M.B.A. (3 Years) (First Semester) (DDE)
EXAMINATION July 2003
QUANTITATIVE METHODS
Time 3 hours Maximum Marks: 70
Note: Attempt five questions in all selecting at least one question from each unit.
All questions carry equal marks
UNIT I
Q1
a) A firm sells a product at the rate of Rs 200 per unit. Variable costs
are Rs 120 per unit and fixed costs are Rs 40,000. How many units
the company should produce
a. For break even
b. For earning the profit of Rs 21,440
Solution
Let the company produce x units.
Total variable cost = Rs 120x
Revenue = Rs 200x
Fixed Costs = Rs 40000
a) For break even point revenue = investment
∴ 200 x = 120 x + 40000
∴ 80 x = 40000
40000
x=
80
∴ x = 500
Hence the company should produce 500 units for break -even point for
given fixed cost
b) For earning the profit of 21440
∴ 200 x = 120 x + 40000 + 21440
∴ 80 x = 61440
61440
x=
80
∴ x = 786
Hence for earning a profit of 21440 also the company should produce 768
units.

b) A man saved Rs 16500 in ten years. In each year after first he saved Rs 100
more than he did in the proceeding year. How much did he save in the first year?
Solution: Let the man saved Rs x in first year. There after he saved 100+x,
200+x, 300+x, ….900+x
His savings can be seen to be in Arithmetic Progression with first term as
x and common difference as 100
We know that
n
Sum of n terms of an AP is give by (a + l )
2
Here a = x
n = 10
l = 900x
Sum of total savings = 16500 (given)
10
16500 = ( x + 900 + x)
2
16500
= 900 + 2 x
5
16500
2x = − 900
5
2 x = 3300 − 900
2400
x=
2
x = 1200
Hence the man saved Rs 1200 in first year
Q2 Using matrix notation solve the following matrix
Solution
x − 2 y + 3z = 4
2 x + y − 3z = 5
− x + y + 2z = 3
According to matrix notation the above set of equations can be represented in
matrix form as
A.X =B
 1 −2 3   x  4
 1 − 3 and X =  y  
Where A =  2    and B = 5
− 1 1
 2 z
   3
 

Therefore X = A-1 B
Finding the inverse of Matrix A
Cofactor of Cell 1,1 = 5
Cofactor of Cell 1,2 = -1
Cofactor of Cell 1,3 = 3
Cofactor of Cell 2,1 = 7
Cofactor of Cell 2,2 = 3
Cofactor of Cell 2,3 = 1
Cofactor of Cell 3,1 = 5
Cofactor of Cell 3,2 = 9
Cofactor of Cell 3,3 = 5
Value of the matrix =16
 5 7 5
 
Therfoe A-1 = − 1 3 9 / 16
 3 1 5
 
 4
 
Multiplying A-1 .B we get the matrix 3
 2
 
 4
 
Therefore X = 3
 2
 
Therefore x=4 y=3 and z=2
UNIT II
Q 3 What do you mean by measures of central tendency and measures of
dispersion? How do you differ from one another? Discuss the merits
and limitations of the following measures:
i) Median
ii) Range
Answer: Measure of Central Tendency: An average value is a single value
within a range of the data that is used to represent all of the values in the series.

They are the typical values around which other items of the distribution
congregate. They are very useful:
a) For describing the distribution in concise manner
b) For comparative study of different distribution
Characteristics of a good measure of central tendency are:
a) It should be rigidly defined
b) It should be easy to understand and calculate even for a non-
mathematical person
c) It should be based on all the observations
d) It should be suitable for further mathematical treatment
e) It should be affected as little as possible by fluctuations
Different measures of central tendency are:
a) Arithmetic Mean
b) Median
c) Mode
d) Geometric Mean
e) Harmonic Mean
Measure of Variations: The degree to which numerical data tend to spread
about an average value is called variation or dispersion of the data.
Characteristics for an ideal Measure of variation are:
a) It should be rigidly defined
b) It should be easy to understand and calculate
c) It should be based on all observations
d) It should be amenable to further mathematical treatment
e) It should be affected as little as possible by fluctuations
Different measures of variations are:
a) Range
b) Quartile Deviation
c) Percentile Range
d) Mean Deviation
e) Standard Deviation
Merits of Median are:
a) It is rigidly defined
b) It is easy to understand and calculate
c) It is based on all observations
d) It is suitable to further mathematical treatment
e) Of all averages Mean is least affected by fluctuations
Limitations of Mean are:
a) It is very much affected by extreme observations
b) Cannot be used in case of open end classes

