Fundamentals of business mathematics and statistics ICAI

FOUNDATION : PAPER - FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS STUDY NOTES The Institute of Cost Accountants of India CMA Bhawan, 12, Sudder Street, Kolkata - 700 016 FOUNDATION First Edition : January 2013 Second Edition : September 2014 Published by : Directorate of Studies The Institute of Cost Accountants of India (ICAI) CMA Bhawan, 12, Sudder Street, Kolkata - 700 016 www.icmai.in Printed at : Repro India Limited Plot No 02, T.T.C MIDC Industrial Area, Mahape, Navi Mumbai 400 709, India Website : www.reproindialtd.com Copyright of these Study Notes is reserved by the Institute of Cost Accountants of India and prior permission from the Institute is necessary for reproduction of the whole or any part thereof Syllabus PAPER 4: FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS (FBMS) Syllabus Structure A B Fundamentals of Business Mathematics Fundamentals of Business Statistics 40% 60% A 40% B 60% ASSESSMENT STRATEGY There will be written examination paper of three hours OBJECTIVES To gain understanding on the fundamental concepts of mathematics and statistics and its application in business decisionmaking Learning Aims The syllabus aims to test the student’s ability to: Understand the basic concepts of basic mathematics and statistics    Identify reasonableness in the calculation Apply the basic concepts as an effective quantitative tool    Explain and apply mathematical techniques Demonstrate to explain the relevance and use of statistical tools for analysis and forecasting   Skill sets required Level A: Requiring the skill levels of knowledge and comprehension CONTENTS Section A: Fundamentals of Business Mathematics Arithmetic Algebra Calculus Section B: Fundamentals of Business Statistics Statistical representation of Data Measures of Central Tendency and Dispersion Correlation and Regression Index Numbers Time Series Analysis- basic applications including Moving Average Probability 10 Theoretical Distribution SECTION A: FUNDAMENTALS OF BUSINESS MATHEMATICS [40 MARKS] Arithmetic (a) (b) Simple and Compound interest including application of Annuity Ratios and Proportions (c) Bill Discounting and Average Due Date (d) Mathematical reasoning – basic application Algebra (a) Set Theory and simple application of Venn Diagram (b) Variation, Indices, Logarithms (c) Permutation and Combinations – basic concepts 40% 60% (d) Linear Simultaneous Equations ( variables only) (e) Quadratic Equations (f) Solution of Linear inequalities (by geometric method only) (g) Determinants and Matrices Calculus (a) Constant and variables, Functions, Limit & Continuity (b) Differentiability & Differentiation, Partial Differentiation (c) Derivatives – First order and Second order Derivatives (d) Maxima & Minima – without constraints and with constraints using Lagrange transform (e) Indefinite Integrals: as primitives, integration by substitution, integration by part (f) Definite Integrals: evaluation of standard integrals, area under curve SECTION B: FUNDAMENTALS OF BUSINESS STATISTICS [60 