SUBMITTED TO: Mrs. BIKRAMJEET KAUR SUBMITTED BY: KARINA ROLL NO. 170 CLASS: B.COM 1st SUBJECT: BUSINESS MATHEMATICS AND STATISTICS

2012 APRIL QUESTION PAPER SECTION - A QUESTION NO.1 (a) Difference between Idempotent Matrix and Involutory Matrix. ANS: 1. IDEMPOTENT MATRIX A square matrix A is said to be idempotent if A² = A. 2. INVOLUTORY MATRIX A square matrix A is said to be involutary if it is suare is equal to the identity matrix i.e. A² = I.

QUESTION NO.1 (b) Discuss the functions of STATISTICS. ANS: FUNCTIONS EXPRESSION OF FACTS IN NUMBERS: One of the principal function of statistics is to express facts relating to different phenomena in numbers. Statement is vasted with certainty when facts are expressed in numbers. SIMPLE PRESENTATION: Another function of statistics is to present complex data in a simple form, so that it becomes easy to comprehend. Statstics renders comple data very simple by expressing it in terms of aggregate, average. Percentage, graphs and diagrams.

ENLARGES INDIVIDUAL KNOWLEDGE AND EXPERIENCE: The principal function of statistics is that it enlarges individuals knowledge and expertise. One can understand clearly and precisely such concepts as national income, population, industrial production, and the like. IT COMPARES FACTS: Another function of statistics is to compare the data relating to facts. Data have no meaning unless these are compared and inter-related.

IT FACILITATES POLICY FORMULATION: Precise nature of each problem can be ascertained from the analysis and interpretation of data. As a result of it, some policy may be formulated. IT HELPS OTHER SCIENCES IN TESTING THEIR LAWS: Many laws of economics, namely, law of demand. Law of supply have been verified with the help of statistics. IT ESTABLISHES RELATIONSHIP BETWEEN FACTS: Statistics also establish relationship between two or more facts. Tools of correlation of statistics tell if two facts have any relation between them or not and what kind of relation it has?

IT HELPS IN FORECASTING: Statistics helps in forecasting changes in future with regard to a problem. IT ENABLES REALISATION OF MAGNITUDE: Statistics enables realisation of magnitude of a problem. 10. PRESENTATION OF DATA IN CONDENSED FORM: Primary data are very much complex and haphazard. Such complex data make it impossible to draw any conclusion.

QUESTION NO.1 (c) Explain properties of COEFFICIENT OF CORRELATION. ANS: PROPERTIES OF COEFFICIENT OF CORRELATION CHANGE OF ORIGIN AND SCALE: Shifting the origin or scale does not affect in any way the value of correlation coefficient. Coefficient of correlation is independent of the change of origin and scale. GEOMETRIC MEAN OF REGRESSION COEFFICIENTS: Correlation of coefficient is the geometric mean of the regression coefficients byx and bxy. Symbolically: r = √byx . bxy

If X and Y are independent varible, than coefficient of correlation is zero but the converse is not necessary true. PURE NUMBER: ‘r’ is a pure number and is independent of the units of measurements. This implies that even if the two varibles are expressed in two different units of measurements. SYMMETRIC: The coefficient of correlation between two varibles X and Y is symmetric i.e. Rxy = Ryx. It means that either we compute the values of correlation coefficient between X and Y or between Y and X, the coefficient of correlation remains the same.

QUESTION NO. 1 (e) The difference between Compound Interest and Simple Interest on a certain sum of money at 10% per annum for 2 years is Rs.40. Find sum. ANS: Suppose the sum = Rs.100 P = 100, n = 2, r = 10% S.I. at 10% for 2 years = p×r×n÷100 = 100×10×2÷100 = Rs. 20 Let A be the C.I, then A = P(1+r/100)n A = 100(1+10/100)2 A = Rs. 121 C.I = A – P C.I = 121-100 C.I = 21 Difference b/w C.I & S.I on Rs.100 = 21-20= 1P=100 if the difference b/w C.I & S.I is Rs.100 then sum = Rs.100 if the difference b/w C.I & S.I is 40, then sum = 100×40 = Rs. 4000

QUESTION NO.1 (f) The mean and standard deviation of a set of 100 observations were worked out as 40 and 5 respectively by a computer which by mistake took the value 50 in place of 40 for one observation. Find the correct mean and variance. ANS: Given: N =100, Mean = 40, S.D =5 Calculation of Correct Mean Mean = ∑X/N 40 = ∑X/100 Incorrect ∑X = 4000 Correct ∑X = 4000+Correct Item- Incorrect item Correct ∑X = 4000+40-50 = 3990 Correct Mean = 3990/100 = 39.9 Calculation of Correct S.D S.D = √∑X²/N –(mean)² ∑X² = 162500 Correct ∑X ² = 1625+(Correct item) ² - (Incorrect item)² Correct ∑X² = 1625+(40)² - (50)² = 161600 Correct S.D = √16100/100 – (39.9) ² = 4.89 Variance = 4.89 × 4.89 = 23.91

SECTION - B QUESTION NO. 2 Explain how would you identify the cases of redundant constraints, no solution, multiple solution and unbounded solution of a LINEAR PROGRAMMING. ANS: 1. INFEASIBLE SOLUTION OR NO SOLUTION Sometimes the system of constraints in a LPP has no common point which satisfies all the constraints. In such cases, the LPP is said to have no feasible solution or infesible solution.

