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Unformatted text preview: Appendix Quantitative Methods This Appendix on Quantitative Methods has been designed for students
who do not have exposure to mathematics and statistics. It is aimed to
provide students an insight into the basics of mathematics, business
statistics, probability, and probability distributions. This knowledge
will enable the student to apply these tools and techniques in the study
of Operations Management. Operations Management Contents
Section 1
Section 2 Measures of Central Tendency 16 – 25 Section 3 Measures of Dispersion 26 – 33 Section 4 Concepts of Probability 34 – 40 Section 5 2 Basics of Mathematics 3 – 15 Probability Distributions 41 – 46 Quantitative Methods: Appendix Section 1 Basics of Mathematics
In this section, we will discuss:
• Number Systems • Simultaneous Equations • Polynomials • Theory of Indices • Permutations and Combinations • Progressions • Functions It is important to revise the decimal, binary and octal number systems before
proceeding on to study the quantitative methods. Also, it is important to understand
the concepts of Simultaneous Equations, Polynomials, and the Theories of Indices to
solve problems which are complex in nature.
Some of the other important concepts that have to be understood are Permutations and
Combinations and the three progressions – arithmetic, geometric, and harmonic. NUMBER SYSTEMS
Apart from the two most commonly used number systems – decimal
(0,1,2,3,4,5,6,7,8,9) and Roman (I,II,III,IV,V,VI....), we have other systems such as
octal number system represented by (0,1,2,3,4,5,6 and 7) and hexadecimal number
system based on the digits 0 to 9 and also the letters from A to F. Decimal System
• The decimal system has 10 digits (0 to 9). • Each digit is a number. • If there are two or more than two digits in a number, each digit has a value that is
ten times greater than the digit just to its right. For example, if 9628 is a number
that has four digits. We can say that 8 is in the units place, 2 is in the tens place, 6
is in the hundreds place and 9 is in the thousands place. • Each digit in a number has two values one is called the absolute value and the
other is called as the place value. If 9628 is taken, 9 has an absolute value which
is 9 itself and place value which is 9000, for 6, absolute value is 6 itself and place
value is 600, similarly for 2 and 8 the absolute values are 2 and 8 again and the
place values are 20 and 8 respectively. • A decimal system has number 10 as its base or radix. • A number in any number system can be represented as
N = d n +1 × r n + ...............d 3 r 2 + d 2 r 1 + d 1 r 0 Where, N = any number of any base system,
d n = digit in the nth position,
r n = the exponent for (n+1)th digit position.
3 Operations Management Binary Number System
• The binary number system is based on only two variables 0 and 1. • In the binary number system each digit in the number is twice the value of the
digit to its right. • The binary number system has 2 as its base or radix. • A binary number 11111 can be expressed in decimal system as
4 3 2 1 0 (11111) 2 = 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2 = (31) 10
• In binary system a carryover occurs every time 2 is reached • Any number having ‘n’ bits in binary representation can be expressed as
P1 × 2 n −1 + P2 × 2 n − 2 + P3 × 2 n −3 + ............. + Pn −1 × 21 + Pn × 2 0 Where, P 1 & P n are called as most significant bit and least significant bit
respectively.
Conversion from decimal to binary
To convert a number from decimal to binary, the decimal number is divided by 2 until
it is reduced to zero.
Example A1.1:
To convert the decimal number 35 to binary
Solution: ÷ 2 = 17
17 ÷ 2 = 8
8 ÷ 2=4
4 ÷ 2=2
2 ÷ 2=1
1 ÷ 2=0 35 remainder = 1
remainder = 1
remainder = 0
remainder = 0
remainder = 0
remainder = 1 Therefore the binary number is 100011
Conversion from binary to decimal
The above given number can be converted back into decimal number system in the
following manner
(100011) 2 = 1× 2 5 + 0 × 2 4 + 0 × 2 3 + 0 × 2 2 + 1× 21 + 1× 2 0 = (35)10 Fractional numbers
• A fraction number, for instance 0.2 in the decimal number system can be written
as 2/10 or 2 × 101 • We can convert a fractional binary into a decimal number in the following way,
(0.100101)2 =1× 2−1 + 0 × 2−2 + 0 × 2−3 + 1× 2−4 + 0 × 2−5 + 1× 2−6 = 0.5 + 0 + 0 + 0.0625 + 0 + 0 + 0.015625
= (0.578125) 10
• 4 Similarly a fractional decimal can be converted into binary number in the
following way, Quantitative Methods: Appendix
(0.758215) 10 = 0.758215 × 2 ×
0.03286 ×
0.06572 ×
0.13144 ×
0.26288 ×
0.52576 × 0.51643 2 remainder = 1 2 remainder = 1 2 remainder = 0 2 remainder = 0 2 remainder = 0 2 remainder = 0 The number is obtained by writing the integer part of all remainder numbers up to 6
digits. Thus the number is given by (0.110000) 2 Octal Number System
• The octal number system consists of digits from 0 to 8 • The base or radix of the octal number system is 8 • A number in the decimal number system can be converted into octal number
system by dividing the number with 8 successively. Example A1.2:
To convert 957 into octal system
Solution:
957 ÷ 8 remainder = 5 ÷8
14 ÷ 8
1÷8
0÷8 remainder = 7 119 remainder = 6
remainder = 1 Thus the number is (1675) 8
Example A1.3:
To convert octal into decimal
Solution:
Let us consider the same example given above.
(1675) 8 = 1 × 8 + 6 × 8 + 7 × 8 + 5 × 8 = 512 + 384 + 56 + 5 = 957
3 2 1 0 Note: Similarly, we can convert fractional decimal to octal and fractional octal to
decimal. SIMULTANEOUS EQUATIONS
Simultaneous Equations are a set of equations which have more than one unknown
variable. When the equations are solved, we obtain the values of these variables and
they satisfy both the equations simultaneously. The equations can be solved in two
ways:
5 Operations Management Solving through Substitution
Simultaneous Equations can be solved by simply substituting the value of the
unknown variable (x or y) obtained from one of the two equations in the other
equation.
Example A1.4:
Find out the values of x and y from the equations: x – 2y = 4 and 2x + 3y = 1, using
substitution approach.
Solution:
From the equation, x – 2y = 4, we can obtain the value of x = 4 + 2y. Substituting this
in the second equation, 2x + 3y = 1, we have
2 (4 + 2y) + 3y = 1
8 + 4y + 3y = 1
7y = 7
y = 1.
By substituting the value of y in any of the given equations, we get
x = 4 + 2 (1)
x=42
x = 2.
Therefore, the values x = 2 and y = 1 satisfy both the given equations simultaneously. Solving through Elimination
Simultaneous Equations can be solved by eliminating any one of the variables which
has the same coefficient in both the equations. If the values are different, we need to
make them equal to facilitate such elimination. This can be done by multiplying or
dividing the entire equation by a number.
Example A1.5:
Find out the values of x and y from the equations: 2x – 3y = 1 and x + y = 3 using
elimination approach.
Solution:
To solve the given equations, we need to eliminate one of the variables, x or y, which
has the same value in both the equations. Let us eliminate x by making the
coefficients the same in both equations. This can be done by multiplying the second
equation throughout by 2. The second equation then becomes 2x + 2y = 6. Now, we
can eliminate x as the coefficients are the same in both equations.
2x – 3y = 1
(–) 2x + 2y = 6
0–5y=–5
– 5y = –5
y = –5/–5
y=1
6 Quantitative Methods: Appendix
Now, by substituting y = 1 in any one of the equations, we can arrive at the value of
x., Let us take x + y = 3
x + (1) = 3
x=3–1
x=2
Therefore, x = 2 and y = 1
Note: If there are three equations with three unknown variables, we need to solve
them by taking any two equations at a time. Thereafter, the process is the same as that
of solving two simultaneous equations. POLYNOMIALS
A polynomial is a mathematical expression that consists of one or more unknown
variables. The variables are unknown and often have coefficients that are raised to the
power. For example, p2, x2 + 2y, ab2 + c are all examples of polynomials.
Example A1.6:
What is the value of x in the equation x + y2 = 2, if y = 2?
