{"id":12214,"date":"2019-12-17T14:56:35","date_gmt":"2019-12-17T09:26:35","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=12214"},"modified":"2021-07-02T15:36:31","modified_gmt":"2021-07-02T10:06:31","slug":"1st-puc-statistics-question-bank-chapter-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-statistics-question-bank-chapter-5\/","title":{"rendered":"1st PUC Statistics Question Bank Chapter 5 Analysis of Univariate Data"},"content":{"rendered":"

You can Download Chapter 5 Analysis of Univariate Data Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers<\/a>\u00a0Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Statistics Question Bank Chapter 5 Analysis of Univariate Data<\/h2>\n

1st PUC Statistics Analysis of Univariate Data Textbook Questions and Answers<\/h3>\n

Question 1.
\nWhat do you mean by \u2018Central tendency\u2019 and \u2018Measure of central tendency\u2019?
\nSolution:
\nIn any frequency distribution most of the frequencies are concentrated at the centre, this property of concentration of the frequencies around the centre is called Central tendency.
\nThe central value representing the majority of the frequencies\/values is called Measure of central tendency or Average.<\/p>\n

Question 2.
\nWhat is an average?
\nSolution:
\nThe central value representing the majority of the values\/frequencies is called Average.<\/p>\n

Question 3.
\nWhat are the various measures of central tendency?
\nSolution:
\na. Arithmetic mean<\/p>\n

    \n
  • Median<\/li>\n
  • Mode<\/li>\n
  • Geometric mean<\/li>\n
  • Hormonic mean<\/li>\n<\/ul>\n

    Question 4.
    \nWhat are the desired characteristics\/properties\/essentials of good measure of central tendency?
    \nSolution:<\/p>\n

      \n
    • It is rigidly defined.<\/li>\n
    • It is simple to understand and easy to calculate<\/li>\n
    • It should be based on all values of the data<\/li>\n
    • It should not be affected by the extreme values of the data<\/li>\n
    • It should have sampling stability<\/li>\n
    • It should be capable of further statistical analysis<\/li>\n<\/ul>\n

      Question 5.
      \nDefine Arithmetic mean and give the formula
      \nSolution:
      \nArithmetic mean is defined as the sum of values divided by the number of values. Denoted by-
      \n\u2234 \\(\\overline{\\mathrm{X}}=\\frac{\\Sigma \\mathrm{X}}{\\mathrm{n}}\\)<\/p>\n

      Question 6.
      \nWrite the properties of a good measure of central tendency.
      \nSolution:<\/p>\n

        \n
      • It is rigidly defined<\/li>\n
      • It is based on all values of the data<\/li>\n
      • It should not be affected by the extreme values of the data<\/li>\n
      • It should have the sampling stability<\/li>\n
      • It should be capable of further statistical analysis<\/li>\n<\/ul>\n

        Question 7.
        \nWrite any two merits and demerits of A.M Sol.
        \nSolution:
        \nMerits:<\/p>\n

          \n
        • It is simple to calculate and easy to understand<\/li>\n
        • It is rigidly defined\/It is based on all values.<\/li>\n<\/ul>\n

          Demerits:<\/p>\n

            \n
          • It is.affected by extreme values of the data.<\/li>\n
          • It cannot be easily calculate for the distribution with open-end class interval.<\/li>\n<\/ul>\n

            \"KSEEB<\/p>\n

            Question 8.
            \nDefine Median and give the formula.
            \nSolution:
            \nMedian is defined as that value of the data, which divides the data into two equal parts, when the data is arranged in ascending order.
            \nDenoted by \u2018M\u2019
            \nM = size of the \\(\\left(\\frac{n+1}{2}\\right)^{t h}\\) item \/ term; n – Number of items
            \nFor continuous data :Median class = Size of the \\(\\left(\\frac{\\mathrm{N}}{2}\\right)^{\\mathrm{th}}\\) item \/ term
            \n\"1st
            \nc.f- cum. frequency up to median Class;
            \nf-freq. med. Class;
            \ni-width of med. class<\/p>\n

            Question 9.
            \nGive any two merits and demerits of Median.
            \nSolution:
            \nMerits:<\/p>\n

              \n
            • It is rigidly defined<\/li>\n
            • It is not affected by extreme values<\/li>\n<\/ul>\n

              De-merits:<\/p>\n

                \n
              • It is not based on all observations<\/li>\n
              • It has no sampling stability<\/li>\n<\/ul>\n

                Question 10.
                \nDefine Mode and give the formula.
                \nSolution:
                \nMode is defined as that value of the data, which occurs more frequently, OR Which occurs greatest number of times. Denoted by \u2018Z\u2019
                \nFor Ungrouped data: Z= Size of the data which more frequently
                \nFor Discrete data: Z= Size of the highest frequency
                \nFor Continuous Data: Modal class = Size of the highest frequency
                \n\"1st
                \nf0<\/sub>-freq. preceding class; f1<\/sub>– frequency of the modal class;
                \nf2<\/sub>-frequency, succeeding to Mod. C1ass i- width of Modal class<\/p>\n

                Question 11.
                \nDefine Harmonic mean with formula.
                \nSolution:
                \nHarmonic mean is defined the reciprocal of the Arithmetic mean of the reciprocal alues of the – data
                \n\"1st<\/p>\n

                \"KSEEB<\/p>\n

                Question 12.
                \nDefine Geometric mean with formula.
                \nSolution:
                \nGeometric mean is defined as the nth root of the product of ‘n\u2018 if number of observations. If x2<\/sub>,x1<\/sub>,………xn<\/sub> are the\u2018n\u2019number of observations then,
                \nie. GM =\\(\\sqrt[n]{\\left(x_{1} \\cdot x_{2} \\dots \\dots x_{n}\\right)}\\)
                \nUsing Logarithms, GM= Antilog {\\(\\frac{\\Sigma \\mathrm{Logn}}{\\mathrm{n}}\\)}<\/p>\n

                Question 13.
                \nGive any two uses of G.M and H.M
                \nSolution:
                \nGM and HM are used to calculate the averages, where very accurate values are needed, in cases of Time, Velocity’, Speed, Bank rates, Interest rates etc.<\/p>\n

                Question 14.
                \nWhat is the relation between AM, GM and HM.
                \nSolution:
                \nx\u0304\u00a0\u2265 GM \u2265 HM<\/p>\n

                Question 15.
                \nWrite down Empirical relation between mean, median and mode
                \nSolution:
                \nMode = [3 Median – 2 Mean]<\/p>\n

                Question 16.
                \nGive any two uses of Mode
                \nSolution:
                \nMode is used to find the true characteristics in any manufacturing industries, such as to find the most common size in ready made garments, Shoes, any items etc.<\/p>\n

                Question 17.
                \nCalculate G.M and 1I.M:4,16
                \nSolution:
                \n\"1st
                \nQuestion 18.
                \nFind GM. and H.M:27,64,125
                \nSolution:
                \n\"1st<\/p>\n

                Question 19.
                \nWhat is the weighted A.M and mention its uses
                \nSolution:
                \nWhen the relative importance\/weights\/weightages of the various items used in finding the averages are required, Weighted averages are used.
                \nLet X1<\/sub>, X2<\/sub>, ….. Xn<\/sub> are the number of values with respective weights assigned as W1<\/sub>, W1<\/sub>………..Wn<\/sub>
                \nThen Weighted Arithmetic Mean is – x\u0304w<\/sub> = \\(\\frac{\\Sigma \\mathrm{WX}}{\\Sigma \\mathrm{W}}\\)<\/p>\n

                Question 20.
                \nGive the formula for calculating combined A.M for two set of values.
                \nSolution:
                \n\"1st<\/p>\n

                \"KSEEB<\/p>\n

                Question 21.
                \nWhat do you mean by partition values?
                \nSolution:
                \nIf the data is divided into various parts, the dividing points are called partition Values. The partition values are Median, Quartiles, Deciles and Percentiles.<\/p>\n

                Question 22.
                \nWhat are partition values?
                \nSolution:
                \nThe partition values are Median, Quartiles, Deciles and Percentiles<\/p>\n

                Question 23.
                \nDefine Quartiles, Deciles and Percentiles.
                \nSolution:
                \nQuartiles: If the data is divided into four equal parts with three dividing points, then the dividing points are called Quartiles. The first dividing point is Q1<\/sub> the second dividing point is Q2<\/sub>\/Median and the third dividing point is Q3<\/sub>
                \nDeciles: If the data is divided into ten equal parts with nine dividing points, the dividing points are called Deciles. They are denoted by D1<\/sub>, D2<\/sub>…D9<\/sub>
                \nSimilarly Percentiles can be defined.<\/p>\n

                Question 24.
                \nWrite down formula for calculation of partition values in general, for disrete and continuous data.
                \nSolution:
                \nI. For Ungrouped data
                \nArrange the data in ascending order (Array).
                \nFor Quartiles: Qj<\/sub> = size of \\(\\frac{n+1}{4}\\) th value of the data; Where j= 1,2,3.
                \nFor Deciles: Dj<\/sub> = size of j(n+l )\/10th<\/sup> th value of the data Where j = 1,2, 3,….9.
                \nFor Percentiles : Pj<\/sub> = size of j[latex]\\frac{n+1}{4}[\/latex]th<\/sup> Value of the data Where j = 1,2,3,…..9.<\/p>\n

                II. For Discrete data:
                \nCalculate Less than Cumulative Frequency.
                \n\"1st<\/p>\n

                III. For Continuous Data:
                \nCalculate the Continuous frequency.
                \n\"1st<\/p>\n

                \"KSEEB<\/p>\n

                Problems<\/strong><\/p>\n

                Question 1.
                \nFollowing mark of 8 students compute mean marks: 84, 75,45,42,59,62,55,50.
                \nSolution:
                \nDirect method : X = \\(\\frac{\\sum x}{n}\\)
                \nLet x be the marks : No. of students = 8
                \nMean Marks = X = \\(\\frac{472}{8}\\) = 59<\/p>\n

                Question 2.
                \nFind mean weight (kgs) : 53, 52.4, 54 50.8 + 48.4 + 44.4
                \nSolution:
                \nDirect method: X = \\(\\frac{\\sum x}{n}\\)
                \nLet X be the weight (in kgs), n- No. of students = 6
                \n:. Mean weight (in kgs) x\u0304 = \\(\\frac{\\sum x}{n}\\) = \\(\\frac{303}{6}\\) = 50.5 kgs<\/p>\n

