 2.012

= 4970 (Approx.)

Hence, the present value = ` 4,970.

(b) A man wishes to buy a house valued at ` 50,00,000. He is prepared to

pay ` 20,00,000 now and the balance in 10 equal instalments. If the

interest is calculated at 8% p.a. compound annually, what should he

pay annually?

 1 = 
Given 0.4634
(1.08)10

Solution:
Cash down price = ` 20,00,000
Balance amount to be paid in 10 equal instalments = ` (50,00,000 – 20,00,000)

= ` 30,00,000.
Let ` A be the required amount of each instalment.
We know that the present value

P = A 1 − (1 + i )−n 
i 

Here, P = ` 30,00,000, i = 8% = 0.08, n = 10
Putting these values, we get

30, 00, 000 = A 1 − (1 + 0.08)−10 
0.08

or 2, 40,000 = A 1 − (1.08)−10 

or 2,40,000 = A(1 – 0.4634)

A.20 Business Mathematics and Statistics

or 2, 40,000 = A × 0.5366
or A = 2, 40,000 = ` 4,47,260.53

0.5366

MODULE – II

Group – A

5. Answer the following question:

(c) Karl Pearson’s coefficient of correlation between two variables x and y is

0.52, their covariance is 7.8. If the variance of x is 16, find the S.D. of y.

Solution: Given that, rxy = 0.52, cov ( x, y) = 7.8

= s 2 = 16 or sx = 4
x

We are to calculate the value of σy
cov(x, y)
we know that, rxy =
sx ×sy

or 0.52 = 7.8 or sy = 7.8 = 3.75 .
4×sy 2.08

(d) If the two regression coefficients are bxy = –0.4 and byx = –0.9, find the
value of correlation coefficient rxy.

Solution: We know that, r = bxy × byx

= −0.4 × −0.9 = 0.36

= 0.6

Since, bxy and byx are negative,
Therefore, r = –0.6

Group – B

6. Answer the following questions:
(b) Calculate the correlation coefficient of the following data;
x 63 60 67 70 61 69
y 61 65 64 63 63 68

Solution: Calculation of correlation coefficient

x y U=x– x V=x– x u2 v2 u.v
63 4 9 6
60 61 –2 –3 25 1 –5

65 –5 1

Solution of C.U. Question Paper–2017 (New Syllabus) A.21

x y U = x – x V = x – x u2 v2 u.v

67 64 2 0 4 0 0

70 63 5 –1 25 1 –5

61 63 –4 –1 16 1 4

69 68 4 4 16 16 16

390 384 0 0 90 28 16

x = Σx = 390 = 65
n6

y = Σy = 384 = 64
x 6

rxy = ruv = Σuv = 16
Σu2 Σv2 90 28

= 16 = 16 = 0.32
2520 50.2

or

You are given that variance of x = 36. The regression equations are
60x – 27y = 321 are 12x – 15y + 99 = 0. Find the correlation coefficient between
the variables; the average values of x and y; the S.D. of y.

Solution: Given that σx2 = 36 or σx = 6
Let 60x – 27y = 321 be the regression equation of x on y and 12x – 15y + 99 = 0

be the regression equation of y on x.
60x – 27y = 321

or 60x = 27y + 321

or x = 27 y + 321
60 60

\ 27 9
bxy = 60 = 20

Again, 12x – 15y + 99 = 0

or 15y = 12x + 99

or 12 99
y = 15 x + 15

Therefore,  byx = 12 = 4
15 5

Now, r2 = bxy × byx = 9 × 4 = 9
20 5 25

A.22 Business Mathematics and Statistics

or r= 9 = 3 = 0.6 <1
25 5

Since the value of r is less than unity, our assumption is correct.