Solution:
Given that x + y2 = 2. By substituting the value of y with its value, 2, in the equation,
we get x + (2)2 = 2
i.e., x + 4 = 2
x=2–4
x = –2 Factorization
Factorization is a process of finding the factors in a given expression. The factors can
be obtained by substituting all the possible values for which the value of the
expression would be zero.
Example A1.7:
Find the factors of (2a2 – 8)
Solution:
Let us start substituting ‘a’ with values starting from 1
a = 1, 2a2 – 8 = 2(1)2 – 8 = 2×1 – 8= 2 – 8 = 6 ≠ 0
a = 2, 2a2 – 8 = 2(2)2 – 8 = 2×4 – 8= 8 – 8 = 0
Since a is raised to the power of 2, let us check whether the expression would be zero
when a = 2
a = 2, 2a2 – 8 = 2(2)2 – 8 = 2×4 – 8= 8 – 8 = 0
Therefore, the factors of expression are a = 2 or 2
Example A1.8:
Factorize 9a2  49
7 Operations Management
Solution:
The given expression is 9a2  49. It can be written as (3a)2  (7)2
The above expression is in the form of a2  b2 which is equal to (a  b) (a + b).
Therefore, the expression can be further written as (3a  7) and (3a + 7). Therefore the
factors of the expression are (3a  7) and (3a + 7) and a = 7/3 or 7/3 Quadratic Equations
A quadratic equation is referred to as an equation involving a quadratic polynomial. It
is an equation in the second degree with a single variable. The standard form of a
2
Quadratic Equation is ax + bx + c = 0 , where a, b, and c are real numbers and
constants, and a ≠ 0. The equation is in the second degree as the highest power of x is
two. The values of x which satisfy the equation are called the roots of the equation.
2
The roots of equation ax + bx + c = 0 are;
− b − b − 4ac 2 − b + b − 4 ac 2 2a 2a x= , and x = For the above,
• −b
2
, and the equation will have only one root.
If b − 4 ac = 0, x =
2a • 2
If b − 4 ac < 0, then the equation has no real roots. Example A1.9:
Find the roots of the equation 5x2 – 7x – 12 = 0
Solution:
For the given equation, a = 5, b = 7 and c = 12
The roots of the equation will be,
− b − b − 4ac 2 = = = 8 x1 = i. (7 ) − (
7− ( 2a 49 − (− 240 )
10
289 ) 10
7 − 17
10 = = − 10
10 = 1 ) (− 7 )2 − 4(5)(− 12 ) − (− 7 ) − 2(5 ) Quantitative Methods: Appendix
−b+ b ii. x2 = = = = (7 ) + (
7+ 2 − 4 ac = 2a 49 − (− 240 ) (− 7 )2 − 4(5)(− 12 ) − (− 7 ) + 2(5 ) ) 10 ( 289 ) 10
7 + 17 24 = 10 10 = 12
5 Therefore, the roots of x = 12/5 or x = 1 THEORY OF INDICES
The theory of indices provides rules or laws for problems which have complex or
difficult quantities. The rules simplify the complex quantities and help to solve the
given problem.
Rule I x m × x n = x m+n
Example A1.10:
2
2+1
2 ×2 = 2 Solution:
2
LHS = 2 ×2 = (2 × 2) × 2 = 8 RHS = 2 2 +1 3 = 2 = 2× 2 × 2 = 8 Therefore, the rule holds good.
Rule II xm
= x m−n
n
x
Example A1.11:
3
3 4
2 = 34− 2 Solution:
LHS = 3
3 RHS = 3 4
2 4− 2 =
= 3× 3× 3× 3
3×3 = 81
9 =9 32 = 3 × 3 = 9
9 Operations Management
Therefore, the rule holds good.
Rule III (x ) mn = x mn Example A1.12: (2 ) 23 = 2 2×3 Solution: () LHS = 2
RHS = 2 23 2×3 2 2 2 = ( 2 × 2 × 2 ) = 4 × 4 × 4 = 64
6 = 2 = 2 × 2 × 2 × 2 × 2 × 2 = 64 Therefore, the rule holds good.
Rule IV (xy )n = x n y n
Example A1.13: (4 × 5)2 = 4 2 × 5 2
Solution:
LHS = (4 × 5)2 = (4 × 5) × (4 × 5) = 20 × 20 = 400 RHS = 4 × 5 = (4 × 4) × (5 × 5) = 16 × 25 = 400
2 2 Therefore, the rule holds good. PERMUTATIONS AND COMBINATIONS
Each of the arrangements that can be made by taking some or all of a number of
things or objects is called a permutation. If we want to find the number of ways to
arrange the three letters in the word BAT, in different twoletter groups, where AT is
different from TA without repeating the letters, then we can write them as BA, BT,
AB, AT, TB & TA. Here, since we just had four letters it was easier for us to write the
permutations. But imagine a 10 letter word where we want to make groups of 4 letters
each. To write all the permutations, it is very difficult since a large number is
involved.
Therefore we use the formula to calculate the permutations given by,
n Pr = n!
(n − r )! Where, the number of ‘n’ dissimilar things taken ‘r’ at a time is denoted by n Pr n and r are positive integers and n ≥ r
n! is the continued product of first n natural numbers.
Example A1.14:
How many different groups can be formed by using four letters out of the 10 letters
A,B,C,D,E,F,G,H,I,J.
10 Quantitative Methods: Appendix
Solution:
The number of dissimilar things =10, the number of permutations taken is 4.
Therefore we can substitute n = 10 and r = 4 in the formula.
10 P4 = 10!
10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
=
= 5040
(10 − 4)!
6 × 5 × 4 × 3 × 2 ×1 Suppose, if we want to find the number of combinations of size 2 without repeated
letters from the word BAT, where order does not matter and BA is same as AB. We
can thus write the three combinations as BA, BT and AT.
The formula for number of combinations on ‘n’ dissimilar things taken ‘r’ at a time is
given by,
n n!
r!(n − r )! Cr = Example A1.15:
In how many ways can 4 letters be selected from the 10 letters given above?
Solution:
10 C4 = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
10!
=
= 210 ways.
4!(10 − 4)!
( 4 × 3 × 2 × 1)(6 × 5 × 4 × 3 × 2 × 1) PROGRESSIONS
Progressions are of three types – Arithmetic progression, Geometric progression and
Harmonic progression. Arithmetic Progression
A series of numbers is said to be in arithmetic progression, if they increase or decrease
by a common difference.
For example: 5, 10, 15, 20, 25, 30….
66, 55, 44, 33, 22…..
The above two series of numbers are said to be in arithmetic progression, since the
first one increases with a common difference of 5 and the second one decreases with a
common difference of 11.
The general form of arithmetic progression is given by,
a, (a+d), (a+2d), (a+3d), (a+4d), ……….[a+(n1)d]
Where, ‘a’ is the first term of the equation, d is the common difference.
The nth term in arithmetic progression is given by,
T n = a + (n1) d
The sum to n terms of a series in Arithmetic progression is given by,
Sn = n
[2a + (n1) d]
2 11 Operations Management
Example A1.16:
Find the sum of the series 2, 5, 8, 11…up to 46 terms.
Solution:
The first term in the given series is 2 and the common difference is 3. Using the
formula,
Sn = n
[2a + (n1) d],
2 S 46 = 46
[2(2) + (461) 3]
2 = 23[4 + (45)3] = 3197.
Note: We can also find the number of terms and the common difference if the first
term, the last term and the sum is given.
Arithmetic mean: If a 1 , a 2 , a 3 , a 4 ,........a n are real numbers and are in arithmetic
progression then the arithmetic mean is calculated by the formula,
Arithmetic mean = a 1 + a 2 + a 3 + a 4 + ........ + a n
n Geometric Progression
Quantities or a number series is said to be in geometric progression, when they
increase or decrease by a constant factor. The constant factor is the common ratio and
is calculated by dividing any term that immediately precedes it. For example,
2, 6, 18, 54, 162, 486………….
96, 48, 24, 12……………
In the above example, each term in the first series is multiplied by 3, and in the second
series each term is multiplied by ½.
The general form of a geometric progression is given by,
a , ar, ar 2 , ar 3 , ar 4 , ar 5 ...................
Where, ‘a’ is the first term of the series and ‘r’ is the common ratio.