MARKS] Statistical Representation of Data (a) Diagrammatic representation of data (b) Frequency distribution (c) Graphical representation of Frequency Distribution – Histogram, Frequency Polygon, Ogive, Pie-chart Measures of Central Tendency and Dispersion (a) Mean, Median, Mode, Mean Deviation (b) Quartiles and Quartile Deviation (c) Standard Deviation (d) Co-efficient of Variation, Coefficient of Quartile Deviation Correlation and Regression (a) Scatter diagram (b) Karl Pearson’s Coefficient of Correlation (c) Rank Correlation (d) Regression lines, Regression equations, Regression coefficients Index Numbers (a) (b) Problems involved in construction of Index Numbers (c) Uses of Index Numbers Methods of construction of Index Numbers Time Series Analysis – basic application including Moving Average (a) Moving Average Method (b) Method of Least Squares Probability (a) Independent and dependent events; Mutually exclusive events (b) Total and Compound Probability; Baye’s theorem; Mathematical Expectation 10 Theoretical Distribution (a) Binomial Distribution, Poisson Distribution – basic application (b) Normal Distribution – basic application I FUNDAMENTALS OF LAWS AND ETHICS Content SECTION - A BUSINESS MATHEMATICS Study Note : Arithmetic 1.1 Ratio & Proportion 1.1 1.2 Simple & Compound Interest (Including Application of Annuity) 1.6 1.3 Discounting of Bills and Average Due Date 1.19 1.4 Mathematical Reasoning - Basic Application 1.37 Study Note : Algebra 2.1 Set Theory 2.1 2.2 Inequations 2.15 2.3 Variation 2.18 2.4 Logarithm 2.23 2.5 Laws of Indices 2.33 2.6 Permutation & Combination 2.39 2.7 Simultaneous Linear Equations 2.51 2.8 Matrices & Determinants 2.56 Study Note : Calculus 3.1 Function 3.1 3.2 Limit 3.6 3.3 Continuity 3.17 3.4 Derivative 3.23 3.5 Integration 3.57 SECTION - B STATISTICS Study Note : Statistical Representation of Data 4.1 Diagramatic Representation of Data 4.1 4.2 Frequency Distribution 4.5 4.3 Graphical Representation of Frequency Distribution 4.10 Study Note : Measures Of Central Tendency and Measures of Dispersion 5.1 Measures of Central Tendency or Average 5.1 5.2 Quartile Deviation 5.42 5.3 Measures of Dispersion 5.44 5.4 Coefficient Quartile & Coefficient variation 5.61 Study Note : Correlation and Regression 6.1 Correlation & Co-efficient 6.2 Regression Analysis 6.1 6.20 Study Note : Index Numbers 7.1 Uses of Index Numbers 7.1 7.2 Problems involved in construction of Index Numbers 7.2 7.3 Methods of construction of Index Numbers 7.2 7.4 Quantity Index Numbers 7.12 7.5 Value Index Number 7.13 7.6 Consumber Price Index 7.13 7.7 Aggregate Expenditure Method 7.14 7.8 Test of Adequacy of the Index Number Formulae 7.15 7.9 Chain Index Numbers 7.19 7.10 Steps in Construction of Chain Index 7.20 Study Note : Time Series Analysis 8.1 Definition 8.1 8.2 Components of Time Series 8.1 8.3 Models of Time Series Analysis 8.2 8.4 Measurement of Secular Trend 8.3 8.5 Method of Semi Averages 8.3 8.6 Moving Average Method 8.3 8.7 Method of Least Squares 8.6 Study Note : Probability 9.1 General Concept 9.1 9.2 Some Useful Terms 9.1 9.3 