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Quantitative analysis july 2003


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  • 1. M.B.A. (3 Years) (First Semester) (DDE) EXAMINATION July 2003 QUANTITATIVE METHODSTime 3 hours Maximum Marks: 70Note: Attempt five questions in all selecting at least one question from each unit.All questions carry equal marks UNIT IQ1 a) A firm sells a product at the rate of Rs 200 per unit. Variable costs are Rs 120 per unit and fixed costs are Rs 40,000. How many units the company should produce a. For break even b. For earning the profit of Rs 21,440Solution Let the company produce x units. Total variable cost = Rs 120x Revenue = Rs 200x Fixed Costs = Rs 40000 a) For break even point revenue = investment ∴ 200 x = 120 x + 40000 ∴ 80 x = 40000 40000 x= 80 ∴ x = 500 Hence the company should produce 500 units for break -even point for given fixed cost b) For earning the profit of 21440 ∴ 200 x = 120 x + 40000 + 21440 ∴ 80 x = 61440 61440 x= 80 ∴ x = 786 Hence for earning a profit of 21440 also the company should produce 768 units.
  • 2. b) A man saved Rs 16500 in ten years. In each year after first he saved Rs 100more than he did in the proceeding year. How much did he save in the first year?Solution: Let the man saved Rs x in first year. There after he saved 100+x,200+x, 300+x, ….900+x His savings can be seen to be in Arithmetic Progression with first term asx and common difference as 100We know that nSum of n terms of an AP is give by (a + l ) 2Here a = x n = 10 l = 900xSum of total savings = 16500 (given) 10 16500 = ( x + 900 + x) 2 16500 = 900 + 2 x 5 16500 2x = − 900 5 2 x = 3300 − 900 2400 x= 2 x = 1200 Hence the man saved Rs 1200 in first yearQ2 Using matrix notation solve the following matrixSolution x − 2 y + 3z = 4 2 x + y − 3z = 5 − x + y + 2z = 3According to matrix notation the above set of equations can be represented inmatrix form as A.X =B  1 −2 3   x  4  1 − 3 and X =  y   Where A =  2    and B = 5 − 1 1  2 z    3  
  • 3. Therefore X = A-1 B Finding the inverse of Matrix A Cofactor of Cell 1,1 = 5 Cofactor of Cell 1,2 = -1 Cofactor of Cell 1,3 = 3 Cofactor of Cell 2,1 = 7 Cofactor of Cell 2,2 = 3 Cofactor of Cell 2,3 = 1 Cofactor of Cell 3,1 = 5 Cofactor of Cell 3,2 = 9 Cofactor of Cell 3,3 = 5Value of the matrix =16  5 7 5  Therfoe A-1 = − 1 3 9 / 16  3 1 5    4  Multiplying A-1 .B we get the matrix 3  2    4  Therefore X = 3  2  Therefore x=4 y=3 and z=2 UNIT IIQ 3 What do you mean by measures of central tendency and measures of dispersion? How do you differ from one another? Discuss the merits and limitations of the following measures: i) Median ii) RangeAnswer: Measure of Central Tendency: An average value is a single valuewithin a range of the data that is used to represent all of the values in the series.