The nth term in a geometric progression is given by, Tn = ar n −1 The sum to n terms of a series in geometric progression is given by, S= Tn −
1− a
r 1
r The geometric mean is given by, G = ab
n The product of n geometric means is given by (ab) 2
12 Quantitative Methods: Appendix
Example A1.17:
Find the sum of the series 3, 6, 12, 24…up to 25 terms.
Solution:
The sum to ‘n’ terms of a series in geometric progression is given by, S= a Tn −
1− r 1
r The given series is: 3, 6, 12, 24…up to 25 terms.
Here,
n = 25
a=3
r=2
T25 = 3 (2)251 = 3 × 16,777,216 = 50,331,648 S= 3
T25 −
2
1− 1 50,331,648 −
= 1 2 3
2= 2 100,663, 296 − 3
2
1 100,663,293
= 2 2
1 = 100,663,293 2 Harmonic Progression
Quantities are said to be in harmonic progression, if the series obtained by taking
reciprocals of the corresponding terms of the given series is an arithmetic progression.
For example, the series 1/3, 1/6, 1/9…….. is a harmonic progression, since the series
obtained by taking reciprocals of its corresponding terms, 3,6,9…. is an arithmetic
progression.
A general form of a harmonic progression is given by, 1/a, 1/(a+d), 1/(a+2d)…….
The nth term of a harmonic progression is given by, 1/ [a + (n1) d]
The harmonic mean of the positive real numbers a1…….an is given by,
H= n
1
a1 + 1
a2 + 1
a3 + ........... 1
an The three means can be represented by A.M. ≥ G.M. ≥ H.M.
Example A1.18:
Find out the 16th term in the series: 1/5, 1/9, 1/13, 1/17……....
Solution:
The given series is: 1/5, 1/9, 1/13, 1/17……... 13 Operations Management
The reciprocal of the values in the given series form arithmetic progression, i.e., 5, 9,
13, 17, ……. The 16th term in the series in arithmetic progression can be calculated by
using the formula: T n = a + (n1) d
Here,
a=5
d=4
n = 16
Substituting the values, we have, T16 = 5 + (16 – 1) 4 = 5 + 15 × 4 = 65
In harmonic progression, T16 of the given series would be the reciprocal of 65, i.e.,
1/65. FUNCTIONS
• Many decision making situations are dealt with by constructing mathematical
models. • A model can be considered a set of relationships among economic variables. • Functions can be easily used to state such relationships • A function expresses the relationship of one variable or a group of variables
(called the domain) with another variable (called the range) by associating every
member in the domain to a unique member in range. • Suppose a variable Y is related to a variable X as Y = 5X + 3, then we can say
that, for any given value of X, the value of Y can be calculated using the above
relationship. Thus, the equation enables us to construct a range of Y values for a
given table of X values. • If we have more than one independent variable that determines the value of a
dependent variable say, x. The dimension of the function is determined by the
number of independent variables in the domain of a function. For instance, x = f
(y, z) is a two dimension function,
x = f ( X1,X 2 ,X3.......X n ) is an ndimension function.
A linear function is depicted in the following manner,
y = a 0 + a 1 x 1 + a 2 x 2 + ......... + a n x n • In two dimensional space, a linear function is a straight line and is written as,
y = a + mx
In the two dimensional case, the form of a linear function can be obtained, if the
coordinates of the two points on the straight line are known. Consider x ′ and x′′
are two given values of x and the corresponding y values are y ′ and y ′′ then the
slope of the line is: m = y '− y ' '
x '− x ' ' The intercept is: a = y '−[ y '− y ' '
x'− x ' ' x' In the ‘n’ dimensional case, we need the coordinates of at least ‘n’ points for
finding the linear functional form.
14 Quantitative Methods: Appendix SUMMARY
• This unit provides a basic knowledge of mathematics. • The types of number systems discussed in this unit include decimal system,
binary system and octal system. • Simultaneous equations can be solved by substitution or by elimination. • 2
ax + bx + c = 0 is the standard form of a quadratic equation of the second
− b ± b − 4 ac 2 degree and the roots are 2a . • The Theory of Indices is applied to simplify complex quantities to solve the given
problem. • Each of the arrangements that can be made by taking some or all of a number of
things or objects is called a permutation. • The progressions are of three types – Arithmetic progressions, Geometric
progressions and Harmonic progressions. • The different types of functions are linear, inverse, exponential and logarithmic. 15 Operations Management Section 2 Measures of Central Tendency
In this section, we will discuss:
• Mathematical Averages • The Median • The Mode The aim of statistical analysis is to calculate the average point of data that represents
the characteristics of the entire raw data. This average point is located centrally where
all others values of the data cluster. This central value is called as the measure of
central tendency. In this unit we will discuss the various measures of central tendency
apart from discussing mathematical averages and different types of averages.
The primary objective of averaging is to calculate the average value that represents the
characteristics of the entire raw data. Other objectives of averaging are,
• To find out the average value that helps in comparison of two or more sets of data
(for example, the principal of a school can compare the marks of two different
sections of a class to analyze their performance). • Averages help in establishing relationships between separate groups in
quantitative terms. (relationship between two sets of data can be established on a
similar parameter like ‘marks’ in the previous point) • To draw valuable conclusions about a population by studying a small
representative sample of that population. • To help in effective decisionmaking. Averages are of two types,
1. Mathematical averages – Arithmetic mean, Geometric mean, and Harmonic
mean. 2. Positional averages – Median and mode MATHEMATICAL AVERAGES
The mathematical averages are  arithmetic mean, geometric mean, and harmonic
mean. Arithmetic Mean
The arithmetic mean is easier to calculate and is most frequently used. It is
represented as x .
Calculating the mean from ungrouped data
1n
x = n ∑ xi
i =1
Where,
x is the sample mean i is the set of natural numbers
16 Quantitative Methods: Appendix
n
∑ x is the sum of values of all observations.
i =1
If the mean is calculated for the entire population, then it is called population mean
( µ ), given by
x µ =∑ N
Where, N is the number of elements in the population.
Example A2.1: Table A2.1: Students’ Roll Numbers and their Weights
Student’s Roll No. 1 2 3 4 5 6 7 8 9 10 Weight 34 45 41 33 39 47 54 39 46 38 Calculate the average weight of the students.
Solution:
To calculate the average weight of the students given above, we can make use of the
formula,
x=∑ x 34 + 45 + 41 + 33 + 39 + 47 + 54 + 39 + 46 + 38
=
= 41.6
n
10 Calculating mean from grouped data
x=∑ fx
n Where,
∑ is the notation for sum
f is the number of observations in each class
x is the mid point of each class and
n is the number of observations in each class
Example A2.2: Table A2.2(a): Table Representing the Marks of 50 Students
Class (Marks) Frequency 2130 4 3140 6 4150 7 5160 11 6170 10 7180 7 8190 5 Total 50 Calculate the mean for the above given data.
17 Operations Management
Solution:
To calculate the mean for the above grouped data we use the formula x = ∑ fx
n Mid point = x = (lower limit + upper limit)/2 Table A2.2(b): Calculation of Arithmetic Mean for Grouped Data
Class (Marks)
2130
3140
4150
5160
6170
7180
8190 Mid Point (x)
25.5
35.5
45.5
55.5
65.5
75.5
85.5 Frequency (f)
4
6
7
11
10
7
5 (f) × (x)
102
213
318.5
610.5
655
528.5
427.5
∑ fx = 2855 x=∑ 2855
fx
=
= 57.1
n
50 The average marks of the class are 57.1.
There is a shortcut method for calculating the mean (this is especially useful if the
number of classes are more and are of equal size)
In this method, an assumed mean is located (usually from the centre of the classes)
and negative integers are assigned to all the values smaller to the assumed mean and
positive integers are assigned to all the values that are greater than the assumed mean
(assumed mean is given a value of zero).