Therefore, r = 0.6

60x – 27y = 321 … (i)

12x – 15y = –99 … (ii)

Multiplying equation (ii) by 5 and then subtracting from equation (i) we get

60x – 27y = 321

60x – 75y = – 495

–+ +

48y = 816

or y = 816 = 17
48

Putting the value of ‘y’ in equation (ii) we get
12x – 15 × 17 = –99

or 12x – 255 = –99 or 12x = 255 – 99 = 156
or x = 156 = 13

12

Hence, x = 13 and y = 17

We know that, byx = r × s y
s x

or 4 = 0.6 × sy or × 4.8 = s y
5 6 0.6

or σy = 8
Therefore, S.D. of y(σy) = 8

(e) Construct seasonal indices from the following time series data:

Year/quarter I II III IV

2014 90 75 87 70

2015 75 80 78 75

2016 80 75 75 72

Solution: Calculation of seasonal indices

Year/quarter I II III IV Total
2014
2015 90 75 87 70 –
2016
Total 75 80 78 75 –
Average
80 75 75 72 –

245 230 240 217 932

81.7 76.67 80 72.33 310.7

Year/quarter Solution of C.U. Question Paper–2017 (New Syllabus) A.23
Grand average
Seasonal indices I II III IV Total
– – – – 77.675
Average ´100 105.18 98.71 103 93.11 400
Grand Average

Group –C

7. (a) Fit a least squares trend line to the following data:

Year 2008 2009 2010 2011 2012 2013 2014
28
Average production per 20 22 21 24 25 23
month (’0000 tons)

Hence, find the average production per month in the year 2017.

Solution: Let y = a + bt be the equation of the straight line trend. Here the
number of years is odd, so mid year 2011 is taken as origin and one year as unit.

Fitting a straight line trend

Year Average production per t = year – 2011 t2 y.t
month (’0000 tons) (y)
2008 –3 9 –60
2009 20 –2 4 –44
22

2010 21 –1 1 –21
2011 24 0 0 0

2012 25 1 1 25
2013 23 2 4 46
2014 28 3 9 84
Total ∑y = 163 ∑t = 0 ∑t2 = 28 ∑yt = 30

Normal equations are:

    ∑y = na + b∑t … (i)

and ∑yt = a∑t + b∑t2 … (ii)

From (i), 163 = 7a + b. 0 or 7a = 163 or a = 23.28

From (ii), 30 = a. 0 + b. 28 or 28b = 30 or b = 1.07

Therefore, the trend equation is

y = 23.28 + 1.07t with origin at 2011 and t unit = 1 year.

The value of t for the year 2017 will be 6. Hence, the average production

(’0000 tons) per month for the year 2017

y = 23.28 + 1.07 × 6

= 23.28 + 6.42 = 29.7

or

A.24 Business Mathematics and Statistics

For the following series of observations, verify that the 4-year centred moving
averages are equivalent to a 5-year weighted moving average with weights 1, 2,
2, 2, 1 respectively:

Year 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016

Sale 2 6 1 5 3 7 2 6 4 8 3
(` ’0000)

Solution: Calculation of 4-year centred moving average

Year Sales (` ’0000) 4-year moving total 2-item moving 4-year moving
average
(not centered) total (centered)
–
2006 2 ––
–
2007 6 – 3.625
1 – 3.875
2008 5 4.125
2009 3 14 4.375
2010 7 29 4.625
2011 2 4.875
2012 6 15 5.125
2013 4 16 31
2014 8 17 33 –
2015 3 –
2016 18 35

19 37
20 39
21 41
––

––

Calculation of 5-year weighted moving average

Year(a) Sales (b) 5 – year weighted 5–year weighted moving
(` ’0000) moving total (c) average (d = c / 8)
2006 –
2007 2 – –
2008 6 3.625
1 –
2009 3.875
5 2×1+6×2+1×2+5×2+
2010 3 × 1 = 29 4.125
3
2011 6×1+1×2+5×2+3×2+ 4.375
7 7 × 1 = 31
2012 4.625
2 1×1+5×2+3×2+7×2+
2 × 1 = 33

5×1+3×2+7×2+2×2+
6 × 1 = 35

3×1+7×2+2×2+6×2+
4 × 1 = 37

Solution of C.U. Question Paper–2017 (New Syllabus) A.25

Year(a) Sales (b) 5 – year weighted 5–year weighted moving
(` ’0000) moving total (c) average (d = c / 8)
2013 4.875
6 7×1+2×2+6×2+4×2+
2014 8 × 1= 39 5.125
4
2015 2×1+6×2+4×2+8×2+ –
2016 8 3 × 1 = 41 –
3
–

–

From the last column of the above two tables it is clear that 4-year centered
moving averages are equivalent to a 5-year weighted moving average with
weights 1, 2, 2, 2, 1 respectively.