c) Cannot be determined by inspection nor graphically
d) Cannot be used when dealing with qualitative characteristics
e) Cannot be calculated even if a single observation is not known
f) It is extremely asymmetrical distribution
Merits of Range are:
a) It is the simplest method to find variation of data
b) It is rigidly defined
c) Readily comprehensible
Limitations of Range are:
a) It is not based on entire set of data
b) Very much affected by extreme observations
c) If smallest and largest value are unaltered and all other values are
replaced the range still remains same.
d) Cannot be used with open end classes
e) It is not suitable for any further mathematical treatment
f) It is too indefinite to be used as a practical measure of dispersion
Q 4 Calculate the values of
i) Arithmetic Mean
ii) Mode
iii) Quartile deviation
iv) Standard Deviation
Solution:
C.I. f x d=(x-A)/100 f.d c.f
(A=500)
200-300 20 250 -2.5 -50 20
300-400 39 350 -1.5 -58.5 59
400-500 86 450 -5 -430 145
500-600 35 550 5 175 180
600-700 14 650 1.5 21 194
700-800 6 750 2.5 15 200
Total 200 -327.5
Here Assumed Mean (A) = 500
Class Width (h) = 100
i) Arithmetic Mean = 500 + 100 * (-327.5)/200 = 336.25
ii) Mode: Since the distribution is regular the class corresponding to the
maximum frequency i.e. 86 is the modal class. Hence modal class is 400-500
and using the mode formula we get

h( f 1 − f 0)
Mo = l +
2 f 1− f 0 − f 2
100(86 − 39)
= 400 + = 447 (approx)
2 * 86 − 39 − 35
iii) Quartile Deviation Here N= 200 and N/4 = 50
3N/4 = 3 * 200/4 = 150
Q1 First Quartile lies in 300-400
Q3 Third Quartile lies in 500-600
100
Q1 = 300 + (50 − 20) = 376.92
39
100
Q3 = 500 + (150 − 86) = 682.86
35
Hence Quartile Deviation = (682.86 – 376.92 ) / 2 = 152.97
UNIT III
Q5 a) Define the term correlations and regressions. What are the points of
difference between the two?
b) Describe various components of time series
Answer
a) Correlation - Correlation is an analysis of the co-variation between
two or more variables
Correlation analysis enables us to have an idea about the degree and
direction of the relationship between the two variables under study
Regression – Regression analysis is a mathematical measure of the
average relationship between two or more variables in terms of the
original units of the data.
In regression analysis there are two types of variables. The variable
whose value is influenced or is to be predicted is called dependent
variable and the variable, which influences the values or is used for
prediction is called independent variable
Correlation analysis vs. regression Analysis

It means relationship between two or It means stepping back or returning to the
more variables average values and is a mathematical
measure expressing the relationship
between the two variables
Correlation coefficient between 2 It aims at providing functional relationship
variables is a measure of direction and between 2 variables under study and then
degree of linear relationship using this relationship to predict or
estimate the value of dependent variable
Correlation need not imply cause and It clearly indicates the cause and effect
effect relationship relationship
It is independent of the units of Regression coefficient are absolute
measurement. It is a pure number measures representing the change in the
between +1 and –1 value of the variable for a unit change in
the value of the other variable
There may be non-sense correlation due No such thing like that
to pure chance
It has limited applications It has wider applications as it can deal
with both linear and non-liner
relationships
b) Components of time series are:
a. Secular trend or Long Term Movement: The general tendency
of the time series data to increase or decrease or stagnate
during a long period of time is called the secular trend or simply
trend. This phenomenon is mostly observed in most of the
series relating to Economics and Business. e.g. an upward
tendency is usually observed in time series relating to
population, production and sales of products.
b. Periodic Movements or Short Term fluctuations: They can be
classified into two parts
i. Seasonal Fluctuation: These variations in time series are
due to the rhythmic forces, which operate in regular and
periodic manner over a span of less than a year, i.e.
during a period of 12 months and have the same or
almost same pattern year after year.
ii. Cyclical Variations: The oscillatory movements in a time
series with period of oscillation greater than one year are
termed as cyclical variations. These variations in time
series are due to ups and downs recurring after a period
greater than one year. These cyclical fluctuations are
more or less regular, are not necessarily periodic. One
complete period, which normally lasts from 7 to 9 years,
is termed as a ‘cycle’.