In the above table we can assign the positive and negative values by considering 5160 class as the assumed mean as shown below: Table A2.3(a)
Class Code 2130 1 3140 2 4150 3 5160 0 6170 1 7180 2 8190 3 The formula for calculating the sample mean using the shortcut method is given by,
x = x 0 + w∑ (u × f )
n Where, x is the mean
x 0 is the value of the class mark assigned the code 0
18 Quantitative Methods: Appendix
w is the numerical width of the class interval
u is the code assigned to each class
f is the frequency of the class
n is the total number of observations in the sample
Example A2.3: Table A2.3(b): Calculation of Arithmetic Mean for Grouped Data
(ShortCut Method)
Class (Marks) Mid Point (x) Code (u) Frequency (f) (u) × (f) 2130 25.5 1 4 4 3140 35.5 2 6 12 4150 45.5 3 7 21 5160 55.5 0 11 0 6170 65.5 1 10 10 7180 75.5 2 7 14 8190 85.5 3 5 15
∑ uf = 2 x = x 0 + w∑ (u × f )
n = 55.5+10(2/50)
= 55.9 which is approximately equal to 57.1 (the result in the previous method)
Weighted Arithmetic Mean
The weighted arithmetic mean is calculated by considering the relative importance of
each of the values to the total value. The formula for calculating weighted average is:
x= ∑ (w × x)
sw Where, x = the symbol for weighted mean
w = weight allocated to each observation
∑ ( w × x ) = the sum of each weight multiplied by that element
s w = sum of all the weights
Example A2.4:
Calculate the weighted arithmetic mean from the following data.
Table A2.4(a): Quantity and Price Details
Items Quantity Price per unit (in Rs.) Books 5000 25 Pencils 2500 5 19 Operations Management
Items Quantity Price per unit (in Rs.) Erasers 2300 3 Pens 6500 15 Solution:
In the given example, prices
Table A2.4(b): Calculation of Weighted Arithmetic Mean
Items
Books
Pencils
Erasers
Pens Price (x)
25
5
3
15 w×x
125,000
12,500
6,900
97,500 Total x= Quantity (w)
5,000
2,500
2,300
6,500
16,300  241,900 ∑ (w × x)
sw x = the symbol for weighted mean
w = weight allocated to each observation
∑ ( w × x ) = the sum of each weight multiplied by that element = Rs. 241,900
s w = sum of all the weights = 16,300
x= ∑ (w × x)
sw = Rs. 241,900
16,300 = Rs. 14.84049. The weighted arithmetic mean is calculated when it is difficult to calculate the average
on common parameters. For instance, if a company employs three types of laborers –
skilled, semiskilled and unskilled, their productivity cannot be calculated uniformly,
as it is inappropriate. As a result the above formula is used to calculate the weighted
arithmetic mean. Geometric Mean
Geometric mean is used to calculate the average rate of change of certain quantities
over a period of time, which cannot be accurately measured with the help of
arithmetic mean.
Geometric mean = n product of all the values . Where, n is the number of values.
Example A2.5:
Calculate the growth rate of the company from the data given below: Table A2.5: Data Regarding the Growth Rate of a Company
Year 2 3 4 5 6 Growth Rate %
20 1
3 4 6 6 7 8 Quantitative Methods: Appendix
Solution:
Geometric mean = n product of all the values
The growth factor is equal to 1+ (rate/100) = 6 1.03 × 1.04 × 1.06 × 1.06 × 1.07 × 1.08 = 1.0565 is the average growth factor.
The growth rate is then calculated as
= 1.05651 = 0.0565
= 0.0565 × 100 = 5.65 percent per year. Harmonic Mean
Harmonic mean is considered as the reciprocal of the arithmetic mean of the
reciprocal of the given individual observations.
Harmonic mean = N
1
1
1
+
+ ......... +
X1 X 2
Xn Where, X 1 , X 2 …. refer to various items of the variable and N refers to the total
number of items.
Example A2.6:
If you travel from a place A to B with a speed of 50 km/hr and back to A from B with
60 km/hr speed, calculate the average speed of the entire journey. by substituting the
values in the above formula.
Solution:
Here X1 = 50 and X 2 = 60 and N = 2
Substituting these values in the above given formula, we have
2 Average speed =
[ 1
50 + 1
60 = 2 × 50 × 60 6000
=
= 54.5 km / hr
50 + 60
110 THE MEDIAN
The median is the middle value of any distribution, that is, half the values are above
the median and half are below it. Calculating Median from Ungrouped Data
If the given distribution has odd number of values, then the middle value of the
distribution becomes the median and if the given distribution has even number of
values then the average of the two middle values becomes the median. If the
distribution has frequencies that are odd in number then (n+1)/2th value gives the
median and when total of the frequencies is even say 2n, then the arithmetic mean of
nth and (n+1)th gives the median.
21 Operations Management
Example A2.7:
Given below is the data relating to the profits of certain companies during the year
20022003. Table A2.7(a): Data Regarding the Profits of Different Companies
Companies Profits (Rs. in million) ABC Solutions Ltd 16 Power Projects International 23 Hyperactive Systems Pvt. Ltd 34 Infinite Telesystems Ltd 33 Imprint Technologies 21 United Technologies Pvt. Ltd 65 Excel Media Ltd 45 Supersoft 73 Data Links Ltd 32 SDN Telenetworks 22 Solution:
To obtain the median, the data given above should be arranged in an order (ascending
or descending). Table A2.7(b): Assigning Ranks (in descending order)
Companies
Supersoft
United Technologies Pvt. Ltd
Excel Media Ltd
Hyperactive Systems Pvt. Ltd
Infinite Telesystems Ltd
Data Links Ltd
Power Projects International
SDN Telenetworks
Imprint Technologies
ABC Solutions Ltd Profits (Rs. in million)
73
65
45
34
33
32
23
22
21
16 Rank
1
2
3
4
5
6
7
8
9
10 The median for the above data will be the 5th and 6th values
= (33+32)/2
= 32.5.
Therefore the median profit value for the ten companies is Rs 32.5 million. Calculating Median from Grouped Data
For finding the median for the grouped data, the median class has to be identified and
subsequently interpolation has to be used.
The formula for calculating the median is given by 22 Quantitative Methods: Appendix
N +1
) − (F + 1)
[2W + L m
fm
Where,
( L m = lower limit of the median class
f m = frequency of the median class F = Cumulative frequency up to the lower limit of the median class
W = width of the class interval
N = total frequency
Example A2.8:
Calculate the median for the following data. Table A2.8(a): Data Regarding Students Attendance
Percentage of Attendance
010
1020
2030
3040
4050
5060
6070
7080
8090
90100 No. of Students
3
12
16
15
18
22
21
13
6
4 Solution:
The above data represents the attendance percentage of 130 students in a college.
We can calculate the cumulative frequency as given below: Table A2.8(b): Cumulative Frequency Calculation
Percentage of
Attendance No. of
Students (f) Cumulative
Frequency 010 3 3 1020 12 15 2030 16 31 3040 15 46 4050 18 64 5060 22 86 6070 21 107 7080 13 120 8090 6 126 90100 4 130 The total frequency = 130
23 Operations Management
Median can be calculated as (n+1)/2th term
= (130+1)/2 = 65.5
65.5 lies in the 5060 class interval. Therefore 5060 is the median class, of which the
lower limit is 50.
Thus, L m = 50, N=130, f m = 22, F = 64 ,W = 10
N +1
) − (F + 1)
Substituting these values in the formula [ 2W + L m
fm
( 130 + 1
) − (64 + 1)
2
[10 + 50 = 50.227%
22
( Thus 50.23% is the median attendance of the students. THE MODE
The value of the variable that occurs most frequently in a dataset is called as a mode. Calculating Mode from Ungrouped Data
Mode can be calculated from the ungrouped data by identifying the most frequent
values that are occurring in the data set.
Example A2.9:
The data given below consists of marks of 20 students. Table A2.9: Data Regarding Marks of 20 Students
67 65 45 58 48 39 27 37 45 50 20 45 67 78 45 77 30 60 45 68 Solution:
Here 45 is the mode as it has been repeated five times (which is the highest number of
times any number has been repeated in the list) Calculating Mode from Grouped Data
To calculate mode from the grouped data, we use the following formula,
Mode M o = L mo + [ d1w
d1 + d 2 Where,
L mo = lower limit of the modal class
d 1 = frequency of the modal class – frequency of the class just below it
d 2 = frequency of the modal class – frequency of the class just above it
w = width of the modal class. 24 Quantitative Methods: Appendix
Example A2.10:
Calculate the mode for the following data. Table A2.10: Cumulative Frequency Calculation
Marks Secured
010
1020
2030
3040
4050
5060
6070
7080
8090
90100
Solution: Frequency (f)
2
10
8
5
24
30
25
9
8
6 Cumulative Frequency
2
12
20
25
49
79
104
113
121
127 L mo = 50, d 1 = 3025 = 5, d 2 = 3024 = 6, w = 10
Mode M o = L mo + [
= 50+ M o = 50 + [ d1w
d1 + d 2 510 = 50 + 4.545 = 54.545
11 The mode of the data is 54.545 marks.
Note: In cases where the distribution is moderately asymmetrical, the relation between
mean, median and mode is given by the following formula.