                Question 3.
                \nFollowing \u00a1s em penses of 8 workers:
                \nEmpenses ( \u20b9 ): 12500,8500, 10,400,6000,8000,5000,4400,9200. compute mean empcnsc.
                \nSolution:
                \nThe data is big in size, so use
                \nShortcut method: X = A + \\(\\frac{\\Sigma d}{n}\\) ;d=X – A
                \nA – Arbitrary assumed mean
                \nLet X be the monthLy expenditure<\/p>\n\n\n\n\n\n
                X<\/td>\nD= X \u2013 A(8000)<\/td>\n<\/tr>\n
                \n

                12500<\/p>\n

                8500<\/p>\n

                10400<\/p>\n

                6000<\/p>\n

                8000<\/p>\n

                5000<\/p>\n

                4400<\/p>\n

                9200<\/p>\n<\/td>\n

                		4500<\/p>\n

                500<\/p>\n

                2400<\/p>\n

                -2000<\/p>\n

                0<\/p>\n

                -3000<\/p>\n

                -3600<\/p>\n

                1200<\/td>\n<\/tr>\n

                n = 8<\/td>\n\u03a3d = 8600 \u2013 8600 = 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                A = 8000 (Assume any value within the data)
                \n\"1st
                \n\u2234 Mean.expenditure : x\u0304 = \u20b9 8000<\/p>\n

                Question 4.
                \nThe priers of gold obseredover 10 days gave the following data. Compute \u2019mean prime of gold per 10 gram:
                \n2750,2600,2580,2700,2850,2680,2900,2575,2950,2720
                \nSolution:
                \nUse shortcut method :\\(\\overline{\\mathrm{X}}=\\mathrm{A}+\\frac{\\sum d}{n}\\)
                \nLet X denote the mile (x\u0304).
                \n\"1st
                \nA = 2800 (Assume within highest and lowest value in the data)
                \n\"1st
                \nMean price of gold (x\u0304) = \u20b9 2830.5<\/p>\n

                Question 5.
                \nFollowing is the T.V sets sold over number of day. Comput mean No. of sets sold per day.
                \n\"1st
                \nSolution:
                \nLet X and f be the no. of t.v. sets sold and no. of days.
                \nThe data is small in size.
                \nUse direct method : x\u0304 = \\(\\frac{\\sum f x}{\\mathrm{N}}\\)
                \n\"1st<\/p>\n

                \"KSEEB<\/p>\n

                Question 6.
                \nNumber of students attended over a month is reconded as below. Compute mean number of students attendence per day.
                \n\"1st
                \nSolution:
                \nLet X and f be the attendance of students and no. of students.
                \n\"1st
                \n\u2234 Mean number of students attended (x\u0304) = 71.5357<\/p>\n

                Question 7.
                \nFind mean:
                \n\"1st
                \nSolution:
                \nThe data having common difference so, usr deviation method
                \n\"1st
                \nC – Common difference
                \nA – Arbitrary assumed mean.
                \nLet X and\/be number of passengers and number of trips.
                \n\"1st
                \nQuestion 8.
                \nFind the Arithmetic mean:
                \n\"1st
                \nSolution:
                \nLet X and f be the values and frequency
                \n\"1st
                \n\"1st
                \nQuestion 9.
                \nFrom the following frequency distribution compute mean.
                \n\"1st
                \nSolution:
                \nLet class intervals (Cl) and\/be the marks and number of students.
                \n\"1st<\/p>\n

                \"KSEEB<\/p>\n

                Question 10.
                \nSolution:
                \n\"1st<\/p>\n

                Question 11.
                \nThe number of workers and duty wages are given below compute mean
                \n\"1st
                \nSolution:
                \nLet Cl and f be the daily wages and no. of workers
                \n\"1st<\/p>\n

                Question 12.
                \nThe number of worker and daily expenses are recorded below: compute mean
                \n\"1st
                \nSolution:
                \nLet Cl and f be the daily expenses and no. of workers
                \n\"1st
                \n\"1st<\/p>\n

                Question 13.
                \nCalculate the mean for 2, 4, 6, 8, 10
                \nSolution:
                \n\"1st<\/p>\n

                Question 14.
                \nCalculate the geometric mean of the following series of monthly income of a batch of families: 180,250,490, 1400, 1050. Using logarithmic method.<\/p>\n\n\n\n\n\n\n\n\n\n
                X<\/td>\nlogx<\/td>\n<\/tr>\n
                180<\/td>\n2.2553<\/td>\n<\/tr>\n
                250<\/td>\n2.3979<\/td>\n<\/tr>\n
                490<\/td>\n2.6902<\/td>\n<\/tr>\n
                1400<\/td>\n3.1461<\/td>\n<\/tr>\n
                1050<\/td>\n3.0212<\/td>\n<\/tr>\n
                \u03a3 logx =\u00a0\u00a013.5107<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                Solution:
                \nGM=Antilog [Xlogx\/n] = Antilog [ 13.5107\/5] = Antilog(2.7021) = 503.6<\/p>\n

                \"KSEEB<\/p>\n

                Question 15.
                \nCalculate the average income per head from the data given below .Use geometric mean.
                \n\"1st
                \nSolution:
                \nLet X and f the income (Rs) and Number of families.
                \nGM = Antilog {\\(\\frac{\\Sigma f \\log x}{N}\\)}
                \n\"1st
                \nG.M = Antilog \\(\\left[\\frac{\\Sigma f \\log x}{N}\\right]\\)= Antilog \\(\\left[\\frac{482.257}{186}\\right]\\) = Antilog (2.5928)
                \nGM=Rs 391.50<\/p>\n

                Question 16.
                \nFrom the given data calculate H.M 5,10,17,24,30
                \nSolution:<\/p>\n\n\n\n\n\n\n\n\n\n
                X<\/td>\n1\/X<\/td>\n<\/tr>\n
                5<\/td>\n0.2000<\/td>\n<\/tr>\n
                10<\/td>\n0.1000<\/td>\n<\/tr>\n
                17<\/td>\n0.0588<\/td>\n<\/tr>\n
                24<\/td>\n.0.0417<\/td>\n<\/tr>\n
                30<\/td>\n0.0333<\/td>\n<\/tr>\n
                Total<\/td>\n0.4338<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                Question 17.
                \nThe marks secured by some students of a class are given below. Calculate the harmonic mean.
                \n\"1st
                \nSolution:
                \n\"1st
                \nH.M = N\/(f.1\/x) = 18\/0.1968 = 21.91<\/p>\n

                \"KSEEB<\/p>\n

                Question 18.
                \nFrom the following take of literacy rate and number of cities compute median
                \n\"1st
                \nSolution:
                \nLet C.I and f be the literary rate (%) and no. of cities
                \n\"1st
                \nLCF (Less than cumulative frequency)
                \nMedian class = size of Nth<\/sup> term = \\(\\frac{50}{2}=\\) = 25th<\/sup> term
                \n(25th<\/sup> term includes in LCF of 27 group i.e., all 13 to 27 terms)
                \n\u2234 Median class = 75-85
                \n\"1st
                \nL – Lower limit of median class = 75
                \n\\(\\frac{N}{2}=\\) = 25; C.f – Cumulative frequency upto median class = 12
                \nf – Simple frequency of median class =15 i
                \ni – Width of the median class = 10
                \nM = 75 + [\\(\\frac{25-12}{15}\\)+ x 10] = 75 + 8.667 = 83.667<\/p>\n

                Question 19.
                \nThe age of students are given below compute median
                \n\"1st
                \nSolution:
                \nLet Cl and f be the age (years) and number of students
                \n\"1st
                \nMedian class = Size of term = \\(\\left(\\frac{\\mathrm{N}}{2}\\right)^{\\mathrm{th}}\\) term = \\(\\frac{50}{2}\\) = 25th<\/sup> term = 7-8
                \n(25th<\/sup> term is in the group of 32 of LCF i.e., all values from 16 to 32 lies in the group)
                \n\"1st<\/p>\n

                Question 20.
                \nFrom the following data compute median
                \n\"1st
                \nSolution:
                \nLet Cl and f be the Life and number of bulbs
                \n\"1st
                \nMedian class = Size of term = \\(\\left(\\frac{\\mathrm{N}}{2}\\right)^{\\mathrm{th}}\\) term = \\(\\frac{190}{2}\\) = 95th term = 17-18
                \n\"1st<\/p>\n

                Question 21.
                \nFind the median
                \n\"1st
                \nSolution:
                \nLet Cl and\/be the marks and number of students
                \nHere class interval is (Cl) are inclusive, so convert into exclusive C.I use, correction factor
                \n\"1st
                \nSubtract C.F. in all lower limits and add to all upper limits in all inclusive class intervals.
                \n\"1st
                \nMedian class = Size of term = \\(\\left(\\frac{\\mathrm{N}}{2}\\right)^{\\mathrm{th}}\\) term = \\(\\frac{50}{2}\\) = 25th term = (29.5 – 39.5)
                \n\"1st<\/p>\n

                \"KSEEB<\/p>\n

                Question 22.
                \nFind median : 23, 27, 15, 18, 25, 28, 30, 20
                \nSolution:
                \nArrange in ascending order
                \n15, 18, 20, 23,25, 27, 28, 30 ;n = 8
                \n\"1st
                \nQuestion 23.
                \nFind the median : 75,80,81,75,75,88,75,84,74,70
                \nSolution:
                \nArrange in ascending order
                \n70, 74, 75, 75, 75, 75, 80, 81, 82, 84
                \nn = 10<\/p>\n

                Question 24.
                \nFind the median :
                \n\"1st
                \nSolution:
                \n(Discrete data)
                \nFirst calculate LCF (Less than cumulative frequency); add frequencies from top
                \nMedian = size of \\(\\left(\\frac{\\mathrm{N}+1}{2}\\right)^{\\mathrm{th}}\\) term<\/p>\n

                \"KSEEB<\/p>\n

                Median<\/strong><\/p>\n

                Question 1.
                \nCompute median for the following data
                \nNo. of apples per box : 22,18,25,32,28,26,30
                \nSolution:
                \nUngrouped\/Raw data
                \nArrange in ascending order n = 7
                \nNo. of apples per box: 18,22,25, [26], 28, 30, 32
                \nMedian = Size of \\(\\left[\\frac{\\mathrm{N}+1}{10}\\right]^{\\mathrm{th}}\\) term = \\(\\frac{7+1}{2}\\) = 4th<\/sup> term
                \n\u2234 Median = 26 apples per box.
                \n\"1st
                \nMedian = \\(\\left(\\frac{40+1}{2}\\right)^{\\text {th }}\\) term = 25.5th <\/sup>term.
                \nHere 20.5 lies in the group of 29 of LCF., its value at X is median.
                \n\u2234 Median 15.<\/p>\n

