{"id":17464,"date":"2020-01-03T15:04:38","date_gmt":"2020-01-03T09:34:38","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17464"},"modified":"2021-07-02T15:31:24","modified_gmt":"2021-07-02T10:01:24","slug":"1st-puc-statistics-previous-year-question-paper-march-2019-north","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-statistics-previous-year-question-paper-march-2019-north\/","title":{"rendered":"1st PUC Statistics Previous Year Question Paper March 2019 (North)"},"content":{"rendered":"

Students can Download 1st PUC Statistics Previous Year Question Paper March 2019 (North), Karnataka 1st PUC Statistics Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Statistics Previous Year Question Paper March 2019 (North)<\/h2>\n

Time: 3 Hrs 15 Min
\nMax. Marks: 100<\/p>\n

Note :<\/p>\n

    \n
  1. Graph sheets and statistical tables will be supplied on request<\/li>\n
  2. Scientific calculators are allowed<\/li>\n
  3. All working steps should be clearly shown.<\/li>\n<\/ol>\n

    Section-A<\/span><\/p>\n

    I. Answer any ten of the following questions: (10 \u00d7 1 = 10)<\/span><\/p>\n

    Question 1.
    \nDefine statistics according to Croxton and Cowden’s.
    \nAnswer:
    \nStatistics is the science of collection, presentation, analysis and interpretation of numerical data.<\/p>\n

    Question 2.
    \nDefine population.
    \nAnswer:
    \nThe total number of units under consideration is called population.<\/p>\n

    Question 3.
    \nWhat is sampling?
    \nAnswer:
    \nIt is method of collection of the data based on a part or a portion of representative items from the population.<\/p>\n

    Question 4.
    \nMention One objective of classification.
    \nAnswer:
    \nClassification reduces the size of the data.<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nWrite the formula to find the mid-point of a class.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 6.
    \nWhich graph is used to locate median.
    \nAnswer:
    \nOgives are used to locate median.<\/p>\n

    Question 7.
    \nFind the geometric mean of 3 and 27.
    \nAnswer:
    \nG.M = \\(\\sqrt{3 \\times 27}=\\sqrt{81}=\\sqrt{9^{2}}\\) = 9<\/p>\n

    Question 8.
    \nDefine Kurtosis.
    \nAnswer:
    \nKurtosis means peakedness or steepness of a frequency curve as compared to a normal curve.<\/p>\n

    Question 9.
    \nFor data, if M = 25 then what is value of P50<\/sub>?
    \nAnswer:
    \nP50<\/sub> = 25<\/p>\n

    Question 10.
    \nGive the range of correlation co-efficient.
    \nAnswer:
    \n\u03b3 = \u00b1 1 or – 1 \u2264 \u03b3 \u2264 + 1<\/p>\n

    Question 11.
    \nWhat is interpolation?
    \nAnswer:
    \nInterpolation is the procedure of estimating the unknown value of the dependent variable for a given value of the independent variable which is within the limits of independent variable.<\/p>\n

    Question 12.
    \nIf P (A) = \\(\\frac { 1 }{ 4 }\\) then find P (A1<\/sup>).
    \nAnswer:
    \nP(A1<\/sup>) = l – P(A) = l – \\(\\frac { 1 }{ 4 }\\) = \\(\\frac { 3 }{ 4 }\\)<\/p>\n

    Section-B<\/span><\/p>\n

    II. Answer any ten of the following questions: (10 \u00d7 2 = 20)<\/span><\/p>\n

    Question 13.
    \nWrite the limitations of statistics.
    \nAnswer:<\/p>\n

      \n
    1. Statistics does not deal with qualitative data. It deals with only quantitative data.<\/li>\n
    2. Statistical results are riot exact.<\/li>\n<\/ol>\n

      Question 14.
      \nWhat is primary data? Mention one method of collection of primary data.
      \nAnswer:
      \nPrimary data are those which are directly collected by the investigator himself initialls for his own purpose.
      \nMethod of collection of primary date is direct personal interview method.<\/p>\n

      Question 15.
      \nMention two parts of a table.
      \nAnswer:
      \nThe parts of table are
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 16.
      \nDefine range and class limits.
      \nAnswer:
      \nRange is the difference between the highest and the lowest value in the data.
      \nThe lowest and the highest values which are taken to define the boundaries of a class are called class limits.<\/p>\n