Mode = 3Median – 2Mean SUMMARY
• We analyze the data statistically to calculate the average point of the data. • The average point of the data that is located centrally is called as the measure of
central tendency. • There are two types of averages mathematical averages – Arithmetic mean,
Geometric mean and Harmonic mean and Positional averages – Median and
mode. 25 Operations Management Section 3 Measures of Dispersion
In this section, we will discuss:
• Range • Interquartile Range and Quartile Deviation • Mean Deviation • Variance • Standard Deviation • Coefficient of Variation ‘Measures of dispersion’ helps in measuring how spread out a set of data is. That is, it
helps in analyzing how a data set is distributed, or how far each element is from some
measure of central tendency. Some of the important methods of measuring the
dispersion of data are range, interquartile range and quartile deviation, mean
deviation, variance and standard deviation. RANGE
The difference between the value of the smallest observation and the value of the
largest observation present in the distribution is called as the range.
Range = L (largest value)  S (smallest value)
In case of a grouped or continuous frequency distribution, the range is given as the
difference between the upper limit of the highest class and the lower limit of the
lowest class. Coefficient of the Range
For comparing the observations that are in different units, coefficient of the range is
used.
Coefficient of the range = largest value  smallest value
largest value + smallest value = L S
L+S Example A3.1:
Given below are the details regarding the marks secured by students in a class. We
will now try to compute range and coefficient of range for the following data. Table A3.1: Frequency of Marks
Marks
1019
2029
3039
4049
5059 Frequency (f)
6
20
26
34
14 Solution:
Range = 5910 = 49, coefficient of range =
26 L S
L+S = 49/69 = 0.7101 Quantitative Methods: Appendix INTERQUARTILE RANGE AND QUARTILE DEVIATION
Range has its own limitations for measuring the dispersion as it is based on two
extreme observations. It fails to explain the pattern of the data inbetween. The range
calculated based on the middle 50 percent of the observations is called as interquartile
range.
The interquartile range is calculated from the observations obtained after discarding
one quartile of the observations at the lower end and another quartile of the
observations at the upper end of the distribution. If the entire distribution is divided
into four parts (quartiles), the interquartile range is the difference between the first
quartile and the third quartile.
Interquartile range = Q 3 − Q1
Quartile deviation is defined as one half of the interquartile deviation.
Quartile deviation Q.D. = Q 3 − Q1
2 Coefficient of Quartile Deviation
The relative measure of quartile deviation is called coefficient of quartile deviation
given by,
Coefficient of Quartile Deviation = Q 3 − Q1
Q 3 + Q1 Calculation of quartile deviation for ungrouped data is given by,
Lower quartile Q1 =
Upper quartile Q 3 = ( N + 1)
th observation
4
3( N + 1)
th observation, where N = total number of observations.
4 Calculation of quartile deviation for ungrouped data is given by,
1
( N − C)
Q1 = L 1 + 4
×h
f
3
( N − C)
Q3 = L3 + 4
×h
f
Where,
L1 = lower boundary of the first quartile class Q1
L3 = lower boundary of the third quartile class Q 3
N = total cumulative frequency
f = frequency of the quartile class
h = class interval (width)
C = cumulative frequency of the class just above the quartile class
27 Operations Management
Example A3.2:
We will now try to calculate the quartile deviation for the data given below: Table A3.2: Age of Employees in an Organization
Age Frequency (f) Cumulative frequency 2530 30 30 3035 25 55 3540 33 88 4045 34 122 4550 25 147 5055 34 181 5560 19 200 Solution:
Q1 = ¼ (N) = ¼ (200) = 50
This observation will fall in the class (3035)
L1 = 30, C = 30, f = 25, h = 5
Q1 = 30 + 50 − 30
× 5 = 34
25 Q 3 = ¾(200) = 150
It falls in the class (5055)
L3 = 50, C =147, f = 34, h = 5
Q 3 = 50 + 150 − 147
× 5 = 50.44
34 Quartile deviation = Q 3 − Q1
50.44 − 34
=
= 8.22
2
2 Coefficient of Quartile Deviation = Q 3 − Q 1 50.44 − 34
= 0.1946
=
Q 3 + Q1 50.44 + 34 MEAN DEVIATION
Mean deviation is obtained by calculating the absolute deviation of each observation
from mean.
Absolute mean deviation = Absolute mean deviation =
Where,
28 ∑ x −µ
N ∑ x−x
n (in case of total population) (in case of a sample) Quantitative Methods: Appendix
x = value of observation µ = the mean of population
N = number of observations in the population x = sample mean
n = number of observations in the sample
Example A3.3:
Let us now calculate the absolute mean deviation of the number of days worked by 10
laborers during a year.
S. No. 1 Observation
in days 2 3 4 5 6 7 8 9 10 150 200 185 176 189 134 178 185 165 198 Solution:
Table A3.3: Calculation of Absolute Deviation
Observation in days (x) Deviation from mean Absolute Deviation (x x ) S. No x−x 1 134 42 42 2 150 26 26 3 165 11 11 4 176 0 0 5 178 2 2 6 185 9 9 7 185 9 9 8 189 13 13 9 198 22 22 10 200 24 24 N=10 ∑ x = 1760 Mean ( x ) = ∑ x − x = 158 ∑x
= 1760/10 = 176
n Absolute mean deviation = ∑ x−x
n = 158/10 = 15.8 days. Mean Deviation for Grouped Data
Mean deviation for grouped data is given by x = 1
N ∑f x − x = 1
N ∑f d 29 Operations Management
Where,
x = mid value of the class interval
f = corresponding frequency
N = total cumulative frequency
x − x = the absolute value of the deviations d= difference of the values of x from the average x .
Example A3.4:
Calculation of mean deviation for the data given below:
Table A3.4(a): Age Details
Age Frequency (f) 2530 30 3035 25 3540 33 4045 34 4550 25 5055 34 5560 19 Solution:
Table A3.4(b): Calculation of Absolute Mean Deviation
Class
Interval f× x Frequency
(f) Mid Value of
Class
Interval (x) 2530 30 27.5 825 14.425 432.75 3035 25 32.5 812.5 9.425 235.625 3540 33 37.5 1237.5 4.425 146.025 4045 34 42.5 1445 0.575 19.55 4550 25 47.5 1187.5 5.575 139.375 5055 34 52.5 1785 10.575 359.55 5560 19 57.5 1092.5 15.575 295.925 Total 200 297.5 8385 x= ∑f ×x
= 8385/200 = 41.925
N Absolute mean deviation = x = 30 x−x 1
∑ f x − x = 1628.8/200 = 8.144.