                Question 2.
                \nFind the median for the following frequency distribution.
                \n\"1st
                \nSolution:
                \n\"1st
                \nMedian = Size of \\(\\left[\\frac{\\mathrm{N}+1}{10}\\right]^{\\mathrm{th}}\\) term = \\(\\frac{110 +1}{2}\\) = = 55.5th<\/sup> term
                \n\u2234 Median = 120.<\/p>\n

                \"KSEEB<\/p>\n

                Mode<\/strong><\/p>\n

                Question 1.
                \nFind the value made for the following ungroupcd data: 25,14,33,40,35,28,42,40,62,38,40,35
                \nSolution:
                \n(Ungrouped\/Raw data)
                \nMode = Size of most repeated term
                \nMode(Z) = 40 [It is more no. of times]<\/p>\n

                Question 2.
                \nFollowing is the data regarding the no. of childern per couple. Find medal number of children:
                \n4,1,0,2, 5,2, 5, 2,1,2,1,5, 4,2,1,2
                \nSolution:
                \nMode = Size of most repeated term
                \n\u2234 Mode (Z) = 2<\/p>\n

                Question 3.
                \nFind the value of mode :
                \n\"1st
                \nSolution:
                \n(Discrete data)
                \nMode = Size of highest frequency
                \nHighest frequency is 25, its size (X) is 40.
                \n\u2234 Mode (Z) = 40<\/p>\n

                Question 4.
                \nFind mode number of children per family
                \n\"1st
                \nSolution:
                \nMode = Size of highest frequency.
                \nLet X and f be the number of children and number of families.
                \nHighest frequency is 35, its size (X) is Z.
                \n\u2234 Mode(Z) = 2 children per family.<\/p>\n

                \"KSEEB<\/p>\n

                Question 5.
                \nFind the mode:
                \n\"1st
                \nSolution:
                \n(Continuous data)
                \nFirst find the medal class :
                \nMedal class = Size of highest frequency
                \nApply the formula:
                \n\"1st
                \nLet X and f be size and no. of items
                \n\"1st
                \nMedal clas = 6 – 8 [It has highest frequency]
                \nL = 6 \u00ae Lower limit of medal class
                \nf0<\/sub> = 1 \u00ae Frequency preceeding to model class
                \nf1<\/sub> = 15 \u00ae Frequency of the model class
                \nf2<\/sub> = 10 \u00ae Frequency succeeding to model class
                \ni = 2 Width of the medal clas [i.e., 8 minus 6] = 2
                \n\"1st
                \n\u2234 Z = 11.2307 is the modal size of items.<\/p>\n

                Question 6.
                \nOn 30 days of business hours, the number of T.V sets sold in a certain show room arc as given below.
                \n\"1st
                \nSolution:
                \nLet X and f be the number of t.v\u2019s sold and no. of days
                \n\"1st
                \nModal class = Size of highest frequency
                \n= 30 – 40 [This class has highest frequency]
                \n\"1st
                \nZ = 33.3333 is the modal no. of T.V\u2019s sold in a shop.<\/p>\n

                \"KSEEB<\/p>\n

                Question 7.
                \nCalculate the production of sugar from the following data
                \n\"1st
                \nSolution:
                \nLet X and \/be the production and number of days.
                \nThe frequency distribution is of inclusive class intervals. So convert into exclusive class intervals. Using correction factor (CF)
                \n\"1st
                \nSubtract CF in all lower limits and add to all upper limits.
                \nMedal class = Size of highest frequency = (29.5 – 34.5)
                \n\"1st
                \nQuestion 8.
                \nFind the mode:
                \n\"1st
                \nSolution:
                \nConvert inclusive class into exclusive class intervals
                \nC.F = \\(\\frac{\\text { LL II class -UL I class }}{2}=\\frac{15-4}{2}\\) = 0.5
                \nSubtract CF = 0.5 in all lower limits and add to all upper limits.<\/p>\n\n\n\n\n\n\n\n\n
                X<\/td>\nf<\/td>\n<\/tr>\n
                (-0.5) \u2013 14.5<\/td>\n4<\/td>\n<\/tr>\n
                14.5-29.5<\/td>\n9 \u2190 f0<\/sub><\/td>\n<\/tr>\n
                29.5-44.5<\/td>\n15 \u2190 f1<\/sub><\/td>\n<\/tr>\n
                44.5-59.5<\/td>\n8 \u2190 f2<\/sub><\/td>\n<\/tr>\n
                55.5-74.5<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                Medal class = Size of highest frequency = (29,5 – 44.5)
                \n\"1st
                \nZ = 29.5+ 6.9230 = 36.423.<\/p>\n

                Question 9.
                \nCalculate mode :
                \n\"1st
                \nSolution:
                \nIn the above frequency distribution there are two highest frequencies, the data is called Bi-medal. To find mode use empirical relation between mean, median and mode:
                \nZ = 3 median – 2 mean
                \nZ = 3m – 2x\u0304
                \n\\(\\overline{\\mathrm{x}}=\\frac{\\Sigma f_{m}}{\\mathrm{N}}\\); m- midpoints
                \nAfter calculating LCF
                \n\"1st
                \nLCF = Frequencies added from top
                \n\\(\\overline{\\mathrm{X}}=\\frac{\\sum f_{m}}{\\mathrm{M}}=\\frac{1955}{65}\\) = 30.0769
                \nMedian class = Size of \\(\\left(\\frac{\\mathrm{N}}{2}\\right)^{\\mathrm{th}}\\) term = \\(\\frac{65}{2}\\) = 32.5th<\/sup> term
                \nHere 32.5 is not in the group of 32 of LCF but in the group of 47.
                \nSo, median class = 30-40
                \n\"1st
                \nNow mode (Z) = 3 m – 2x\u0304
                \n= 3(30.3333) – 2(30.0769)
                \n= 90.9999 – 60.1538
                \nZ = 30.3333
                \nNote : Calculation of mode for indefined data we can also use grouping and analysis table method, but for IPUC it is limited study.<\/p>\n

                \"KSEEB<\/p>\n

                Question 10.
                \nCalculate mean, median and mode
                \n\"1st
                \nSolution:
                \nThe distribution is of less than cumulative type, i.e., upper limits and frequencies added from top are given. Convert it to simple frequency distribution.
                \nThe distribution is ill defined for mode use empirical relation.
                \n\"1st
                \nMedian class = Size of \\(\\left(\\frac{\\mathrm{N}}{2}\\right)^{\\mathrm{th}}\\) term = \\(\\frac{150}{2}\\) = 75th term
                \n= 10-12
                \n\"1st
                \nM=11(000’s)
                \n\u2234Mode(Z) = 3m – 2x\u0304 = 3(11) – 2(9.8533) = 33 – 19.7066
                \n= 13.2934 (000\u2019s’).<\/p>\n

                Question 11.
                \nCalculate mean, median and mode for the following frequency distribution
                \n\"\"
                \nSolution:
                \nThe given table is more than cumulative frequency distribution. Convert into simple frequency distribution (i.e., lower limits and frequencies added from below are given)
                \n\"1st
                \nMedian class = Size of \\(\\left(\\frac{\\mathrm{N}}{2}\\right)^{\\mathrm{th}}\\) term = \\(\\frac{100}{2}\\) = 50th term = (100-110)
                \n\"1st
                \nM= 104.5454IQ
                \nMedian class = Size of highest frequency = 100 – 110
                \n\"1st
                \nZ = 100 + 3.3333 = 103.3333IQ<\/p>\n

                \"KSEEB<\/p>\n

                Question 12.
                \nGiven x\u0304 = 7.3, Fing the missing frequency
                \nSolution:
                \n\"1st
                \nSolution:
                \nLet k be the missing Frequency<\/p>\n\n\n\n\n\n\n\n\n\n
                X<\/td>\nf<\/td>\nfx<\/td>\n<\/tr>\n
                5<\/td>\n4<\/td>\n20<\/td>\n<\/tr>\n
                6<\/td>\n6<\/td>\n36<\/td>\n<\/tr>\n
                7<\/td>\n12<\/td>\n84<\/td>\n<\/tr>\n
                8<\/td>\nK<\/td>\n8K<\/td>\n<\/tr>\n
                9<\/td>\n8<\/td>\n72<\/td>\n<\/tr>\n
                <\/td>\nN = 30 + K<\/td>\n\u03a3fX = 212 + 8K<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                \\(\\overline{\\mathrm{x}}=\\frac{\\Sigma f \\mathrm{x}}{N}\\)
                \n7.3 = \\(\\frac{212+8 K}{30+K}\\)
                \n7.3(30 + K) = 212 + 8K
                \n219 + 7.3K = 212 + 8K
                \n-8K + 7.3K = 212 – 219
                \n– 0.7K = -7
                \nK = \\(\\frac{7}{0.7}\\) = 10
                \nK = lo is the missing frequency<\/p>\n

                Question 13.
                \nFind the missing frequency if the mean is 18.4
                \n\"1st
                \nLet k be the missing Frequency<\/p>\n\n\n\n\n\n\n\n\n\n\n
                X<\/td>\nf<\/td>\nfx<\/td>\n<\/tr>\n
                5<\/td>\n3<\/td>\n15<\/td>\n<\/tr>\n
                10<\/td>\n7<\/td>\n70<\/td>\n<\/tr>\n
                15<\/td>\nK<\/td>\n15K<\/td>\n<\/tr>\n
                20<\/td>\n15<\/td>\n300<\/td>\n<\/tr>\n
                25<\/td>\n7<\/td>\n175<\/td>\n<\/tr>\n
                10<\/td>\n6<\/td>\n180<\/td>\n<\/tr>\n
                <\/td>\nN = 38 + \u00a0K<\/td>\n\u00a0740 + 15K<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                \\(\\overline{\\mathrm{x}}=\\frac{\\Sigma f \\mathrm{x}}{N}\\)
                \n18.4 = \\(\\frac{740+15 K}{38+K}\\)
                \n18.4(38 + K) = 740 + 15K
                \n18.4 \u00d7 38+ 18.4K = 740 + 15K
                \n699.2 + 18.4K = 740 + 15K
                \n18.4K = 740 – 699.2 – 15.K
                \n3.4K = 40.8
                \nK = \\(\\frac{40.8}{3.4}\\)
                \nK = 12 is the missing frequency.<\/p>\n

