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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 × 10-1 • 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=4-2 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 two-letter 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+(n-1)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 + (n-1) d The sum to n terms of a series in Arithmetic progression is given by, Sn = n [2a + (n-1) 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 + (n-1) d], 2 S 46 = 46 [2(2) + (46-1) 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)25-1 = 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 + (n-1) 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 + (n-1) 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 n-dimension 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 co-ordinates 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 decision-making. 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 21-30 4 31-40 6 41-50 7 51-60 11 61-70 10 71-80 7 81-90 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) 21-30 31-40 41-50 51-60 61-70 71-80 81-90 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 short-cut 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 21-30 -1 31-40 -2 41-50 -3 51-60 0 61-70 1 71-80 2 81-90 3 The formula for calculating the sample mean using the short-cut 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 (Short-Cut Method) Class (Marks) Mid Point (x) Code (u) Frequency (f) (u) × (f) 21-30 25.5 -1 4 -4 31-40 35.5 -2 6 -12 41-50 45.5 -3 7 -21 51-60 55.5 0 11 0 61-70 65.5 1 10 10 71-80 75.5 2 7 14 81-90 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 = 1.0565-1 = 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 2002-2003. 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 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 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 0-10 3 3 10-20 12 15 20-30 16 31 30-40 15 46 40-50 18 64 50-60 22 86 60-70 21 107 70-80 13 120 80-90 6 126 90-100 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 50-60 class interval. Therefore 50-60 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 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 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 = 30-25 = 5, d 2 = 30-24 = 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 10-19 20-29 30-39 40-49 50-59 Frequency (f) 6 20 26 34 14 Solution: Range = 59-10 = 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 in-between. 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 25-30 30 30 30-35 25 55 35-40 33 88 40-45 34 122 45-50 25 147 50-55 34 181 55-60 19 200 Solution: Q1 = ¼ (N) = ¼ (200) = 50 This observation will fall in the class (30-35) 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 (50-55) 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) 25-30 30 30-35 25 35-40 33 40-45 34 45-50 25 50-55 34 55-60 19 Solution: Table A3.4(b): Calculation of Absolute Mean Deviation Class Interval f× x Frequency (f) Mid Value of Class Interval (x) 25-30 30 27.5 825 14.425 432.75 30-35 25 32.5 812.5 9.425 235.625 35-40 33 37.5 1237.5 4.425 146.025 40-45 34 42.5 1445 0.575 19.55 45-50 25 47.5 1187.5 5.575 139.375 50-55 34 52.5 1785 10.575 359.55 55-60 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 long-run 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 Non-mutually 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 Sub-urban 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 (A1|B1) = P (A1 and B1)/P (B1) P (A1 and B1) = 365/1297 = 0.2814 P (A1|B1) = 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 (A1|B1) = 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(B|A) 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 (B|A) = (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 (B|A) × P (C|A 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 2-stroke and 4-stroke motorcycles, if the government restricts the sales of 2-stroke 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 non-deterministic 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, non-continuous 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 =1-p = 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. (e-3 = 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 Left-hand tail extends indefinitely but never reaches the horizontal axis 44 Normal probability distribution is symmetrical around a vertical line erected at the mean Right-hand 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 Z-score 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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