N f x−x 1628.8 Quantitative Methods: Appendix VARIANCE
Variance is calculated using the sum of the squared distances between the mean and
each observation divided by the total number of elements in the population. Calculating Variance for Grouped Data
The formula for calculating variance ( σ 2 ) for grouped data is given by,
2 σ= 2
∑ f i (x i − x) N Calculating Variance for Ungrouped Data
The formula for calculating variance ( σ 2 ) for ungrouped data is given by,
σ2 = ∑ (x − x)
N 2 2 = ∑x
− x2
N Where,
σ 2 = Variance
x i = The value of observation
x = Mean N = Total cumulative frequency
f i = frequency of a class STANDARD DEVIATION
Standard deviation or σ is the square root of the average of the squared distances of
the observations from the mean. Calculating Standard Deviation for Ungrouped Data
Standard deviation for ungrouped data is given by,
σ= σ 2 = ∑ (x − x) 2 N Where,
x = the observation x = is the mean
N = ∑ f (total number of observations)
2
2
∑ ( x − x ) = the sum of all values of ( x − x ) σ = standard deviation
31 Operations Management
σ 2 = variance
Properties of standard deviation:
• The value of standard deviation remains constant, if in a series, each of the
observation is increased or decreased by a constant quantity. • In a given distribution if each of the observations is multiplied or divided by a
constant quantity, standard deviation will also be affected likewise. • Standard deviation is the minimum root mean square deviation. Combined Standard Deviation
A combined standard deviation can be computed for two or more groups. It is
calculated using the following formula:
σ12 = n 1 σ1 2 + n 2 σ 2 2 + n 1 d 1 2 + n 2 d 2 2
n1 + n 2 Where,
σ1 = Standard deviation of first group
σ 2 = Standard deviation of second group
d 1 = x1 − x
d2 = x2 − x
x= ( n 1x1 + n 2 x 2 )
( n1 + n 2 ) COEFFICIENT OF VARIATION
Coefficient of variation is a measure of relative dispersion.
Coefficient of variation (%) = Standard deviaton
× 100
Mean Example A3.5:
Let us calculate the variance, standard deviation and coefficient of variance for the
following data:
Table A3.5(a): Companies that have contributed to the society in a particular
year
Contribution (%) x i 10 15 20 25 30 Number of Companies. f i 34 45 55 43 32 Solution:
x= 32 (10 × 34) + (15 × 45) + (20 × 55) + (25 × 43) + (30 × 32)
= 19.856
34 + 45 + 55 + 43 + 32 Quantitative Methods: Appendix
Table A3.5(b): Calculation of Standard Deviation
(x i − x) xi fi 10 34 340 9.856 97.14 3302.76 15 45 675 4.856 23.58 1061.1 20 55 1100 0.144 0.020 1.1 25 43 1075 5.144 26.46 1137.78 30 32 960 10.144 102.9 3292.8 Total 209 4150 x= ∑ fi x i
∑ fi σ2 = fi x i (x i − x )2 f i (x i − x )2 8795.54 = 4150/209 = 19.856 2
∑ fi (x i − x ) ∑ fi = 8795.54/209 = 42.08 Variance of social responsibility of among 209 companies is 42.08.
2 The standard deviation = (σ) =
Coefficient of variation (%) = ∑ fi (x i − x )
=
N 42.08 = 6.48 Standard deviaton
× 100
Mean = 6.48/19.856
= 0.3267 = 32.67%. SUMMARY
• Range is the difference between the value of the smallest observation and the
value of the largest observation present in the distribution. • The interquartile range is the difference between the first quartile and the third
quartile. • Quartile deviation is defined as one half of the interquartile deviation. • Mean deviation is obtained by calculating the absolute deviation of each
observation from mean. • Variance is calculated using the sum of the squared distances between the mean
and each observation divided by the total number of elements in the population. • Standard deviation is the square root of the average of the squared distances of
the observations from the mean. 33 Operations Management Section 4 Concepts of Probability
In this section, we will discuss:
• Basic Probability Concepts • Types of Probability • Probability Rules • Bayes’ Theorem When an event can occur in a finite number of discrete outcomes, the probability of an
event is the ratio of the number of ways in which the event can occur to the total
number of possibilities, assuming that each of them is equally likely can be defined as
probability. Simply putting it, probability is the measure of how likely it is, that some
event will occur. For instance, ‘what is the probability that India will win the cricket
match today’ is a simple case of probability. Managers often have to take decisions
under complex and uncertain situations. The theory of probability reduces the
uncertainty of decision making situations. Today, the concepts of probability are
widely used by business managers for effective decision making. We will discuss
some of the probability concepts, the different types of probabilities, the probability
rules and finally the Bayes’ theorem in this unit. BASIC PROBABILITY CONCEPTS
Certain basic concepts of probability are:
•
• Random experiment: Any act that may be repeated under similar conditions
resulting in a trial which yields an outcome. For example drawing a blue ball
from a bag containing 6 blue 7 red and 9 black balls. • Outcome: The result of a random experiment is called as an outcome • Event: One or more possible outcomes of an experiment • Elementary event: A single possible outcome of an experiment • Compound event: If two or more events occur in relation with each other, then
their simultaneous occurrences is called a compound event. • Favorable event: The number of cases favorable to a trial is called as a favorable
event. For instance the number of favorable cases of drawing a spade from a pack
of 52 cards is 13. • Mutually exclusive events: If the happening of an event precludes the happening
of all other events. For instance, when a die is being thrown, the event of getting
each of the six faces numbered 1 to 6 is mutually exclusive. • 34 Experiment: A process that results in two or more outcomes is called as an
experiment, for example tossing a fair coin. Dependent or independent events: Two events A and B are independent if the fact
that A occurs does not affect the probability of B occurring. For instance, in
tossing a coin, the trial is not affected by the previous or subsequent trials. While
in case of dependent event it is not the case. For instance if there are six blue and
six yellow balls in a bag and if we pick one yellow ball and do not replace it, then Quantitative Methods: Appendix
it will affect the probability of second ball is affected and the event is dependent
on the previous trial.
• Exhaustive events: The total number of possible outcomes. For instance in tossing
a fair coin there are two possible outcomes – Head and a Tail. • Equally likely events: When an event does not occur more often than the others it
is said to be equally likely. For instance in tossing a coin the event of getting a
head or a tail is equal. • Complementary events: The number of unfavorable cases in an experiment. For
instance in throwing a die the favorable event of getting 1 is 1 and the
unfavorable event of getting it is 5. TYPES OF PROBABILITY
There are three basic approaches to probability – classical approach, relative
frequency approach and subjective approach.
Classical approach: The classical approach to probability is to count the number of
favorable outcomes, the number of total outcomes and express the probability as a
ratio of these two numbers. This approach is based on the assumption that each event
is equally likely to occur.
In this approach the probability of happening of an event is given by,
Event (E) = Number of favorable outcomes
Total number of outcomes Probability of an event not happening = 1 probability of happening of an event.
Relative frequency of occurrence approach: In the relative frequency approach, the
probability of an event is proportion of times an event occurs in the longrun when the
conditions are stable.
Probability = Number of trials in which the event occurs
Total number of trials Subjective approach: Subjective probabilities are those assigned to events by the
researchers or managers based on their past experiences or evidences available. In
simple terms the educated guesses of managers in decision making situations is said to
be based on subjective probability. PROBABILITY RULES
Considering A as an event and the probability denoted by P (A) certain rules of
probability exist.
Rule (1): The probability of the entire sample space S is 1 that is P(S) = 1.
Rule (2): The probability of an event A, must be greater than or equal to 0 and less
than or equal to 1.
Rule (3): If A and B are mutually exclusive events, then the probability of (A or B) is
equal to the sum of probabilities of A and B. P (A or B) = P (A) + P (B), since P (A
and B) = 0 as A and B are mutually exclusive. 35 Operations Management Addition Rule
Mutually exclusive events
The addition rule of probability for mutually exclusive events is given by,
P (A or B or C) = P (A) + P (B) + P (C)
As there are 52 possible equally likely outcomes in picking a card from 52 cards
P (A or B or C) = 3
1
1
1
=
+
+
= P (A) + P (B) + P (C)
52 52 52 52 Example A4.1:
Excel electronics produces refrigerators of two colors –white and blue in 150 liter and
350 liter capacities. Currently the company has about 500 refrigerators ready for sale.
The company may receive orders for refrigerators with specific features as mentioned.
The company wants to successfully meet the customers demand so that they do not
lose them to competition. For this the company wants to find out the probability that
the finished goods inventory has products with desirable features. Table A4.1: Inventory of Excel Electronics
Features
150 liters (C)
350 liters (D)
Total White (A)
150
150
300 Blue (B)
125
75
200 Total
275
225
500 Here the sample space S is the set of all the refrigerators in the inventory. We will try
to find the probability of a random selection of a white refrigerator from the inventory
or P (A). We will also find out the probability of randomly selecting a white
refrigerator with a 150 liter capacity.