                Question 14.
                \nFind the missing frequency from the following frequency distribution if x\u0304 = 24.
                \n\"1st
                \nSolution:
                \nLet K be the missing frequency
                \n\\(\\overline{\\mathrm{X}}=\\frac{\\sum f_{m}}{\\mathrm{N}}\\)
                \n\"1st
                \n\\(\\overline{\\mathrm{X}}=\\frac{\\sum f_{m}}{\\mathrm{N}}\\)
                \n24 = \\(\\frac{510+35 K}{24+K}\\)
                \n24(24 + K) = 5 10 + 35K
                \n576 + 24K = 5 10 + 35K
                \n24K – 35K = 510 – 576
                \n-11K = -66
                \nK = \\(\\frac{66}{11} \\) =6
                \n\u2234 Missing frequency = K = 6<\/p>\n

                Question 15.
                \nFind the missing frequency if x\u0304 = 15.587
                \n\"1st
                \nSolution:
                \nLet K be the missing frequency
                \n\\(\\overline{\\mathrm{X}}=\\frac{\\sum f_{m}}{\\mathrm{N}}\\)
                \n\"1st
                \n\\(\\overline{\\mathrm{X}}=\\frac{\\sum f_{m}}{\\mathrm{N}}\\)
                \n15.587 = \\(\\frac{513+17 K}{34+K}\\)
                \n15.587(34 + K) =+ 17K
                \n529.958 + 15.587K = 513 + 17K
                \n15.587K – 17K = 513 – 529.958
                \n-1.413K = -16.958
                \n16.958
                \nK = \\(\\frac{16.587}{1.413} \\) = 12
                \n\u2234 Missing frequencyK= 12.<\/p>\n

                Question 16.
                \nFind the missing frequencies from the following data.
                \n\"1st
                \nSolution:
                \nGiven that N = 200 and x\u0304 = 83.875
                \nLet K1 <\/sub>and K2<\/sub> be the missing frequencies
                \n\"1st
                \n\\(\\overline{\\mathrm{X}}=\\frac{\\sum f_{m}}{\\mathrm{N}}\\)
                \nGiven = N = 200
                \n\"1st
                \n83.875 \u00d7 200 = 10962.5 + 62.5K1<\/sub> + 137.5K2<\/sub>
                \n62.5K1<\/sub> + 137.5K2<\/sub> = 16775 – 10962.5
                \n62.5K1<\/sub> + 137.5K2<\/sub> = 5812.5 …………..(1)
                \nIt is given that M = 200
                \ni.e., 137 + K1<\/sub> +K2<\/sub>= 200
                \nK1 <\/sub>+ K2 <\/sub>= 63…………….(2)
                \nFrom (1) and (2): (1)-(2) x 62.5
                \n\"1st
                \nPut K2<\/sub> = 25 in (1)
                \nK1<\/sub> + 25 = 63
                \nK1<\/sub>= 63 – 25 =38
                \n\u2234 K1<\/sub> = 38, K2<\/sub> = 25 are the missing frequencies.<\/p>\n

                \"KSEEB<\/p>\n

                Question 17.
                \nFind missing frequencies if x\u0304 = 4.8286
                \n\"1st
                \nSolution:
                \nLet K1<\/sub> and K2<\/sub> be the missing frequencies
                \n\"1st
                \n\\(\\overline{\\mathrm{X}}=\\frac{\\sum f_{m}}{\\mathrm{N}}\\),Given N = 35
                \n4.8286 = \\(\\frac{82+3 K_{1}+7 K_{2}}{35}\\)
                \n169 = 82 + 3K1<\/sub> + 7K2<\/sub>
                \n3K1<\/sub> + 7K2<\/sub> = 87 ……….. (1)
                \nAnd given : N = 35
                \ni. e, 18 + K1<\/sub> + K2<\/sub> = 35
                \nK1<\/sub> + K2<\/sub> = 17
                \nFrom equations (1) and (2)
                \n\"1st
                \nPut K2<\/sub> = 13 in(l) : K1<\/sub> + 9 = 17
                \nK1<\/sub> = 8
                \n\u2234 K1<\/sub> = 8, K2<\/sub> = 9 are missing frequencies.<\/p>\n

                Question 18.
                \nFind the missing frequency if the median M = 14.75
                \n\"1st
                \nSolution:
                \nLet K be the missing frequency<\/p>\n\n\n\n\n\n\n\n\n\n
                Cl<\/td>\nf<\/td>\nLCF<\/td>\n<\/tr>\n
                0 – 5<\/td>\n3<\/td>\n3<\/td>\n<\/tr>\n
                5 – 10<\/td>\n5<\/td>\n8<\/td>\n<\/tr>\n
                10-15<\/td>\n–<\/td>\n8+K<\/td>\n<\/tr>\n
                20-25<\/td>\n12<\/td>\n20+K<\/td>\n<\/tr>\n
                25-30<\/td>\n5<\/td>\n25+K<\/td>\n<\/tr>\n
                <\/td>\nN = 25+K + K<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                Since M= 14.75
                \n\u2234 Median class =10 – 15
                \n\"1st
                \n2 \u00d7 4.75K = (9 + K)J
                \n9.5K = 45 + 5K
                \n9.5K- 5K= 45
                \n4.5K = 45
                \n\u2234 K=10<\/p>\n

                Question 19.
                \nCalculate the missing frequency if median M = 10.3
                \n\"1st
                \nSolution:
                \nLet K be the missing frequency
                \n\"1st
                \nSince median M = 10.3
                \n\u2234 Median class = 9 – 12
                \n\"1st
                \n19.0667 x 2 = 68 – K
                \nK = 68 – 38.1334
                \n\u2234 K = 29.8664 \u2245 30
                \nGeometric mean and harmonic mean<\/p>\n

                \"KSEEB<\/p>\n

                Question 20.
                \nFind the G.M (i) 4.9 (ii) 8.27.64
                \nSolution:
                \n\"1st<\/p>\n

                Question 21.
                \nCalculate GM for the following data: 23,450,180,42.5, 2.7,38.2,3.58,0.20
                \nSolution:
                \nUsing logarithmic method : G.M = AL\\(\\frac{\\sum \\log X}{n}\\)<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
                X<\/td>\nlogX<\/td>\n<\/tr>\n
                23<\/td>\n1.3617<\/td>\n<\/tr>\n
                450<\/td>\n2.6532<\/td>\n<\/tr>\n
                180<\/td>\n2.2553<\/td>\n<\/tr>\n
                42.5<\/td>\n1.6284<\/td>\n<\/tr>\n
                2.7<\/td>\n1.4314<\/td>\n<\/tr>\n
                38.2<\/td>\n1.5821<\/td>\n<\/tr>\n
                3.58<\/td>\n0.5539<\/td>\n<\/tr>\n
                0.20<\/td>\n-1.3010<\/td>\n<\/tr>\n
                n = 8.<\/td>\n9.7670<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                \"1st
                \nAdd mantissa part and subtract the characteristic part
                \nGM = AL\\(\\frac{10.7670}{8}\\) = AL[1.3459] = 22.1768<\/p>\n

                Question 22.
                \nCalculate M for the following frequency distribution of diameter of P.V.C pipes supplied to a Agrahara Municipality.
                \n\"1st
                \nSolution:
                \n(Discrete data)
                \n\"1st
                \nLet X and f be the diameter (mm) and number of pipes
                \n\"1st
                \nGM = \\(\\frac{121.5286}{80}\\) = Antilog[1.5191]
                \nG.M = 33.04227<\/p>\n

                Question 23.
                \nCalculate Geometric Mean:
                \n\"1st
                \nSolution:
                \n\"1st
                \nGM = AL(1.1602)
                \nGM = 14.4617<\/p>\n

                Question 24.
                \nCalculate the GM from the following data
                \n\"1st
                \nSolution:
                \n(continuous data)
                \n\"1st
                \nm = midpoint
                \n\"1st
                \nGM = Antilog {\\(\\frac{27.8123}{40}\\)} = Antilog [0.6953]
                \nGM = 4.958<\/p>\n

                \"KSEEB<\/p>\n

                Question 25.
                \nCalculate harmonic mean of 1.05,2,1.53,2.6,1.75,1.5
                \nSolution:
                \n\"1st
                \n\"1st<\/p>\n

                Question 26.
                \nCalculate H.M from the following data : 4, 6, 8, 10, 12
                \nSolution:
                \n\"1st
                \nH.M = 6.8962<\/p>\n

                Question 27.
                \nFind Harmonic for the following frequency distribution
                \n\"1st
                \nSolution:
                \n(Discrete data)
                \n\"1st
                \nH.M = 5.0386<\/p>\n

                Question 28.
                \nCalculate the harmonic mean for the following data
                \n\"1st
                \nSolution:
                \n(Continuous data)
                \n\"1st
                \n\"1st
                \nH.M = 16.8123<\/p>\n

                Question 29.
                \nA man travels at a speed of 20kms per hour to reach a destination and return at a speed of 16 kms per hour. Find his average speed
                \nSolution:
                \n\"1st
                \nHM = \\(\\frac{2}{0.1125}\\) = 17.77 kms\/hour 0.1125<\/p>\n

                Question 30.
                \nUsing the following data show that AM >GM>HM 3,4,5,6,7.
                \nSolution:
                \nAM = \\(\\frac{\\Sigma \\mathrm{X}}{n}=\\frac{25}{5}\\) = 5
                \nGM = Antilog \\(\\frac{\\sum \\log X}{n}\\) = \\(\\frac{3.4014}{5}\\)
                \nGM=Antilog {0.6803} =4.7894
                \n\"1st
                \nHM = \\(\\mathrm{HM}=\\frac{n}{\\Sigma\\left(\\frac{1}{\\mathrm{X}}\\right)}\\)
                \nHM = \\(\\frac{5}{1.0928} \\) = 4.5754
                \nHere.AM 5; GM = 4.7894 and \u00a1-1M 4.5754
                \n\u2234 AM > GM > HM<\/p>\n

                \"KSEEB<\/p>\n

                Measures of dispersion<\/strong><\/p>\n

                Question 1.
                \nWhat is Diapcrskrn\u00edVnrlalion?
                \nSolution:
                \nThe deviation\/variation of actual salues fions any measure of central tendency is Dispersion\/ Variation. The degree\/the extent of deviation is the measure ofdispe&on. The tneaswes of dispersion are Range, Quartile deviation, Mean Deviation and the Standard deviation<\/p>\n