Measurement of Probability 9.3 9.4 Theorems of Probability 9.8 9.5 Bayes’ Theorem 9.11 9.6 Odds 9.15 Study Note 10 : Theoretical Distribution 10.1 Theoretical Distribution 10.1 10.2 Binomial Distribution 10.1 10.3 Poisson Distribution 10.13 10.4 Normal Distribution 10.19 Section - A BUSINESS MATHEMATICS p(x)= l0Cx(l/5)x(4/5)1–x (i) Probability that all 10 are defective is p(10)=l0C10(l/5)10(4/5)°= (ii) 510 Probability that all 10 are good = 1– P (all are defective) = 1− 510 (iii) Probability that at least one is defective is given by the sum of probabilities, viz p(1) + p(2) + p(3)+ +p(10) or – p(0) = 1– l0C10(1/5)0(4/5)10 = 1– (4/5)10 (iv) Probability of at the most three defective items is p(X < 3) = p(X = 0)+ p(X = 1) + (X = 2) + p(X = 3) = p(0)+p(1)+p(2)+p(3) = l0C0(1/5)0(4/5)10+ l0C1(1/5)l (4/5)9 + l0C2(1/5)2(4/5)8 + l0C3(1/5)3(4/5)7 = 1(.107)+10(.026)+45(.0067)+120(.0016) = 0.107 + 0.26 + 0.30 + 0.192 = 0.859 Example : The incidence of occupational disease in an industry is such that the workmen have a 20% chance of suffering from it What is the probability that out of six workmen, or more will contact the disease ? Solution : Let X represent the number of workers suffering from the disease Then the possible values of X are 0, 1, 2, p = p ( worker suffer from a disease) = 20/100 = 1/5 q = – (1/5) = 4/5, n = Using the formula for binomial distribution, we have p (X) = 6Cx(1/5)x(4/5)6–x the probability that or more workers contact the disease is p(X>4) = p(4) + p(5) + p(6) = 6C4(1/5)4 (4/5)2 + 6C5(1/5)5 (4/5)+ 6C6(1/5)6 = 15 ×16 6× 265 + + = 15625 15625 15625 15625 = 0.016 FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 10.7 Theoretical Distribution Example : The probability that an evening college student will graduate is 0.4 Determine the probability that out of students (a) none, (b) one, and (c) atleast one will graduate Solution : ⎛ ⎞⎟ ⎛ ⎞⎟ ⎜ ⎜ n = p = 0.4 ⎜⎜ ⎟⎟ ; q = 0.6 ⎜⎜ ⎟⎟ ⎝10 ⎠ ⎝10 ⎠ r ⎛4⎞ P(r ) = Cr pr qn−r = Cr ⎜⎜ ⎟⎟ ⎜⎝10 ⎠⎟ n−r ⎛ ⎞⎟ ⎜⎜ ⎟ ⎝⎜10 ⎠⎟ (a) The probability of zero success ⎛ ⎞⎟ = 5C0 ⎜⎜⎜ ⎟⎟ ⎝10 ⎠ ⎛ ⎞⎟ ⎜⎜ ⎟ ⎝⎜10 ⎠⎟ ⎛ ⎞⎟ = 1×⎜⎜⎜ ⎟⎟ ⎝10 ⎠ = 0.078 (b) The probability of one success ⎛ ⎞⎟ ⎛ ⎞⎟ = C1 ×⎜⎜⎜ ⎟⎟×⎜⎜⎜ ⎟⎟ ⎝10 ⎠ ⎝10 ⎠ = × ×(6)4 = 0.259 (c) The probability of atleast one success = 1– probability of no success = 1– 0.078 = 0.922 Example : 12 coins are tossed What are the probabilities in a single tossing getting : (1) or more heads, (2) less than heads, (3) atleast heads If the 12 coins are tossed 4096 times, or (4) how many occasions would you expect these to be : (a) less than heads, (b) atleast heads, (c) exactly heads Solution : (1) The probability of getting or more heads is : P(9) + P(10) + P(11) + P(12) P(9) = 12 ⎛ 1⎞ C9 ⎜⎜ ⎟⎟ ⎝⎜ ⎠⎟ ⎛ 1⎞⎟ ⎜⎜ ⎟ ⎝⎜ ⎠⎟ 10.8 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS 10 P(10) = 12 P(11) = 12 P(12) = 12 P( ≥ 9) = 12 ⎛ 1⎞ C10 ⎜⎜ ⎟⎟⎟ ⎝⎜ ⎠ 11 ⎛ 1⎞ C11 ⎜⎜ ⎟⎟ ⎜⎝ ⎠⎟ ⎛ ⎞⎟ ⎜⎜ ⎟ ⎝⎜ ⎠⎟ ⎛ 1⎞⎟ ⎜⎜ ⎟ ⎝⎜ ⎠⎟ 12 ⎛ 1⎞ C12 ⎜⎜ ⎟⎟⎟ ⎜⎝ ⎠ ⎛ 1⎞ C9 ⎜⎜ ⎟⎟ ⎜⎝ ⎠⎟ 10 ⎛ ⎞⎟ ⎛ ⎞ ⎜⎜ ⎟ + 12 C ⎜⎜ ⎟⎟ 10 ⎜ ⎟ ⎜⎝ ⎠⎟ ⎝2⎠ 11 ⎛ 1⎞⎟ ⎛ ⎞ ⎜⎜ ⎟ + 12 C ⎜⎜ ⎟⎟ 11 ⎜ ⎟ ⎜⎝ ⎠⎟ ⎝2⎠ 12 ⎛ 1⎞⎟ 12 ⎛ ⎞ ⎜⎜ ⎟ + C ⎜⎜ 1⎟⎟ 12 ⎜ ⎟ ⎜⎝ ⎠⎟ ⎝ 2⎠ ⎧ ⎫ ⎪⎛ ⎞12 12 ⎪ ⎪ ⎜ ⎟⎟ ( C + 12 C + 12 C + 12 C )⎪ ⎨ ⎜ 10 11 12 ⎬ = ⎪⎜⎝ ⎠⎟ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ = (220 + 66 + 12 + 1) 4096 299 or7.3% 4096 The probability of getting less than heads: P(>3) = P(0) + P(1) + P(2) = (2) 12 = 12 12 12 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ C0 ⎜⎜ ⎟⎟⎟ + 12 C1 ⎜⎜ ⎟⎟⎟ + 12 C2 ⎜⎜ ⎟⎟⎟ ⎝⎜ ⎠ ⎝⎜ ⎠ ⎝⎜ ⎠ ⎧⎛ ⎞12 ⎪ ⎪⎫ ⎪⎨⎜ ⎟⎟ ( 12 C + 12 C + 12 C )⎪⎬ ⎜ = ⎪⎜⎝ ⎠⎟ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ (3) = (1+ 12 + 66) 4096 = 79 4096 Probability of atleast heads : 12 Probability of heads = Probability of heads = 12 ⎛ 1⎞ 495 C8 ⎜⎜ ⎟⎟⎟ = = 0.1208 ⎜⎝ ⎠ 4096 299 = 0.073 4096 Probability of atleast heads = 0.1208 + 0.073 = 0.1938 (4) (a) Probability of less than heads out of 4096 times 4096 × 79 =79 Times 4096 FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 10.9 Theoretical Distribution (b) probability of atleast heads = l – P(0)+P(l) 12 12 ⎛ 1⎞⎟ ⎛ 1⎞⎟ 12 12 = 1− C0 ⎜⎜⎜ ⎟⎟ + C1 ⎜⎜⎜ ⎟⎟ ⎝2⎠ ⎝2⎠ = 1− 13 4083 = 4096 4096 The number of occasions of getting at least heads in 4083 = 4083 times 4096 4096 tosses 4096 × (c) Probability of heads: 12 ⎛ 1⎞⎟ P(6) = C6 ⎜⎜⎜ ⎟⎟ ⎝2⎠ 12 = 924 4096 The number of occasions of getting exactly heads = 4096 × 924 4096 = 924 times Example : The mean, of a binomial distribution is 20 and standard deviation is Find out n, p and q Solution: Mean = 20 (np) Standard Deviation = npq = npq = 16 npq 16 = or 0.8 np 20 p = – 0.8 = q = n = 20 = 100 0.2 Example 10 : Obtain the binomial distribution for which mean is 10 and the variance is Solution : The mean of binomial distribution m = np = 10 The variance= npq = 10.10 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS q = npq = = 0.5 np 10 p = – q i.e., – = 0.5 np = 10 i.e., n × 0.5 = 10 n = 10 = 20 0.5 Therefore the required binomial distribution is 20 ⎛ 1⎞⎟ (p + q) = (0.5 + 0.5) or ⎜⎜⎜ + ⎟⎟ ⎝2 2⎠ n 20 Example 11 : Obtain the binomial distribution for which the mean is 20 and the variance is 15 Solution : The variance= npq = 15 Mean = np = 20 npq 15 = = np 20 q = q = 1− = 4 np = 20 i.e n× = 20 n = 20 20 × = = 80 1/ The binomial distribution is : 80 (p + q) n ⎛ ⎞⎟ = ⎜⎜⎜ + ⎟⎟ ⎝4 4⎠ Fitting of Binomial Distribution The probability of 0, 1, 2, success would be obtained by the expansion of (q+ p)n Suppose this experiment is repeated for N times, then the frequency of r success is; N × P(r) = N × nCrqn–r pr FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 10.11 Theoretical Distribution Putting r = 0, 1, n, we can get the expected or theoretical frequencies of the binomial distribution as follows : Number of Success(r) Expected or theoretical frequency NP ( r ) Nqn N nC1 qn–1 p N nC2 qn–2 p2 r N nCr qn–r pr n N pn Example 12 : coins are tossed at a time, 256 times Find the expected frequencies of success (getting a head) and tabulate the result obtained Solution : p= 1 ; q = ; n = : N = 256 2 The probability of success r times in n -trials is given by nCr pr qn−r (or) P (r) = nCr pr qn−r r ⎛ ⎞⎟ n = Cr ⎜⎜⎜ ⎟⎟ ⎝2⎠ 8− r ⎛ 1⎞⎟ ⎜⎜ ⎟ ⎝⎜ ⎠⎟ ⎛ ⎞⎟ = Cr ⎜⎜⎜ ⎟⎟ ⎝2⎠ Frequencies of 0, 1, 2, successes