Solution:
After representing white refrigerators with A and blue ones with B and 150 liters
capacity with C and 350 with D. We can calculate,
P (A) = Number of ways in which A occurs
300
= 0.6
=
Number of outcomes in the sample space 500 The probability of randomly selecting a white refrigerator with a 150 liter capacity is:
Here A and C are not mutually exclusive, since a refrigerator may have both the
capacities. Therefore,
P (A or C) = P(A) + P (C) – P (A and C) is
= 300 275 150 17
+
=
= 0.85.
500 500 500 20 Nonmutually exclusive events
If two events are not mutually exclusive, the probability of one of them occurring is
the sum of the marginal probabilities of the two events minus the joint probabilities of
the occurrence of the events.
P (A or B) = P (A) + P (B) – P (A and B). Where, A and B are not mutually exclusive
events.
36 Quantitative Methods: Appendix
Example A4.2:
From a bag containing 6 yellow and 7 green balls what is the probability that the 4
balls randomly drawn by a man without replacement are green.
Solution:
Favorable Events
P (4 balls drawn are green) =
=
Total Events
7 × 6 × 5 × 4 × 3!
3! × 9!
13 ×12 ×11×10 × 9! = 7 C4 13 C4 7 × 6× 5× 4
13 ×12 ×11×10 = 7!
= 13!
7 143 4!(7 − 4)! 7! = 4!(13 − 4)! 3! =
13!
9! = 0.0489. Example A4.3:
If only two balls are drawn without replacement from the above bag containing 6
yellow and 7 green balls. What is the probability that one is yellow and the other is
green?
Solution:
P (one ball drawn is yellow and the other green) =
= 6×7
13! 2! (13 − 2)! = 42
13! × 2!× 11! = 42
13 × 12 ×11! ( 6 C1 )( 7 C1 )
Favorable Events
=
13
Total Events
C2 × 2!×11! = 42 × 2
13 × 12 = 7
13 = 0.5385. Conditional probability: Independent and Dependent Events
Conditional probability is the probability of the occurrence of an event (A), subject to
the occurrence of a previous event B. It is given by, P (A│B) =P (A).
The conditional probability of event A, given that event B occurred when both A and
B are dependent events, as the ratio of number of elements common in both A and B
P(A and B)
.
to the number of elements in B. It is given by, P (A│B) =
P(B)
Example A4.4:
The data regarding the sales of a product is given below. Calculate the conditional
probability that a unit is sold to the urban consumer and that it is to an individual. Table A4.4: Sales Status of Individual and Industrial Consumers
Sales status Individual sales (B1)
(units) Industry sales
(B2) (units) Total Urban consumers (A1) 365 455 820 Suburban consumers (A2) 210 45 255 Rural consumers (A3) 165 57 222 Total 740 557 1297 Solution:
Let us denote urban consumers as A1. The probability of a unit being sold to an urban
consumer is:
P (A1) = 820/1297 = 0.632.
37 Operations Management
The probability of a sale being made to an individual consumer is:
P (B1) = 740/1297 = 0.5705
Now we have to calculate the probability of a unit being sold to an urban consumer
and that he is an individual consumer. We therefore have to calculate the conditional
probability of event A1 occurring given that event B1 has occurred. The formula for
the conditional probability is:
P (A1B1) = P (A1 and B1)/P (B1)
P (A1 and B1) = 365/1297 = 0.2814
P (A1B1) = 0.2814/0.5705 = 0.4932
The probability is 0.4932 that a product is sold to an urban individual consumer.
Note: This probability can also be calculated directly from the given data in the table.
The conditional probability is
P (A1B1) = 365/740 = 0.4932.
The answer is the same as calculated by using the formula for conditional probability. Multiplication Rule
Dependent events
The joint probability of two events A and B which are dependent is equal to the
probability of A multiplied by the probability of B given that A has occurred.
P (A and B) = P (A) × P(B│A) or P (B and A) = P (B) × P (A│B) Independent events
The multiplication rule of probability states that the probability of happening of an
independent event is given by the product or the probabilities of each event.
P (A ∩ B) = P (A). P (B)
Or P (A and B) = P (A). P (B)
Example A4.5:
The probability of a person listening to a radio program on any given day P(A) is 0.1.
Given that he listens to a radio program, the probability of that person listening to it
the second day in succession P(BA) is 0.5. Find the probability of the person listening
to the program on two successive days.
Solution:
Events A and B are dependent events because B cannot occur unless event A
occurred. The probability of a person listening to the program on two successive days
is:
P (A and B) = P (A) × P (BA) = (0.1) × (0.5) = 0.05.
Thus the probability of a person listening to a program on two successive days is 0.05
or 5% of the time.
Note: Joint probability of several dependent events is equal to the product of the
probabilities of occurrences of the preceding outcomes in the sequence.
P (A and B and C) = P (A) × P (BA) × P (CA and B)……..
Marginal probability in case of dependent events is just the addition of the
probabilities of all the events in which the simple event occurs.
Example A4.6:
Let us find out the probability of the following event:
38 Quantitative Methods: Appendix
A woman has two children, what is the probability that both are girls.
Solution:
Let us draw a probability tree. Refer to Figure A4.6. Figure A4.6: Probability Tree
50% Boy
50% Boy
50% Girl
50% Boy
50% Girl
50% Girl
To find the probability that both are girls, we need to just multiply the path that leads
to two girls.
(0.5) × (0.5) = 0.25 = 25 % BAYES’ THEOREM
Probability helps a great deal in managerial decision making. If the manager wants to
forecast the future sales of the company by manipulating the data available to him in
different forms, probability of future sales under different conditions can be
calculated. In some situations according to the changing environmental factors the
probabilities have to be calculated again. For instance, if a company calculates the
future sales probability of its products, say 2stroke and 4stroke motorcycles, if the
government restricts the sales of 2stroke motorcycles, the company has to recalculate
the future sales probability. This recalculation is called as posterior probability. As the
probabilities can be revised and updated accordingly, as new information is obtained,
the study of probability is of great significance in managerial decision making.
The concept of posterior probability has been proposed by Rev. Thomas Bayes. The
formula for obtaining posterior probability under dependence is:
P (A│B) = P(A and B)
P(B) This is known as Bayes’ theorem that helps in finding the conditional probability of
one event occurring (A), given that another (B) has already occurred.
Example A4.7:
An accounts manager in a company classifies 65 % of the customers as ‘good credit’
and the rest as risky credit based on their credit rating. The customers in the risky
category allow their accounts to go overdue 55 % of the time on average. The
customers in the good credit category allow their accounts to go overdue only 10% of
the time. What percentage of overdue accounts is held by customers in the risky credit
category?
39 Operations Management
Solution:
Consider A = an event that a customer is risky credit P (A) = 0.35
B = an event that a customer is not risky P (B) = 0.65
OD = event that a customer’s account is overdue,
P (OD│A) = 0.55
P (OD│B) = 0.1
P (A│OD) = P(OD  A)P(A)
(0.55)(0.35)
=
= 0.7475
P(OD  A)P(A) + P(OD  B)P(B)
(0.55)(0.35) + (0.1)(0.65) Therefore 74.75% of the overdue accounts are held by customers in the risky credit
category. SUMMARY
Probability is the measure of how likely it is, that some event will occur. Certain basic
concepts of probability are experiment, random experiment, outcome, event,
elementary event, compound event, favorable event, mutually exclusive events,
dependent or independent events, exhaustive events, equally likely events,
complementary events. The three approaches to probability are classical approach,
relative frequency approach and subjective approach. The concept of posterior
probability has been proposed by Rev. Thomas Bayes that is popularly known as
Bayes’ theorem. 40 Quantitative Methods: Appendix Section 5 Probability Distributions
In this section, we will discuss:
• Random Variables • Types of Probability Distributions • The Binomial Distribution • The Poisson Distribution • The Normal Distribution Probability distributions are related to frequency distributions and are considered as
theoretical frequency distributions. Probability distribution can be defined as a curve
that shows all the values that the random variable can take and the likelihood that each
will occur. RANDOM VARIABLES
A random variable can be considered as a numerical result obtained by performing a
nondeterministic experiment to generate a random result or it can be considered as a
variable that takes different values as a result of the outcome of a random experiment.
A random variable is said to be discrete if it is allowed to take limited number of
values. For instance, in rolling a die experiment, the outcome is discrete since the
outcome is confined to {1, 2, 3, 4, 5 or 6} limited values.