                Question 2.
                \nWhat are Ahsotntr and Relative measure otdhpers\u0161oits?
                \nSolution:
                \nAn absolute measure ofdispersion is a measure which measures ofdispersion of the given data and are expressed in their respective units of measurements.
                \nThe relative measure ofdispcrsion is used for the comparison ofdispersion ofdifferent data. which is a pure ratio. which does not contain any units of measurements. They are called coefficient.<\/p>\n

                Question 3.
                \nMention the desired qualities of a good measure of dispersion.
                \nSolution:<\/p>\n

                  \n
                • It is rigidly defined<\/li>\n
                • It is based on all salues ofthe dala<\/li>\n
                • It should not bc affected by the extreme alues of the data<\/li>\n
                • It should have the sampling stability<\/li>\n
                • It should be capable of further statistical analysis<\/li>\n<\/ul>\n

                  Question 4.
                  \nGive different measure of dispersion.
                  \nSolution:
                  \nThe measures of dispersion are [a] Range, [b] Quartile deviation, [c] Mean Deviation and [d] The Standard deviation<\/p>\n

                  Question 5.
                  \nDefine Range and its relative measure with their respective formula and give its Demerits Sol.- Range is defined as the difference between the highest value [H] and the smallest value [s] of the data ie. R=[H-S] is an absolute measure,
                  \nThe relative measure of. Range is: Co-efficient Range = \\(\\frac{\\mathrm{H}-\\mathrm{S}}{\\mathrm{H}+\\mathrm{S}}\\)
                  \nIts demerit is that, it is not based on all values and it is affected by extreme values.<\/p>\n

                  Question 6.
                  \nDefine Quartile deviation and relative measure with their respective formula and give its demerits.
                  \nSolution:
                  \nQuartile deviation is also called as Semi inter-Quartile Range, ie. Half of the difference between the upper [Q3<\/sub>] and lower [Q1<\/sub>]
                  \nquartile.i.e QD = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}\\) is an absolute measure.
                  \nThe relative measure of QD is- Co-efficient of QD = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{\\mathrm{Q}_{3}+\\mathrm{Q}_{1}}\\)
                  \nIts demerit is that, it is not based on all values and it \u00a1s affected by extreme values.<\/p>\n

                  Question 7.
                  \nDefine Standard deviation and relative measure with their respective formula.
                  \nSolution:
                  \nMean deviation is defined as the \u2018arithmetic mean of absalute deviations obtained from any mean\u2019 \u00a1e. \\(\\mathrm{MD}=\\frac{\\sum|\\mathrm{x}-\\overline{\\mathrm{x}}|}{\\mathrm{n}}\\), is an absolute measure
                  \nHere the deviation may be taken from mean [x] , med ian[M], mode [Z]. The relative measure
                  \n\\(\\mathrm{MD}=\\frac{\\sum|\\mathrm{x}-\\overline{\\mathrm{x}}|}{\\mathrm{n}}\\) is an absolute measure.
                  \nHere the deviation may be taken from mean [x\u0304], median[M], mode [Z]. The relative measure of Mean deviation is- CMD \\(\\frac{M D[\\bar{x}]}{\\bar{x}}\\) ; similarly for M, and Z.<\/p>\n

                  Its demerit is that, mathematically it suffers from ignoring the algebraic signs. And it gives different results for different measures, hence it is not an accurate method.<\/p>\n

                  Question 8.
                  \nDefine Standard deviation and relative measure with their respective formula.
                  \nSolution:
                  \nStandard deviation is defined as \u2018the square root of the arithmetic mean of the Squares of the deviations obtained from the mean\u2019 OR \u2018it is root mean squared Deviations obtained from mean\u2019 Denoted by
                  \n\"1st
                  \nThe relative measure of SD is-
                  \nThe co-efficient of standard deviation is- C.S.D = \\(\\stackrel{\\sigma}{\\overline{\\mathbf{x}}}\\)<\/p>\n

                  Question 9.
                  \nWhat is a Co-efficient variation\/Give the formula of C.V?
                  \nSolution:
                  \nFor better comparison if C.S.D. is expressed in percentages it is called Co-efficient of coefficient
                  \nof variation ie., co-efficient variation [c.v] = \\(\\stackrel{\\sigma}{\\overline{\\mathbf{x}}}\\) \u00d7 100<\/p>\n

                  Question 10.
                  \nWhich is the measure of dispersion is used to compare variability of two or more distributions?
                  \nSolution:
                  \nTo compare the variability\/dispersion of two or more distributions, co-efficient variation is used<\/p>\n

                  Question 11.
                  \nWhat is meant by Skewness?
                  \nSolution:
                  \nLack of symmetry or asymmetrical distribution is called Skewness.<\/p>\n

                  Question 12.
                  \nWhat are the different measures of Skewness?
                  \nSolution:
                  \n\"1st<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 13.
                  \nDefine Positive and negative Skewness.
                  \nSolution:
                  \nWhen the frequency distribution is plotted on a graph, if the right tail of the curve is Elongated or longer than the left tail, the distribution is called Positively skewed. Where as if the left tail of the curve is elongated or longer than the right tail, the distribution is called Negatively skewed.<\/p>\n

                  Question 14.
                  \nWrite any two difference between the variation and Skewness.
                  \nSolution:
                  \nVariation is the dispersion\/scatteredness of the actual values from any mean, Where as the Skewness is the spread of the frequencies around the centre or not.
                  \n\"1st<\/p>\n

                  Question 15.
                  \nWhat is a symmetric curve?
                  \nSolution:
                  \nIn a frequency distribution if the frequencies are concentrated around the centre, and the curve is non-skew ie look like bell shaped curve and the Mean, Median and Mode are equal\/coincide in the curve and two quartiles Q1<\/sub>and Q3<\/sub> }are equidistant from the median, then the curve is called symmetric curve.<\/p>\n

                  Question 16.
                  \nWrite any two properties of positive and negative Skewness.
                  \nIn Positive Skewness the right tail of the curve is longer than the left tail and here x\u0304 > Z, in Negative Skewness the left tail of the curve is longer than the right tail and x\u0304 < Z<\/p>\n

                  Question 17.
                  \nGive the formula of Karl Pearson\u2019s co-efficient of Skewness.
                  \nSolution:
                  \n\"1st<\/p>\n

                  Question 18.
                  \nGive the formula of Bowley\u2019s co-efficient of Skewness.
                  \nSolution:
                  \nBowley\u2019s Co-efficient of Skewness = \\(\\frac{\\mathrm{Q}_{3}+\\mathrm{Q}_{1}-2 \\mathrm{Q}_{2}}{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}\\)<\/p>\n

                  Question 19.
                  \nDefine the term \u201ckurtosis\u201d
                  \nSolution:
                  \nKurtosis means peaked-ness or steepness of frequency curve as compared to symmetrical curve.<\/p>\n

                  Question 20.
                  \nMention different types of kurtosis.
                  \nSolution:
                  \nKurtosis may be Mesokurtic, Leptokurtic and Platykurtic.<\/p>\n

                  Question 21.
                  \nDefine different types of kurtosis
                  \nSolution:
                  \nA distribution when its frequency curve is as Normal\/Symmetrical curve ie as peaked as Normal curve it is called Mesokurtic.
                  \nA distribution which is more peaked than Normal distribution is called Leptokurtic distribution.
                  \nA distribution which is less peaked than Normal distribution is called Platykurtic distribution<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 22.
                  \nGive the values of co-efficient of kurtosis of different of kurtosis
                  \nSolution:
                  \nFor a Mesokurtic distribution \u03b22<\/sub> = 3,
                  \nFor a Leptokurtic distribution \u03b22 <\/sub>> 3,
                  \nFor a Platycurtic distribution \u03b22 <\/sub>< 3.<\/p>\n

                  Problems<\/strong><\/p>\n

                  Question 1.
                  \nFind Range it’s co-efficient of range for the following data 351,28,40,58,42,50
                  \nSolution:
                  \nHighest value (H) = 58
                  \nLowestValue(L)=28
                  \n\u2234 Range = H – L = 58 – 28 = 30
                  \n\"1st<\/p>\n

                  Question 2.
                  \nIf the highest temperature at North pole region is 14\u00b0C and -4\u00b0C. Find Range and its coefficient.
                  \nSolution:
                  \nH = 14;L = -4
                  \nRange H-L = 14 – (- 4) = 18\u00b0C
                  \n\"1st<\/p>\n

                  Quartile deviation :
                  \nQuestion 3.
                  \nCalculate inter quartile range and semi-inter quartile (Q. D) for the following data. 10,15,8,5,16,25,20,15,18,22,16
                  \nSolution:
                  \nn = 11
                  \nIQR = O3 – Q,and Semi IQR = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{\\mathrm{Q}_{3}+\\mathrm{Q}_{1}}\\)
                  \nArrange in ascending order :
                  \n5,8,10,15,15, 16, 16, 18,20,22,25
                  \n\"1st<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 4.
                  \nCalculate coefficient of Quartile deviation for the following data:
                  \n\"1st
                  \nSolution:
                  \n\"1st
                  \n\"1st<\/p>\n

                  Question 5.
                  \nCalculate Quartile deviation for the following data:
                  \n\"1st
                  \nSolution:
                  \n\\(\\mathrm{Q} \\mathrm{D}=\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}\\)
                  \n\"1st
                  \n\"1st
                  \n\"1st<\/p>\n

                  Question 6.
                  \nFind semi – inter quartile range for the following distribution.
                  \n\"1st
                  \nSolution:
                  \nConvert less than cumcelative frequency distribution into simple (le) apper limits and h.c.fare given)
                  \nSemi mIQR = \\(\\mathrm{Q} \\mathrm{D}=\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}\\)
                  \n\"1st
                  \nQ1<\/sub> = 80 + 0.4545
                  \nQ1<\/sub> = 80.4545
                  \n\"1st
                  \n\"1st<\/p>\n

                  Questi0n 7.
                  \nCalculate Q.D and its relative measure for the following distribution.\u2019
                  \n\"1st
                  \nSolution:
                  \nConvert inclusive class interuals into exclusive C.I
                  \nHere corrcction factor: CF = \\(\\frac{\\mathrm{LLII}-\\mathrm{UII}}{2}=\\frac{20-19}{2}=1 \/ 2=0.5\\)
                  \n\"1st<\/p>\n