  • 4. They are the typical values around which other items of the distributioncongregate. They are very useful: a) For describing the distribution in concise manner b) For comparative study of different distributionCharacteristics of a good measure of central tendency are: a) It should be rigidly defined b) It should be easy to understand and calculate even for a non- mathematical person c) It should be based on all the observations d) It should be suitable for further mathematical treatment e) It should be affected as little as possible by fluctuationsDifferent measures of central tendency are: a) Arithmetic Mean b) Median c) Mode d) Geometric Mean e) Harmonic MeanMeasure of Variations: The degree to which numerical data tend to spreadabout an average value is called variation or dispersion of the data.Characteristics for an ideal Measure of variation are: a) It should be rigidly defined b) It should be easy to understand and calculate c) It should be based on all observations d) It should be amenable to further mathematical treatment e) It should be affected as little as possible by fluctuationsDifferent measures of variations are: a) Range b) Quartile Deviation c) Percentile Range d) Mean Deviation e) Standard DeviationMerits of Median are: a) It is rigidly defined b) It is easy to understand and calculate c) It is based on all observations d) It is suitable to further mathematical treatment e) Of all averages Mean is least affected by fluctuationsLimitations of Mean are: a) It is very much affected by extreme observations b) Cannot be used in case of open end classes
  • 5. c) Cannot be determined by inspection nor graphically d) Cannot be used when dealing with qualitative characteristics e) Cannot be calculated even if a single observation is not known f) It is extremely asymmetrical distributionMerits of Range are: a) It is the simplest method to find variation of data b) It is rigidly defined c) Readily comprehensibleLimitations of Range are: a) It is not based on entire set of data b) Very much affected by extreme observations c) If smallest and largest value are unaltered and all other values are replaced the range still remains same. d) Cannot be used with open end classes e) It is not suitable for any further mathematical treatment f) It is too indefinite to be used as a practical measure of dispersionQ 4 Calculate the values of i) Arithmetic Mean ii) Mode iii) Quartile deviation iv) Standard DeviationSolution: C.I. f x d=(x-A)/100 f.d c.f (A=500)200-300 20 250 -2.5 -50 20300-400 39 350 -1.5 -58.5 59400-500 86 450 -5 -430 145500-600 35 550 5 175 180600-700 14 650 1.5 21 194700-800 6 750 2.5 15 200 Total 200 -327.5Here Assumed Mean (A) = 500Class Width (h) = 100i) Arithmetic Mean = 500 + 100 * (-327.5)/200 = 336.25ii) Mode: Since the distribution is regular the class corresponding to themaximum frequency i.e. 86 is the modal class. Hence modal class is 400-500and using the mode formula we get
  • 6. h( f 1 − f 0) Mo = l + 2 f 1− f 0 − f 2 100(86 − 39) = 400 + = 447 (approx) 2 * 86 − 39 − 35iii) Quartile Deviation Here N= 200 and N/4 = 50 3N/4 = 3 * 200/4 = 150 Q1 First Quartile lies in 300-400 Q3 Third Quartile lies in 500-600 100 Q1 = 300 + (50 − 20) = 376.92 39 100 Q3 = 500 + (150 − 86) = 682.86 35 Hence Quartile Deviation = (682.86 – 376.92 ) / 2 = 152.97 UNIT IIIQ5 a) Define the term correlations and regressions. What are the points of difference between the two? b) Describe various components of time seriesAnswer a) Correlation - Correlation is an analysis of the co-variation between two or more variables Correlation analysis enables us to have an idea about the degree and direction of the relationship between the two variables under study Regression – Regression analysis is a mathematical measure of the average relationship between two or more variables in terms of the original units of the data. In regression analysis there are two types of variables. The variable whose value is influenced or is to be predicted is called dependent variable and the variable, which influences the values or is used for prediction is called independent variable Correlation analysis vs. regression Analysis
  • 7. It means relationship between two or It means stepping back or returning to themore variables average values and is a mathematical measure expressing the relationship between the two variablesCorrelation coefficient between 2 It aims at providing functional relationshipvariables is a measure of direction and between 2 variables under study and thendegree of linear relationship using this relationship to predict or estimate the value of dependent variableCorrelation need not imply cause and It clearly indicates the cause and effecteffect relationship relationshipIt is independent of the units of Regression coefficient are absolutemeasurement. It is a pure number measures representing the change in thebetween +1 and –1 value of the variable for a unit change in the value of the other variableThere may be non-sense correlation due No such thing like thatto pure chanceIt has limited applications It has wider applications as it can deal with both linear and non-liner relationships b) Components of time series are: a. Secular trend or Long Term Movement: The general tendency of the time series data to increase or decrease or stagnate during a long period of time is called the secular trend or simply trend. This phenomenon is mostly observed in most of the series relating to Economics and Business. e.g. an upward tendency is usually observed in time series relating to population, production and sales of products. b. Periodic Movements or Short Term fluctuations: They can be classified into two parts i. Seasonal Fluctuation: These variations in time series are due to the rhythmic forces, which operate in regular and periodic manner over a span of less than a year, i.e. during a period of 12 months and have the same or almost same pattern year after year. ii. Cyclical Variations: The oscillatory movements in a time series with period of oscillation greater than one year are termed as cyclical variations. These variations in time series are due to ups and downs recurring after a period greater than one year. These cyclical fluctuations are more or less regular, are not necessarily periodic. One complete period, which normally lasts from 7 to 9 years, is termed as a ‘cycle’.