8. (a) From the following data, prove that Fisher’s Ideal formula satisfies
both time Reversal and Factor Reversal Tests of index number:

Commodity Base year Current year

A Price Quantity Price Quantity
B
C 4 20 6 18
D
5 15 6 12

2 30 3 30

3 25 5 28

Solution: Let p0, pn denote prices in the base year and current year and q0, qn be
the quantities in the base year and current year respectively.

Commodity p0 q0 pn qn p0q0 pnq0 p0qn pnqn
A 4 20 6 18 80 120 72 108

B 5 15 6 12 75 90 60 72

C 2 30 3 30 60 90 60 90

D 3 25 5 28 75 125 84 140

Total – – – – 290 425 276 410

∑p0q0 = 290, ∑pnq0 = 425, ∑p0qn = 276, ∑pnqn = 410
Fisher’s Ideal index number (omitting the factor 100) is

åå ååI0n = pnq0 ´ pnqn = 425 ´ 410
p0q0 p0qn 290 276

Interchanging ‘0’ and ‘n’, we get

åå ååIn0 = p0qn ´ p0q0
pnqn pnq0

A.26 Business Mathematics and Statistics

= 276 ´ 290
410 425

Now, I0n ´ In0 = 425 ´ 410 ´ 276 ´ 290
290 276 410 425

= 425 ´ 410 ´ 276 ´ 290 = 1 = 1
290 276 410 425

Hence, Fisher’s Ideal formula satisfies Time Reversal Test.
Again, Fisher’s Price Index Number (omitting the factor 100) is

P0n = Spnq0 ´ Spnqn = 425 ´ 410
Sp0q0 Sp0qn 290 276

Fisher’s Quantity Index Number (omitting the factor 100)

Q0n = Sqn p0 ´ Sqn pn = 276 ´ 410
Sq0 p0 Sq0 pn 290 425

Then, P0n ´ Q0n = 425 ´ 410 ´ 276 ´ 410
290 276 290 425

= 425 ´ 410 ´ 276 ´ 410
290 276 290 425

= 410 ´ 410 = 410
290 ´ 290 290

Now, value ratio = Spnqn = 410
Sp0q0 290

Since, P0n × Q0n = value ratio.
Therefore, Fisher’s Ideal Formula satisfies the Factor Reversal Test.

APPENDIX

B

Log Tables

Logarithms

10 0000 0043 0086 0128 0170 5 9 13 17 21 26 30 34 38

0212 0253 0294 0334 0374 4 8 12 16 20 24 28 32 36

11 0414 0453 0492 0531 0569 4 8 12 16 20 23 27 31 35

0607 0645 0682 0719 0755 4 7 11 15 18 22 26 29 33

12 0792 0828 0864 0899 0934 3 7 11 14 18 21 25 28 32

0969 1004 1038 1072 1106 3 7 10 14 17 20 24 27 31

13 1139 1173 1206 1239 1271 3 6 10 13 16 19 23 26 29

1303 1335 1367 1399 1430 3 7 10 13 16 19 22 25 29

14 1461 1492 1523 1553 1584 3 6 9 12 15 19 22 25 28

1614 1644 1673 1703 1732 3 6 9 12 14 17 20 23 26

15 1761 1790 1818 1847 1875 3 6 9 11 14 17 20 23 26

1903 1931 1959 1987 2014 3 6 8 11 14 17 19 22 25

16 2041 2068 2095 2122 2148 3 6 8 11 14 16 19 22 24

2175 2201 2227 2253 2279 3 5 8 10 13 16 18 21 23

17 2304 2330 2355 2380 2405 3 5 8 10 13 15 18 20 23

2430 2455 2480 2504 2529 3 5 8 10 12 15 17 20 22

18 2553 2577 2601 2625 2648 2 5 7 9 12 14 17 19 21

2672 2695 2718 2742 2765 2 4 7 9 11 14 16 18 21

19 2788 2810 2833 2856 2878 2 4 7 9 11 13 16 18 20

2900 2923 2945 2967 2989 2 4 6 8 11 13 15 17 19

B.2 Business Mathematics and Statistics

Logarithms

Log Tables B.3

Antilogarithms

B.4 Business Mathematics and Statistics

Antilogarithms