c. Random or Irregular variations: These fluctuations are purely
random and are result of such unforeseen and unpredictable
forces which operate in absolutely erratic and irregular manner.
Such variations do not exhibit any definite pattern and there is
no regular period or time of their occurrence. These powerful
variations are usually caused by numerous non-recurring factors
like floods, famines, wars, earthquakes etc.
Q 6 In a partially destroyed record, the following data are available:
Two regression lines are
5 x + 3 y = 290
3 x + 2 y = 180
Variance of x is 16
Find:
i) Mean values of x and y
Solution: I) Since both the lines pass through the mean values the point ( x, y )
must satisfy both the equations
Solving both the equations for x and y we get
x= 40 and y= 30
Hence mean of x and y is 40 and 30 respectively
UNIT IV
Q 7 Using suitable examples explain the following terms:
i) Mutually exclusive events
ii) Conditional Probability
iii) Independent Events
Answer
i) Mutually Exclusive Events: Two or more events are said to be
mutually exclusive if the happening of any one of them excludes the
happening of all others in the same experiment. e.g. In toss of a coin
the events ‘head’ and ‘tail’ are mutually exclusive because if head
comes, we can’t get the tail and if tail comes we can’t get head.
ii) Conditional Probability: Conditional probability is the probability of
happening of one event on the condition that other event has already
happened

e.g. Drawing of balls from a bag containing balls (say 8) without
replacement. When first ball is drawn then the sample space will be 8.
But when the second ball is drawn the sample space would reduce
down to 7. So the probability of occurrence of second event gets
affected by the happening of first event
iii) Independent Events: Events are said to be independent of each other
if happening of any one of them is not affected by and does not affect
the happening of any one of others.
e.g. In tossing of a die repeatedly the event of getting ‘5’ in 1st throw is
independent of getting ‘5’ in second, third or subsequent throws
Q8
a) Eight coins are tossed simultaneously. Find the probability of
getting
i. Four Heads
ii. At most 2 tails
Solution If p denotes probability of getting a head = ½ then according to
binomial distribution law
Probability of getting 4 heads is given by
P(r=4) = 8C4 (1/2)4(1/2)8-4
70
=
256
ii) At most 2 tails = Prob of getting 0 tail + Prob of getting 1 tail + Prob of
getting 2 tail
Let p be prob of getting tail =1/2
Therefore Probalitity of getting at most 2 tails is given by
8
C0(1/2)0(1/2)8-0 + 8C1(1/2)1(1/2)8-1 + 8C2(1/2)2(1/2)8-2
37
=
256
(b) Marks obtained by the students of a class are normally distributed with mean
of 58 marks and standard deviation of 4 marks. Find the probability that marks
obtained by a student are
i) Less than 45
ii) Between 55 and 61

Solution
i) Probability that marks obtained are less than 45
µ = 58
σ =2
x = 45
45 − 58
Standard Normal Variate is Z = = -6.5
2
P(z<-6.5) or P(z>6.5) = 0.5 – P(0<z<6.5)
ii) lies between 55 and 61 P(55<x<61)
Calculating SNV for 55
55 − 58
Z1 = = -1.5
2
61 − 58
Z2 = = 1.5
2
In terms of SNV P(-1.5<z<1.5) = 2 P(0<z<1.5) { By Symmetry }
= 2 * .4332
= 0.8664