                  \"KSEEB<\/p>\n

                  Questi0n 8.
                  \nThe sum of lower and upper Quartiles of a set of datais 90 and their difference is 12.5. Find QD and its relative measure.
                  \nSolution:
                  \nGive Q1<\/sub> + Q1<\/sub> = Q3<\/sub> + Q1<\/sub> = 90 and Q1<\/sub> – Q1<\/sub> = 12.5
                  \n\\(\\mathrm{Q.D}-\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}=\\frac{12.5}{2}=6.25\\)
                  \nCOQ = \\(\\frac{\\mathrm{O}_{3}-\\mathrm{Q}_{1}}{\\mathrm{O}_{3}+\\mathrm{Q}_{1}}=\\frac{12.5}{90}\\) = 0.1388<\/p>\n

                  Question 9.
                  \nIf QD=12, Q1<\/sub> = 15 Find Q3<\/sub>
                  \nSolution:
                  \n\"1st<\/p>\n

                  Question 10.
                  \nIf lower and upper quartiles of a data ave 15 & 60. Find Q. D and its coefficient of Q.D.
                  \nSolution:
                  \n\"1st<\/p>\n

                  \"KSEEB<\/p>\n

                  Mean Deviation<\/strong><\/p>\n

                  Question 1.
                  \nCa!culate the mean deviation and its relative measure for the following data based on mean.
                  \n10,20,50, 70,80
                  \nSolution:
                  \n\"1st
                  \nQuestion 2.
                  \nCalculate mean deviation from median.
                  \nX : 5, 8, 12, 6, 10, 3, 10
                  \nSolution;
                  \n\"1st
                  \nHere median = \\(\\left(\\frac{n+1}{2}\\right)^{t h}\\) n= 7
                  \nArrange the ascending order
                  \n\"1st
                  \n\\(\\left(\\frac{7+1}{2}\\right)\\) = 4th ltem
                  \n\u2234 M = 8
                  \nM.D = \\(\\frac{8}{7}\\)= 1.1428
                  \nC.M.D= \\(\\frac{1.1428}{8}\\) = 01428<\/p>\n

                  Question 3.
                  \nCalculate mean deviation from made for the following data.
                  \n10, 12, 15, 18,20, 15, 16, 13, 15
                  \nSolution:
                  \n\"1st
                  \nHere made (Z) = Size of (lie most repeated item\u2019
                  \nZ=15; n=9
                  \n\"1st
                  \n\u03a3 |X – Z| = 19
                  \nMD(z)<\/sub> = \\(\\frac { 19 }{ 9 }\\) = 2.11
                  \nC.M.D =\\(\\frac { 2.11 }{ 15 }\\) = 0.1406
                  \nNote : In practice means deviation will be calculated form median.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 4.
                  \nCalculate mean deviation from mean from the following data:
                  \n\"1st
                  \nsolution:
                  \n\"1st
                  \nNote : In practice mean deviation will be calculated from median.<\/p>\n

                  Question 5.
                  \nCalculate mean deviation from mean from the following data:
                  \n\"1st
                  \nSolution:
                  \n\"1st<\/p>\n

                  Question 6.
                  \nCalculate mean deviation from median and also compute its relative Metascore.
                  \n\"1st
                  \nSolution:
                  \n\"1st
                  \n\"1st<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 7.
                  \nCompute mean deviation from made and its coefficient from the following data.
                  \n\"1st
                  \nHere Z (mode) = Size of highest frequency
                  \n\"1st
                  \nMD(z)<\/sub> = \\(\\frac { 90 }{ 45 }\\) = 2 inches
                  \nC.M.D = \\(\\frac { 2 }{ 28 }\\) = 0.0714<\/p>\n

                  Question 8.
                  \nFind mean deviation and the coefficient of mean deviation from mean for the following frequency distribution.
                  \n\"1st
                  \nSolution:
                  \n\"1st
                  \nx\u0304 = \\(\\frac { 1540 }{ 60 }\\) = 25.67
                  \nM.D = \\(\\frac { 568.04 }{ 60 }\\) = 9.4673
                  \nC.M.D = \\(\\frac { 9.467 }{ 25.67 }\\) = 0.3688<\/p>\n

                  Question 9.
                  \nCompute mean deviation from median and its relative measure from the following data.
                  \n\"1st
                  \nConvert into simple frequency distribution
                  \n\"1st
                  \nHere m – mind point
                  \nCMD = \\(\\frac{\\mathrm{MD}_{(\\mathrm{M})}}{\\mathrm{M}}\\)
                  \nMedian class = \\(\\frac{\\mathrm{N}^{\\mathrm{th}}}{\\mathrm{z}}\\) Item
                  \n\"1st<\/p>\n

                  \"KSEEB<\/p>\n

                  Standard deviation<\/strong><\/p>\n

                  Question 10.
                  \nCalulate S.D and the coefficient of S.D from following runs scored by a batsmen in 8 test matches.
                  \nSolution:
                  \nRuns: 14,28, 12, 18,40,20, 16,36
                  \n\"1st
                  \nHere\u00a0 \\(\\overline{\\mathrm{X}}=\\frac{\\Sigma \\mathrm{X}}{\\mathrm{n}} \\)
                  \nCoefficient of standand deviation:
                  \nCSD = \\frac{\\sigma}{\\bar{x}} , n = 8
                  \n\"1st<\/p>\n

                  Question 11.
                  \nFollowing is the data regarding the temperature recorded over a week in summer at a certain town. Compute standard deviation and its variance.
                  \nSolution:
                  \nTemp (in \u00b0C): 34,40,36,35,37,38,39
                  \n\"1st
                  \nHere \\(\\overline{\\mathrm{X}}=\\frac{\\Sigma \\mathrm{X}}{\\mathrm{n}} \\)
                  \nVariance=\u03c32<\/sup> , n= 7
                  \n\\(\\bar{X}=\\frac{259}{7}\\) = 37
                  \n\u03c3 = \\(\\sqrt{\\frac{28}{7}}\\) 2\u00b0C
                  \nVariance=\u03c32<\/sup> = 22<\/sup>=4\u00b0C2<\/sup>
                  \nNote : Use actual mean method only when mean’s a integer.<\/p>\n

                  Question 12.
                  \nThe following are marks scored by six students in a test. Find S.D and its relative measure marks : 8,15,12,18,10, 9.
                  \nSolution:
                  \n(Data is small size use direct method, Actual mean methed can also be used)
                  \n\"1st
                  \nCoefficient = S.D = \\(\\frac{\\sigma}{\\bar{X}}\\)
                  \n\\(\\overline{\\mathrm{X}}=\\frac{\\Sigma \\mathrm{X}}{\\mathrm{n}} \\)
                  \n\"1st<\/p>\n

                  Question 13.
                  \nFollowing is the data regarding the number accidents occurring in a month. Compute standard deviation graidents.
                  \nSolution:
                  \n\"1st
                  \nData is small in size use direct methead
                  \n\"1st<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 14.
                  \nCalculate S.D for the following data using deviation method :
                  \n15,24,18,30,28,25.
                  \nAnswer:
                  \nUsingAssumed mean method:
                  \n\"1st
                  \n\"1st<\/p>\n

                  Question 15.
                  \nCalculate the S.D for the following frequency distribution regarding weights (in kgs) of students by direct method and step deviation method.
                  \nSolution:
                  \n\"1st
                  \n\"1st<\/p>\n

                  Question 16.
                  \nCalculate S.D and the variane
                  \n\"1st
                  \nSolution:
                  \n(discrete data)
                  \n\"1st
                  \nNo common difference.
                  \n\"1st<\/p>\n

                  Question 17.
                  \nThe following gives the runs made by two Batsnian A and B in 8 innings.
                  \n\"1st
                  \nSolution:
                  \nRaw data : Compute C.V for Batsman A and B: CV \\(=\\frac{\\sigma}{\\bar{X}} \\times 100\\);
                  \n\\(\\overline{\\mathrm{x}}=\\frac{\\Sigma \\mathrm{x}}{\\mathrm{n}}\\)
                  \n\"1st
                  \n\"1st
                  \nHere C.V (A) < C.V (B)
                  \n.’. Batsmen A is more consistent and uniform in scoring runs than batsmen B.<\/p>\n

                  Question 18.
                  \nLives of CFL bulbs of two medels are given below :
                  \n\"1st
                  \nWhich of madel bulbs have more life on an average which madel is more consistent?
                  \nSolution:
                  \nCompute coefficient of variation (C. V) for two madels seperately.
                  \nC. V = \\(\\frac{\\sigma}{\\overline{\\mathrm{x}}} \\times 100\\)
                  \n\"1st
                  \nm = mid point
                  \nFor madel A
                  \n\"1st
                  \nHere fm2<\/sup> = fm.m
                  \n\"1st
                  \nFor Model B
                  \n\"1st
                  \n\"1st
                  \nConclusion : Here x\u0304B<\/sub> > x\u0304A<\/sub>
                  \n\u2234model A bulbs are better life in an average And C.V (A) < C.V (B)
                  \n\u2234Madel B bulbs are more consistentin in life.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 19.
                  \nCalculate combined standard deviation from the following data:
                  \nn1<\/sub> = 30, n2<\/sub> = 50 x\u0304 = 23 x\u03042<\/sub> = 45 ; \u03c31<\/sub>= 8, \u03c32<\/sub>= 12
                  \nSolution:
                  \nFirst compute combined mean
                  \n\"1st
                  \nThe combined standand deviation:
                  \n\"1st
                  \nHere d1<\/sub>2<\/sup> = (x\u03041<\/sub> – x\u0304)2<\/sup> = (23-36.75)2<\/sup> = 189.06
                  \nand d2<\/sub>2<\/sup> = (x\u03042<\/sub> -x\u0304)2<\/sup> = (45-36.75)2<\/sup> = 68.06
                  \n\"1st
                  \n\u03c3 = 15.08<\/p>\n

                  Question 20.
                  \nThe mean and standard deviation of montly salary of factory employees of A and B are as given below :
                  \n\"1st<\/p>\n