A random variable is said to be continuous, if it is allowed to assume any value within
a specified range. For instance if the average height of 1000 children is calculated, the
outcome can be any value in the interval (s) of numbers and is therefore continuous.
Note: Random variables are generally indicated by capital letters and actual values by
small letters. Expected Value of a Random Variable
The expected value of a random variable is obtained by multiplying each value that
the variable can assume by the probability of occurrence of that value and then
summing up these products.
Example A5.1:
Given below is the table containing details regarding the number of students attending
a class in the range between 35 and 40. The number of times each level is reached
during the past 100 days will help the teacher in computing the probability of the
number of students that may be present the next day. 41 Operations Management Table A5.1: Students’ Attendance
No. of Students
(1) No. of Days the Level
was Observed
(2) Probability
Reaching the Level
(3) (1) × (3) 35 21 0.21 7.35 36 11 0.11 3.96 37 21 0.21 7.77 38 13 0.13 4.94 39 22 0.22 8.58 40 12 0.12 4.8
37.4 The expected value of students is 37.4. However, we cannot conclude that 37.4
students will attend the class the next day. This will only help the teacher in analyzing
the student turnout to the class. TYPES OF PROBABILITY DISTRIBUTIONS
There are two types of probability distributions – Discrete probability, Continuous
probability. Binomial and Poisson distributions are discrete probability distributions,
while Normal distribution is a continuous probability distribution. Discrete Probability Distribution
A discrete probability distribution can take a discrete number of values. For example,
if a coin is tossed six times, we can get 2 heads or 3 heads but not 2 ½ heads. Each of
the discrete values has a certain probability of occurrence that is between zero and
one.
The general form of discrete probability distribution for calculating mean is:
µ = x 1 P(X = x 1 ) + x 2 P(X = x 2 ) + ........ + x n P(X = x n )
n µ = ∑ x i P( X = x i )
i =1 Continuous Probability Distribution
In this distribution the variable can assume any value within a given range (continuous
random variable). In this distribution probabilities are measured over intervals and not
single points. That is the area under the curve between two distinct points defines the
probability for that interval. We can therefore say that a continuous probability
distribution is a smooth density curve that models the distribution of a continuous
random variable. THE BINOMIAL DISTRIBUTION
The binomial distribution describes discrete, noncontinuous data resulting from an
experiment that is also known as Bernoulli process. The binomial distribution has an
expected value (or mean, µ ) that can be represented as, µ = n p.
42 Quantitative Methods: Appendix
Variance of the binomial distribution σ 2 = np (1 − p ) = npq Where,
n = total number of Bernoulli trials
p = probability of success
q = probability of failure
Characteristics of Bernoulli process
• Each trail will have only two possible outcomes – success or failure • The probability of the outcome of any trial remains constant over time • The outcome of one trial cannot influence the outcome of another trial The formula for determining ‘r’ successes in ‘n’ trials is given by,
P(X = r ) = n!
(p r × q n −r )
r!(n − r )! Example A5.2:
A fair coin is flipped fifteen times, considering getting heads as success. The
probability of getting 4 heads is given as:
Solution:
p = probability of getting head is 0.5
q = probability of not getting heads =1p = 0.5
r = number of successes = 4
n = number of trials = 15
Probability of getting 4 heads in 15 trials = 15!
4!(15 − 4)! 4
15 − 4
(0.5) (0.5) 15
= (1365) (0.5) = 0.041 Thus the probability of getting 4 heads in 15 trials is 0.041. THE POISSON DISTRIBUTION
The Poisson distribution is applied in situations when an event occurs at random
points in time or space.
The Poisson distribution should meet the following criteria:
(i) Independence: The number of times an event S occurs in any time interval is
independent of the number of times it occurs in any disjoint time interval. (ii) Rate: In a very small time interval, t to t + h (where h is infinitesimally small),
the probability that the event occurs once is approximately λ h (where λ is the
average rate at which the event S occurs per unit of time). (iii) Lack of clustering: The chance of two or more occurrences of S in a very small
interval t to t + h is insignificant in comparison with λ h , the chance of one
occurrence.
The Poisson distribution is given as,
P( X = x ) = λx × e − λ
x!
43 Operations Management
Where,
X = discrete random variable
x = specific value X can take
λ = the mean number of occurrences per interval of time
e = 2.71828
Example A5.3:
Using Poisson distribution, find out the probability that 5 screws are found to be
defective in a box of 100 screws? It is expected that about 3% of the screws in the box
are defective. (e3 = 0.04979)
Solution:
In the given example, let the occurrence of a screw that is defective be success.
Let the number of defective screws be x.
Probability that a screw that is drawn at random is defective, p = 3/100 = 0.03
Total number of screws in the box = 100
λ = n × p = 100 × 0.03 = 3.
x=5
Substituting the values, we get,
P(X = 5) = + 35 × e −3
= 0.1008.
5! THE NORMAL DISTRIBUTION
The normal distribution considers variables like height, weight, marks secured by
students etc. In such cases large number of values seems to cluster near the mean
value and their frequency drops substantially as we move away from the mean. Figure A5: Frequency Curve for the Normal Probability
Distribution
Mean
Median
Mode Lefthand tail
extends indefinitely
but never reaches the
horizontal axis 44 Normal probability
distribution is symmetrical
around a vertical line
erected at the mean Righthand tail
extends indefinitely
but never reaches
the horizontal axis Quantitative Methods: Appendix
The characteristics of the normal probability distribution are:
• The curve has a single peak • The mean of a normally distributed population lies at the center of its normal
curve • The median and the mode are also at the center • The two tails of the normal distribution extend indefinitely and never touch the
horizontal axis. The Standard Normal Distribution
The normal distribution with mean µ = 0 and standard deviation σ = 1 is called as
standard normal distribution. The observation values in a standard normal distribution
are denoted by Z.
For a normal distribution:
80% of the values lie between 1.28 σ and +1.28 σ
95% of the values lie between 1.96 σ and +1.96 σ
98% of the values lie between 2.33 σ and +2.33 σ Standardizing Normal Variables
The standard normal variable is calculated by the formula:
Z= X−µ
σ Where,
X = the value of any random variable
µ = the mean of the distribution of the random
σ = the standard deviation of the distribution
Z = the number of standard deviations from x to the mean of the distribution and is
known as Zscore or standard score.
Example A5.4:
A normal variable X has a mean of 52 and a standard deviation of 14. Find the Z value
corresponding to the X value of 5.
Solution:
Z= X −µ σ = − 5 − 52
14 = 4.07. Example A5.5:
The mean and the standard deviation of a normal variable are 10 and 5, respectively.
What is the probability that the normal variable will take a value in the interval 0.2 to
19.8?
Solution:
Mean (µ) = 10
Standard deviation (σ) = 5
45 Operations Management
Probability (0.2 < X < 19.8) = Probability 0.2 − 10 ≤ Z ≤ 19.8 − 10 5 5 = Probability (1.96 < Z < 1.96)
= 95%
[Because 95% of the area under the standard normal curve lies in the interval 1.96 to
1.96]
We can see this from the Normal Table:
Area under the standard normal curve between 0 and 1.96 is 0.4750.
Due to symmetry of the standard normal distribution, area under the curve between –
1.96 and + 1.96 is twice the area under the curve between 0 and + 1.96.
Probability (–1.96 < Z < + 1.96) = 0.95 or 95%
Any normal variable can be converted into a standard normal variable as illustrated
above. Hence, we can use the standard normal distribution table to find the probability
that the variable will take a value within any given interval. SUMMARY
•
• Random variable is a variable that takes different values as a result of the
outcome of the random experiment. • A random variable is said to be discrete if it is allowed to take limited number of
values. • A random variable is said to be continuous, if it is allowed to assume any value
within a specified range. • 46 Probability distribution can be defined as a curve that shows all the values that the
random variable can take and the likelihood that each will occur. Discrete probability and continuous probability are the two types of probability
distributions, Binomial and Poisson distributions are discrete probability
distributions, while Normal distribution is a continuous probability distribution. ...
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This note was uploaded on 07/20/2010 for the course ICFAI CFA taught by Professor Cfa during the Fall '09 term at Indian School of Business.
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