UNBOUNDED SOLUTION A linear programming problem may have unbounded solution which means that it has no limit on constraints. It simply means that the common feasible region is not bounded in any respect. The primary varibles can take any value in the unbounded region. MULTIPLE OPTIMUM SOLUTION In certain linear programming problems, situation may arise when there is the possibly of more than one optimum solution.

REDUDANT CONTRANT A linear programming problem may have some time a redundant constraint. A redundant constrain is simply one that does not affect the feasible solution region. QUESTION NO. 3 For a group containing 100 observations, the arithmetic mean and standard deviation are 8 and √10.5. For 50 observations selected from these 100 observations the mean and standard deviation are 10 and 2 . Find mean and the S.D of the other half.

ANS: Given: N = 100 , Combined mean = √10. 5 N₁ = 50, Mean₁ = 10, S ANS: Given: N = 100 , Combined mean = √10.5 N₁ = 50, Mean₁ = 10, S.D₁ = 2 N₂ = 100 – N₁ = 100 – 50 = 50 Combined mean₁₂ = N₁mean₁+N₂mean₂/N₁+N₂ Mean₂ = 6 d₁ = mean₁ - Combined mean₁₂ = 10-8= d² ₁ = 4 d₂ = mean₂- Combined mean₁₂ =6-8 = -2 = d² ₂ =4 Combined Mean₁₂ =√N₁S.D²₁+N₂S.D₂²+N₁d²₁+N₂d²₂/N₁+N₂ S.D₂ = 3 , Thus, Mean₂ = 6, S.D₂ = 3

QUESTION NO. 4. Construct a fishers ideal index from the following data and show that it satisfies time reversal and factor reversal tesets: ITEMS 2006 2007 p₀ q₀ p₀ q₀ A 10 40 12 45 B 11 50 11 52 C 14 30 17 30 D 8 28 10 29 E 12 15 13 20

ANS: ITEMS p₀ q₀ p₁ q₁ p₀q₀ p₁ q₀ p₁q₁ p₀q₁ A 10 40 12 45 400 480 540 450 B 11 50 11 52 550 550 572 572 C 14 30 17 30 420 510 510 420 D 8 28 10 29 224 280 290 232 E 12 15 13 20 180 195 260 240 ∑p₀q₀ = 1774,∑p₁q₀ = 2015, ∑p₁q₁ = 2172, ∑p₀q₁=1914 FISHERS IDEAL INDEX P₀₁ = √∑p₁q₀/∑p₀q₀ x∑p₁q₁/∑p₀q₁ × 100 p₀₁ = √2015/1774 x 2172/1914 x 100 p₀₁ = 113.53

TIME REVERSAL TEST TRT is said to be satisfied if p₀₁ × p₁₀ = 1 p₀₁ x p₁₀ = √2015/1774×2172/1914x 1914/2172×1774/2015 p₀₁ × p₁₀ = 1 Thus, Fishers formula satisfies time reversal test. FACTOR REVERSAL TEST FRT is said to be satisfied if P₀₁ x Q₀₁ = ∑p₁q₁/∑p₀q₀ P₀₁xQ₀₁= 2172/1774 = ∑p₁q₁/ ∑p₀q₀ Thus, Fishers formula satisfies factor reversal test.

SECTION - C QUESTION NO. 6. (a) Sampling Error. ANS: SAMPLING ERROR Sampling errors are those which arise due to the method of sampling. Sampling errors arise primarily due to the following reasons; Faulty selection of the sampling method. Substituting one sample for the other sample due to difficulties in collecting the sample. Faulty demarcation of sampling units. Variability of the population which has different characteristics.

QUESTION NO. 6. (b) Importance of BINOMIAL DISTRIBUTION. ANS: IMPORTANCE Theoretical Frequency distribution: The binomial distribution is a theoretical frequency distribution which is based on Binomial Theorem of algebra. With the help of this distribution, we can obtain the theoretical frequencies by multiplying the probability of success by the total number (N). 2. Discrete probability distribution: In which the number of success0,1,2,3,.... n are given in whole numbers and not in fractions.

Shape of Binomial Distribution: The shape of binomial distribution depends on the values of p, q and n. 4. Main parameters: The binomial distribution has two parameters n and p. The entire distribution can be known from these parameters. Uses: It has been found useful in those fields where the outcome is classified into success and failure.

QUESTION NO. 17 Given: Total of the product of deviation of X and Y series = 3044 No. Of pairs of observations = 10 Total Of the deviations of X series = -170 Total Of the deviations of Y series = -20 Total Of the squares of deviations of X series = 8288 Total of the squares of deviations of Y series = 2264 Find out coefficient of correlation when the assumed means of X series and Y series are 82 and 68 respectively.

ANS: COEFFICIENT OF CORRELATION r = ∑dxdy - ∑dx ANS: COEFFICIENT OF CORRELATION r = ∑dxdy - ∑dx.∑dy/N / √∑dx²-(∑dx)²/N x ∑dy²- (∑dy)²/N r = 3044-(-170)(-20)/√8288-(-170)²/10x2264-(-20)²/10 r = 3044-340/√8288-2890x2264-4 r = 2704/3492 r = 0.78

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