                  (i) Which factory pay more wages to the employees \/ greater wage bill?
                  \n(ii) Which factory wages is more uniform in wages.
                  \n(iii) Find combined \/pooled S.D
                  \nAnswer:
                  \nGiven: n1<\/sub>= 100 ; n2<\/sub>=150
                  \nx\u03041<\/sub> =7500 x\u03042<\/sub> = 5500
                  \n\u03c31<\/sub> = 1500 \u03c32<\/sub> = 1200
                  \n(i) Total wages of employees of factory A = n1<\/sub>x\u03041<\/sub> = 100 \u00d7 7500 = \u20b9 750000
                  \nTotal wages of employees of factory B = n1<\/sub>x\u03042<\/sub> = 150 \u00d7 5500 = \u20b9 825000
                  \nFrom the above factory B’s wage bill is more than factory A.
                  \n(ii) C.V (A) = \\(\\frac{\\sigma_{1}}{\\overline{\\mathrm{X}}_{1}}\\) \u00d7 100 = \\(\\frac{1500}{7500}\\) \u00d7 100 = 20%
                  \nCV(B) = \\(\\frac{\\sigma_{2}}{\\overline{\\mathrm{X}}_{2}}\\) \u00d7 100 = \\(\\frac{1200}{5500}\\) \u00d7 100 = 21.82%
                  \nC.V (A) < CV (B)
                  \nFactory A’s wages are more uniform and consistent
                  \n(iii)
                  \n\"1st
                  \nHere x\u0304 combined mean
                  \n\"1st
                  \n\"1st
                  \n\u03c3 = 1650.45 is the combined S.D of wages.<\/p>\n

                  Question 21.
                  \nIf mean and S.D of a distribution are 50 and 15 respectively. Find the coefficient of variation.
                  \nAnswer:
                  \nC.V = \\(\\frac{\\sigma}{\\bar{X}} \\times 100=\\frac{15}{50} \\times 100\\) = 30%<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 22.
                  \nIf mean and C.V of a data are 30 and 18% respectively. Find S.D.
                  \nAnswer:
                  \nC.V = \\(\\frac{6}{\\bar{X}} \\times 100\\)
                  \n15 = \\(\\frac{\\sigma}{30} \\times 100\\)
                  \n\\(\\frac{18 \\times 30}{100}\\) = \u03c3
                  \n\u2234\u00a0 \u03c3 = 5.4<\/p>\n

                  Question 23.
                  \nIf C. V and S. D of a data are 35% and 15. Find the mean.
                  \nAnswer:
                  \nC.V = \\(\\frac{\\sigma}{\\mathrm{X}}\\) \u00d7 100
                  \n35 = \\(\\frac{5}{35}[latex]\\) \u00d7 100
                  \nx\u0304 = \\(\\frac{5}{35}\\) \u00d7 100
                  \nx\u0304 = 42.86<\/p>\n

                  Skewness and Kurtosis<\/strong><\/p>\n

                  Question 1.
                  \nCalculate Karl Pearson’s coefficient of skewness from the following data.
                  \n\"1st
                  \nSolution:
                  \nDiscrete data:
                  \nK.RC.Sk = \\(=\\frac{\\bar{x}-z}{\\sigma}\\)
                  \nHere mean x =\\(\\bar{x}=\\frac{\\Sigma f x}{\\mathrm{N}}\\)
                  \n\"1st
                  \nHere fx2<\/sup> = fx . x
                  \n\\(\\bar{x}=\\frac{\\Sigma f x}{\\mathrm{N}}=\\frac{1445}{70}\\)
                  \nx\u0304 = 20.6428
                  \nz = Heighest frequency is 18, its size is 25
                  \n\u2234\u00a0 z = 25
                  \n\"1st
                  \nKarl pearson’s coefficient of skewness
                  \n= \\(\\frac{x-z}{\\sigma}=\\frac{20.6428-25}{7.1675}\\)
                  \nK.P.C.SK = 0.6078
                  \nThere exists a high degree positive skewness.
                  \nNote: Observe that there are more frquecies at the end of the distribution. So skewness is + ve<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 2.
                  \nCalculate Bowley’s coefficient of skewness:
                  \n\"1st
                  \nSolution:
                  \nDiscrete data:
                  \n\"1st
                  \nHere, quartiles can be calculated:
                  \nAfter calculating less than cumulative frquency (L.C.f)
                  \nQj<\/sub> class = size of \\(j\\left(\\frac{N+1}{4}\\right)^{t h}\\) item ; j = 1, 2, 3
                  \n\"1st
                  \nThe distribution is non-skew i.e., symmetrical.<\/p>\n

                  Question 3.
                  \nFrom the following frequency distribution compute karl pearsons coefficient of skewness
                  \n\"1st
                  \nSolution:
                  \nK.P.C.Sk = \\(\\frac{\\bar{x}-z}{\\sigma}\\)
                  \nHere, for continuous data use:
                  \n\"1st
                  \nModal class = size of highest frequency.
                  \n\"1st
                  \nModal class = 10-20
                  \n\"1st
                  \nThere exists high degree + ve skewness<\/p>\n

                  Question 4.
                  \nCalculate Bowicy\u2019s coefficient of skewness from the following
                  \n\"1st
                  \nSolution:
                  \n\"1st
                  \nTo compute quartiles, Qj<\/sub>class = \\(\\frac{N^{\\text {th }}}{4}\\) item
                  \n\"1st
                  \nBowley’s coefficient of skewness :
                  \n\"1st
                  \nThe distribution is positively skewed.<\/p>\n

                  Question 5.
                  \nCompute B.C.Sk from the following data
                  \n\"1st
                  \nSolution:
                  \nConvert inclusive class intervals into exclusive class intervals. Using the correction factor: LL II class – ULI class
                  \n\\(\\frac{\\text { LL II class -ULI class }}{2}\\)
                  \nCF = \\(\\frac{70-69}{2}\\) = 0.5
                  \nSubtract CF in all lower limits and add to all upper limits, we get,
                  \n\"1st
                  \nBowley’s coefficient of skewness :
                  \n\"1st
                  \nThe distribution is positively skewness.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 6.
                  \nCompute the coefficient of skewness based on Pearson’s method.
                  \n\"1st
                  \nSolution:
                  \nIn the above frequency distribution, the frequencies are asymmetrical. So, to find skewness use:
                  \nK.P.C.S = 3 \\(\\left(\\frac{x-m}{\\sigma}\\right)\\)
                  \nHere x\u0304 = \\(\\frac{\\Sigma f x}{N}\\)
                  \nMedian after finding less than cumulative frequency (L.C.F)
                  \nM = \\(\\left(\\frac{\\mathrm{N}+1}{2}\\right)^{\\mathrm{th}}\\) item; and
                  \n\"1st
                  \nKarl Pearson’s coefficient of skewness:
                  \n\"1st
                  \nNote: Same type of problems can also be solved for continuous data.<\/p>\n

                  Question 7.
                  \nFind the coefficient of skeness if mean = 50, Made = 53 and S.D = 5.
                  \nSolution:
                  \nKarl Pearson’s coefficent of skewness = \\(\\frac{\\bar{x}-z}{\\sigma}\\)
                  \ncoefficent of slewness = \\(\\frac{50-.53}{5}\\) = 0.6<\/p>\n

                  Question 8.
                  \nA frequency distribution given the following results : (i) CV = 5, (ii) SD = 2 (iii) Karl Pearson’s coefficient of skewness = 0.5. Find the mean and the made of the distribution.
                  \nSolution:
                  \nWe know that : C.V = \\frac{\\sigma}{\\bar{x}} \u00d7 100
                  \n0.5 = \\(\\frac{50-z}{2}\\)
                  \n\u2234 x\u0304 = \\(\\frac{2}{5}\\) \u00d7 100 = 40
                  \nK.P.C.Sk = \\(\\frac{\\bar{x}-z}{\\sigma}\\)
                  \n0.5 \u00d7 2 = 40 – z
                  \n5 = \\(\\frac{2}{x}\\) \u00d7 100
                  \n\u2234 z = 40 – 1 =39<\/p>\n

                  Question 9.
                  \nFor a distribution, Bowley’s coefficient of skewness is -0.56, Q1<\/sub> = 16 and Q2<\/sub> = 24. find coefficient of quartile deviation.
                  \nSolution:
                  \n\"1st
                  \n-0.56(Q3<\/sub> – 16) = Q3<\/sub> + 16 – 48
                  \n-0.56Q3<\/sub> +8.96 =Q3<\/sub> – 32
                  \n32 + 8.96 = Q3<\/sub> + 0.56Q3<\/sub>
                  \n40.96 = 1.56Q3<\/sub>
                  \n\u2234 Q3<\/sub> = \\(\\frac{40.96}{1.56}\\) = 26.26
                  \nNow, coeffient of quartile deviation = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{\\mathrm{Q}_{3}+\\mathrm{Q}_{1}}\\)
                  \n= \\(\\frac{26.26-16}{26.26+16}\\) = \\(\\frac{10.26}{42.26}\\) = 0.24<\/p>\n

                  Question 10.
                  \nCalcualte K.P.C.Sk if Mean = 40, Median 45 and \u03c3 = 15.
                  \nSolution:
                  \n\"1st<\/p>\n

                  Question 11.
                  \nThe mean of a distribution is 25. Its S.D is 8 and the coefficient of skewness is -1. Find the median.
                  \nSolution:
                  \nK.P.C.Sk = 3 (\\(\\frac{\\bar{x}-\\mathrm{M}}{\\sigma}\\))
                  \n-1 =3(\\(\\frac{28-\\mathrm{M}}{8}\\))
                  \n\\(\\frac{-8}{3}\\) = 25\u2014M
                  \nM = 25 + 2.67
                  \nM = 27.67<\/p>\n

                  Question 12.
                  \nIn a distribution, sum of Lower and upper quartiles is 60 and their difference is 20. If the median 22, (Then find then coefficient of skeness.
                  \nSolution:
                  \nGiven Q3<\/sub> + Q1<\/sub> = 60 and Q3<\/sub> – Q1<\/sub>= 20
                  \nMedian Q2<\/sub> = 22
                  \n\"1st<\/p>\n

                  Question 13.
                  \nIn a distribution, if Q1<\/sub> = 50 and semi-inter quartile range is 15. If the (listribLltiOn is symmetrical, then find Q2<\/sub>.
                  \nSolution:
                  \nSemi-inter quartile range. Quartile deviation = \\(\\frac{\\mathrm{Q}_{1}-\\mathrm{Q}_{1}}{2}\\) = 15
                  \nQ3<\/sub> – 50 = 15 \u00d7 2
                  \nQ3<\/sub> = 30 + 50 = 80
                  \nFor a symmetrical distribution, the two quartiles are equidistant from the median:
                  \n\u2234 Q3<\/sub> – Q2<\/sub> = Q2<\/sub> – Q1<\/sub>
                  \n2Q2 <\/sub>= Q3<\/sub> + Q1<\/sub>
                  \nMedian\/Q2<\/sub> = \\(\\frac{\\mathrm{Q}_{3}+\\mathrm{Q}_{1}}{2}=\\frac{80+50}{2}\\) = 65<\/p>\n