  • 8. c. Random or Irregular variations: These fluctuations are purely random and are result of such unforeseen and unpredictable forces which operate in absolutely erratic and irregular manner. Such variations do not exhibit any definite pattern and there is no regular period or time of their occurrence. These powerful variations are usually caused by numerous non-recurring factors like floods, famines, wars, earthquakes etc.Q 6 In a partially destroyed record, the following data are available: Two regression lines are 5 x + 3 y = 290 3 x + 2 y = 180 Variance of x is 16 Find: i) Mean values of x and ySolution: I) Since both the lines pass through the mean values the point ( x, y )must satisfy both the equations Solving both the equations for x and y we get x= 40 and y= 30 Hence mean of x and y is 40 and 30 respectively UNIT IVQ 7 Using suitable examples explain the following terms: i) Mutually exclusive events ii) Conditional Probability iii) Independent EventsAnswer i) Mutually Exclusive Events: Two or more events are said to be mutually exclusive if the happening of any one of them excludes the happening of all others in the same experiment. e.g. In toss of a coin the events ‘head’ and ‘tail’ are mutually exclusive because if head comes, we can’t get the tail and if tail comes we can’t get head. ii) Conditional Probability: Conditional probability is the probability of happening of one event on the condition that other event has already happened
  • 9. e.g. Drawing of balls from a bag containing balls (say 8) without replacement. When first ball is drawn then the sample space will be 8. But when the second ball is drawn the sample space would reduce down to 7. So the probability of occurrence of second event gets affected by the happening of first event iii) Independent Events: Events are said to be independent of each other if happening of any one of them is not affected by and does not affect the happening of any one of others. e.g. In tossing of a die repeatedly the event of getting ‘5’ in 1st throw is independent of getting ‘5’ in second, third or subsequent throwsQ8 a) Eight coins are tossed simultaneously. Find the probability of getting i. Four Heads ii. At most 2 tails Solution If p denotes probability of getting a head = ½ then according to binomial distribution law Probability of getting 4 heads is given by P(r=4) = 8C4 (1/2)4(1/2)8-4 70 = 256 ii) At most 2 tails = Prob of getting 0 tail + Prob of getting 1 tail + Prob of getting 2 tail Let p be prob of getting tail =1/2 Therefore Probalitity of getting at most 2 tails is given by 8 C0(1/2)0(1/2)8-0 + 8C1(1/2)1(1/2)8-1 + 8C2(1/2)2(1/2)8-2 37 = 256(b) Marks obtained by the students of a class are normally distributed with meanof 58 marks and standard deviation of 4 marks. Find the probability that marksobtained by a student are i) Less than 45 ii) Between 55 and 61
  • 10. Solution i) Probability that marks obtained are less than 45 µ = 58 σ =2 x = 45 45 − 58 Standard Normal Variate is Z = = -6.5 2 P(z<-6.5) or P(z>6.5) = 0.5 – P(0<z<6.5)ii) lies between 55 and 61 P(55<x<61)Calculating SNV for 55 55 − 58 Z1 = = -1.5 2 61 − 58 Z2 = = 1.5 2 In terms of SNV P(-1.5<z<1.5) = 2 P(0<z<1.5) { By Symmetry } = 2 * .4332 = 0.8664