are : Success Expected Frequency ⎛ ⎞ 256 ⎜⎜ × C0 ⎟⎟⎟ ⎜⎝ 256 ⎠ 1 ⎛ ⎞ × C1⎟⎟⎟ 256 ⎜⎜ ⎜⎝ 256 ⎠ ⎛ ⎞ 256 ⎜⎜ × C2 ⎟⎟⎟ ⎜⎝ 256 ⎠ 28 ⎛ ⎞ 256 ⎜⎜ × C3 ⎟⎟⎟ ⎜⎝ 256 ⎠ 56 ⎛ ⎞ 256 ⎜⎜ × C4 ⎟⎟⎟ ⎜⎝ 256 ⎠ 70 10.12 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS ⎛ ⎞ 256 ⎜⎜ × C5 ⎟⎟⎟ ⎜⎝ 256 ⎠ 56 ⎛ ⎞ 256 ⎜⎜ × C6 ⎟⎟⎟ ⎜⎝ 256 ⎠ 28 ⎛ ⎞ 256 ⎜⎜ × C7 ⎟⎟⎟ ⎜⎝ 256 ⎠ 8 ⎛ ⎞ 256 ⎜⎜ × C8 ⎟⎟⎟ ⎜⎝ 256 ⎠ 10.3 POISSON DISTRIBUTION Poisson distribution was derived in 1837 by a French mathematician Simeon D Poisson (1731-1840) In binomial distribution, the values of p and q and n are given There is a certainty of the total number of events; in other words, we know the number of times an event does occur and also the times an event does not occur, in binomial distribution But there are cases where p is very small and n is very large, then calculation involved will be long Such cases will arise in connection with rare events, for example Persons killed in road accidents The number of defective articles produced by a quality machine, The number of mistakes committed by a good typist, per page The number of persons dying due to rare disease or snake bite etc The number of accidental deaths by falling from trees or roofs etc In all these cases we know the number of times an event happened but not how many times it does not occur Events of these types are further illustrated below : It is possible to count the number of people who died accidently by falling from trees or roofs, but we not know how many people did not die by these accidents It is possible to know or to count the number of earth quakes that occurred in an area during a particular period of time, but it is, more or less, impossible to tell as to how many times the earth quakes did not occur It is possible to count the number of goals scored in a foot-ball match but cannot know the number of goals that could have been but not scored It is possible to count the lightning flash by a thunderstorm but it is impossible to count as to how many times, the lightning did not flash etc Thus n, the total of trials in regard to a given event is not known, the binomial distribution is inapplicable, Poisson distribution is made use of in such cases where p is very Small We mean that the chance of occurrence of that event is very small The occurrence of such events is not haphazard Their behaviour can also be explained by mathematical law Poisson distribution may be obtained as a limiting case of binomial distribution When p becomes very small and n is large, Poisson distribution may be obtained as a limiting case of binomial probability distribution, under the following conditions : p, successes, approaches zero (p → 0) np = m is finite The poisson distribution is a discrete probability distribution This distribution is useful in such cases where the value of p is very small and the value of n is very large Poisson distribution is a limited form of binomial FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 10.13 Theoretical Distribution distribution as n moves towards infinity and p moves towards zero and np or mean remains