                  Question 14.
                  \nWhat is the relation between Q1<\/sub>, Q2<\/sub> and Q3<\/sub> \u00a1n a positively skewed frequency distribution?
                  \nSolution:
                  \nFor a positively skewed distribution:
                  \n(Q2<\/sub> – Q1<\/sub>) < (Q3<\/sub> – Q2<\/sub>),
                  \nSimilarly for a Negatively skewed distribution : (Q2<\/sub> – Q1<\/sub>) > (Q3<\/sub> – Q2<\/sub>)<\/p>\n

                  Question 15.
                  \nThe first four central monientsare, \u2018O\u2019, 20, -50 and loo comment of skewness and kurtosis.
                  \nSolution:
                  \nGiven:
                  \n\u00b51<\/sub>= 0
                  \n\u00b52<\/sup> = 20
                  \n\u00b53<\/sup> = 50
                  \n\u00b54<\/sup>= 100
                  \n\"1st
                  \n\u03b21<\/sub> = 0.3129
                  \nThe distribution is positively skewed.
                  \nKurtosis: \u03b22<\/sub> = \\(\\frac{\\mu_{4}}{\\mu_{2}^{2}}=\\frac{100}{20^{2}}=\\frac{100}{40}\\) = 2.5
                  \nHere \u03b22<\/sub> > 3 the distribution is platykurtic.<\/p>\n

                  Question 16.
                  \nIf \u03b22<\/sub> = 1.25. What type of kurtosis the distribution has?
                  \nSolution:
                  \nHere \u03b23<\/sub> < 3, the distribution is platykurtic.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 17.
                  \nIf \u03b22<\/sub>= 3.723. What type of Kurtosis tue distribution lias?
                  \nSolution:
                  \nHere \u03b22<\/sub> > 3, the distribution is heptokurtic.<\/p>\n

                  Question 18.
                  \nIf \u00b52<\/sub>\u00a0= 32.15 and \u00b54<\/sup> = 4954 find \u03b22<\/sup>.
                  \nSolution:
                  \nKutosis \u03b22<\/sub>\u00a0= \\(\\frac{\\mu_{4}}{\\mu_{2}^{2}}=\\frac{4954}{32.15^{2}}\\) = 4.79<\/p>\n

                  Insert at the end of the unit Xl<\/strong>
                  \nUnit XI Analysis of Bivariate<\/strong><\/p>\n

                  Question 1.
                  \nDraw a slatter diagram for the following data and commen on correlation.
                  \n\"1st
                  \nSolution:
                  \n\"1st
                  \nFrom the diagram, the scatterd points shows raising straightline. There exists a high degree positive correlation between price and supply.<\/p>\n

                  Question 2.
                  \nDraw a scatter diagram for the following data and interpret it.
                  \n\"1st
                  \nSolution:
                  \n\"1st
                  \nFrom the scattered diagram, the spraded fo the points show downward diagram. Hence x and y are negatively correlated.<\/p>\n

                  Question 3.
                  \nThe following data gives the age of a motor cycle and the annual maintenance cost over a period. Find the product moment coefficient of correlation and interpret the same.
                  \n\"1st
                  \nSolution:
                  \nLet x andy be the age and maintenance cost, the data is small in size, so use direct method.
                  \n\"1st
                  \n\u2234 There exists a high degree positive correlation.<\/p>\n

                  Question 4.
                  \nCalculate karl pearson’s coefficient of correlation between .v and y.
                  \n\"1st
                  \nSolution:
                  \nThe data is large in size, so use step deviation or short cut method.
                  \n\"1st
                  \nwhere A, B assumed means and C – common factors from x and y.
                  \n\"1st
                  \nHere \u03b3 > 0.5. The exists high degree or close to perfect positive correlation.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 5.
                  \nFind K.P.C.S of the following data
                  \n\"1st
                  \nSolution:
                  \nLet x and y be ages of husbands and wifes. Here by observation method. Arithmetic means are whole numbers\/ integers. 50, it is convenient to use actual mean method.
                  \n\"1st
                  \n\"1st
                  \nThe correlation between ages of husbands and wifes is high degree.<\/p>\n

                  Question 7.
                  \nCalculate the coefficient fo correlation between the number of children and age of mothers from the following bivariate frequency table.
                  \n\"1st
                  \nSolution:
                  \nLet x and y be the ages of mothers and number of children.
                  \nThe data is discrete nature for ‘y’ and with a common difference in x:
                  \n\"1st
                  \n\"1st
                  \nThere exists a high degree negative correlation between ages of mothers and number of children.<\/p>\n

                  Question 8.
                  \nCalculate the karl pearson’s coefficient of correlation between ages of students and marks obtained in a certain test and interpret.
                  \n\"1st
                  \nSolution:
                  \nLet x and y be the marks and age
                  \n\"1st
                  \n\"1st
                  \nHere, there exists high degree negative correlation between Age of students and their marks.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 9.
                  \nCalculate K.P.C.C from the following date.
                  \n\"1st
                  \nSolution:
                  \nBoth variables are continuous so, use step deviation method.
                  \n\"1st<\/p>\n

                  Question 10.
                  \nIf the covariance between x and y variables is 12.5 and variance of x and y are respectively 16.4 and 13.8. Find the coefficient of correlation between them.
                  \nSolution:
                  \nWe know that:
                  \n\"1st<\/p>\n

                  Question 11.
                  \nGiven COV (x, y) = -30, Var (x) = 225 and Var (y) = 9, find karl pearson’s coefficient of correlation.
                  \nSolution:
                  \n\"1st<\/p>\n

                  Question 12.
                  \nIn a bivariate data, if standard deviations of x and y are 5 and 9 respectively. If the COV (x, y) = 8, find the coefficient of correlation.
                  \nSolution:
                  \n\"1st<\/p>\n

                  Question 13.
                  \nFind the coefficent of correlation:
                  \nSolution:
                  \n\"1st<\/p>\n

                  Question14.
                  \nIn a bivariate data \u03a3xy = 500, \u03a3x = 15, \u03a3y = 200, \u03a3x2<\/sup> = 98, \u03a3y2<\/sup> = 23250 and n = 10. Find the coefficient of correlation.
                  \nSolution:
                  \n\"1st<\/p>\n

                  Question 15.
                  \nThe coefficient of correlation between two variables x and y is 0.65, their covariance is 2.32. The variance of x is 18. Find standard deviation of y.
                  \nSolution:
                  \n\"1st<\/p>\n

                  Question 16.
                  \nIf the coefficient of correlation between x and y is ‘O’ comment.
                  \nSolution:
                  \nIf \u03b3 = 0. the two variables x and y are independent, i.e., Non correlated.<\/p>\n

                  Question 17.
                  \nDefine spurious correlation.
                  \nSolution:
                  \nAbsence of causation between the variables is called spurious correlation.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question19.
                  \nCalculate spearman’s rank correlation for the following data regarding sports and studies of 8 students:
                  \n\"1st
                  \nSolution:
                  \nRanks are given. Let R1<\/sub> and R2<\/sub> be the ranks in sports and studies.
                  \n\"1st
                  \nd = R1<\/sub> – R2<\/sub>, n- No. of observations = 8
                  \n\"1st
                  \nThere exists low degree negative correlation between sports and studies.<\/p>\n

                  Question 20.
                  \nTen competitors in a pointing competition are ranked by two judges. Using rank correlation find out whether the two judges have common table in painting:
                  \n\"1st
                  \nSolution:
                  \nRanks are given. Let R1<\/sub> and R2<\/sub> be the ranks of judgcs I and II
                  \n\"1st
                  \nHere d=R1<\/sub> – R2<\/sub>, n = 10
                  \n\"1st
                  \n\u03c1 = \\(1-\\frac{6 \\times 38}{10^{3}-10}\\)
                  \n\\(\\frac{228}{990}\\) = 1 – 0.2303
                  \n\u03c1 = 0.7097
                  \nthe two Judges have high degree of similar taste in beauty of paintings.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 21.
                  \nThe corresponding values of 2 series are given below.
                  \n\"1st
                  \nFind the coefficient of rank correlation of the above series.
                  \nSolution:
                  \nRanks are not given. We have to assign rank. Assign ranks R1<\/sub> and R2<\/sub> for the variables x and y in descending order as highest value as 1, 2, upto 7.
                  \n\"1st
                  \nThere exists a high degree positive correlation.<\/p>\n

                  Question 22.
                  \nCalculate rank correlation from the following marks in Accountancy and statistics.
                  \n\"1st
                  \nSolution:
                  \nRanks are not given. Assign ranks as R1<\/sub> and R2<\/sub> for the marks in Accountancy and statistics as 1st<\/sup> rank who got highest marks and 2nd <\/sup>to next less marks.
                  \n\"1st
                  \nd=R1<\/sub> – R2<\/sub>, n = 10
                  \n\"1st
                  \n\u03c1 = \\(1-\\frac{6 \\times 106}{10^{3}-10}\\) = 1 – 0.6424
                  \n\u2234 \u03c1 = 0.3576
                  \nThere exists Low degree correlation between marks in Accountancy and Statistics.<\/p>\n

                  Question 23.
                  \nCalculate rank correlation coefficient from the following data:
                  \n\"1st
                  \nSolution:
                  \nHere the values of the variables are repeated.
                  \nSo use, the formula;
                  \n\"1st
                  \nWhere correction factor:CF = \\(\\frac{m^{3}-m}{12}\\); m – No. of times the variable repeated.
                  \nn = 8; Let R1<\/sub> and R2<\/sub> be the ranks of x and y.
                  \n\"1st
                  \nm1<\/sub> = 3 (65 repeated 3 times)
                  \nRanks are averaged as \\(\\frac{6+7+8}{3}\\)= 7
                  \nm2<\/sub> = 2(13 repeated 2 times)
                  \nRanks are averages as = \\(\\frac{4+5}{2}\\)= 4.5
                  \n\"1st
                  \nThere exists high degree positive correlation between x and y.<\/p>\n

                  \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

                  You can Download Chapter 5 Analysis of Univariate Data Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers\u00a0Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Statistics Question Bank Chapter 5 Analysis of Univariate Data 1st PUC Statistics Analysis of Univariate Data …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12214"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=12214"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12214\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=12214"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=12214"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=12214"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}