constant That is, a poisson distribution may be expected in cases where the chance of any individual event being a success is small The Poisson distribution of the probabilities of Occurrence of various rare events (successes) 0, 1, 2, given below: Number of success Probabilities p (X) e–m me–m m2 e−m 2! m3 e−m 3! r m r e−m r! n mne−m n! e = 2.71828 m = average number of occerence of given distribution The Poisson distribution is a discrete distribution with a parameter m The various constants are : Mean =m=p Stanlald Deviation = Variance =m m Example 13 : A book contains 100 misprints distributed randomly throughout its 100 pages What is the probability that a page observed at random contains atleast two misprints Assume Poisson Distribution Solution: Total Number of misprints 100 = =1 Total number of page 100 Probability that a page contain at least two misprints m = P ( r ≥ 2) = – [p (0) + p (1)] p(r) = mr e−m r! 10 e−1 1 = e−1 = = p(0) = 0! e 2.7183 10.14 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS p(1) = 11e−1 11e−1 1 = = e−1 = = 1! 1! e 2.7188 1 + = 736 2.7183 2.7183 p(0) + p(1) = – [p(0)+p(1)] = – 0.736 = 0.264 Example 14 : If the mean of a Poisson distribution’s 4, find (1) S.D (4) m3 (5) m4 Solution : m=4 S.D = µ3 = m = µ4 = m + m2 = + 48 = 52 m= 4=2 Example 15 : Find the probability that at most defective bolts will be found in a box of 200 bolts, if it is known that 2% of such bolts are expected to be defective (e-4=0.0183) Solution : m =n×p = 200 × 02 =4 P(o) = P (0)+P (1)+P (2)+P (3)+P (4)+P (5) ⎛ 42 43 44 45 ⎞⎟ −4 ⎜ = e ⎜⎜⎜1+ + 2! + 3! + 4! + 5! ⎟⎟⎟ ⎝ ⎠ = e–4(1+4+8+10.67+8.53) ⎡ 1⎤ ⎡ 1 ⎤ ⎥ + = 1− ⎢⎢ + ⎥⎥ + = −1⎢⎢ ⎣ e e⎦ ⎣ 2.718 2.178 ⎥⎦ = – 0.736 = 0.264 Example 16 : One fifth per cent of the blades produced by a blade manufacturing factory turn out to be defective The blades are supplied in packets of 10 Use poisson distribution to calculate the approximate number of packets containing no defective, one defective and two defective blades respectively in a consignment of 100,000 packets (Given e02 = 9802) Solution : Here p = , n= 10 500 m= np = ×10 = 0.02 500 FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 10.15 Theoretical Distribution Using the formula for Poisson distribution, the probability of x defective blades is x p(x) = e0.02 (0.02) x! The frequencies of 0, 1, 2, defective blades given by x f(x) = 1,00,000 x e−0.02 (0.20) X! Number of packets with no defective blade = 1,00,000 × e–0.02 = 1,00,000 × 0.9802 = 98,020 Number of packets with one defective blades P (1) = e−0.02 (.02) x ! = 1,00,000 × e–0.02 × 0.02 = 98,020 × 02 = 98,020 × 100 = 1960.4 = 1960 Number of packets with two defective blades is = 1,00,000 × e–0.02× (0.02) = 98020 × 0002 = 19.6040 = 20 Example 17 : It is known from past experience that in a certain plant there are on the average industrial accidents per month Find the probability that in a given year there will be less than aecident Assume Poisson distribution Solution : m =4 p(x = r) = e−m mr e−4 r = r! r! The required probability that there will be less than accidents is given as : p(x