      Question 17.
      \nMention two uses of diagrams and graphs.
      \nAnswer:<\/p>\n

        \n
      1. They give a birds eye view of the entire data at a glance.<\/li>\n
      2. They can be easily understood by the common people.<\/li>\n<\/ol>\n

        Question 18.
        \nIf n1<\/sub> = 100, n2<\/sub> = 150, X\u03041<\/sub> = 10.4 and X\u03042<\/sub> = 10.9 Find the combined mean.
        \nAnswer:
        \n\"1st<\/p>\n

        Question 19.
        \nFor a data, if median is 45 and the mean deviation from median is 10 then find its co\u00ac. efficients.
        \nAnswer:
        \nCo – efficient of mean deviation :
        \n\"1st<\/p>\n

        Question 20.
        \nIf bxy = \\(\\frac { -3 }{ 4 }\\) and byx = \\(\\frac { -1 }{ 3 }\\) find the correlation co-efficient
        \nAnswer:
        \n\u03b3 = \u00b1 \\(\\sqrt{\\mathrm{bxy}, \\mathrm{byx}}\\)
        \n\u03b3 = \\(\\sqrt{\\frac{-3}{4} \\times \\frac{-1}{3}}\\) = \u00b1 0.5<\/p>\n

        Question 21.
        \nGive any two properties of regression lines.
        \nAnswer:<\/p>\n

          \n
        1. Two regression lines coincide at (x\u0304, y\u0304)<\/li>\n
        2. Two regression lines intersect when there is a perfect correlation.<\/li>\n<\/ol>\n

          Question 22.
          \nIn case of two attributes, if N = 100, (AB) = 20, (A) = 56 and (B) = 22, then find the remaining classes and their frequencies.
          \nAnswer:
          \nThe 2 \u00d7 2 contingency table is as shown below.
          \n\"1st<\/p>\n

          Question 23.
          \nDefine sample space and equally likely events.
          \nAnswer:
          \nThe set of all possible outcomes of a random experiment is called sample space.
          \nTwo or more events are said to be equally likely events if they have equal chances of occurrence.<\/p>\n

          Question 24.
          \nIf E(x2<\/sup>) = 74 and E(x) = 5 find the standard deviation.
          \nAnswer:
          \n\"1st<\/p>\n

          \"KSEEB<\/p>\n

          Section-C<\/span><\/p>\n

          III. Answer any eight of the following questions . (8 \u00d7 5 = 40)<\/span><\/p>\n

          Question 25.
          \nWrite a few important functions of statistics.
          \nAnswer:
          \nThe following are the important functions of statistics:
          \n(a) It simplifies and condenses the size of the data.
          \n(b) It facilitates composition.
          \n(c) It enriches our knowledge and widens our experience.
          \n(d) It helps in formulating policies.
          \n(e) It helps in business forecasting.
          \n(f) It presents the facts in a definite form.<\/p>\n

          Question 26.
          \nMention the five rules for drafting a questionnaire.
          \nAnswer:
          \nFollowing are some important points\/rules to be considered while framing a questionnaire.<\/p>\n

            \n
          1. Questions should be simple and easy to understand and get spontaneous answers.<\/li>\n
          2. Questions should not confuse the reader and they should give only one meaning.<\/li>\n
          3. The number of questions should be kept to a minimum.<\/li>\n
          4. Questions should be such that people with little intelligence and educated ones can both answer without much difficults.<\/li>\n
          5. The questions should be arranged in a logical order.<\/li>\n<\/ol>\n

            Question 27.
            \nPrepare a blank table to show the distribution of students of a College according to
            \n(i) Faculty : Arts, Commerce and Science
            \n(ii) Sex : Male and Female
            \n(iii) Year : 2009 and 2010
            \nAnswer:
            \nBlank table showing the distribution of the college students according to faculty, sex and the year.
            \n\"1st<\/p>\n

            \"KSEEB<\/p>\n

            Question 28.
            \nFollowing are the marks obtained by two students A and B in an annual examination. Represent the data by percentage bar diagram.
            \n\"1st
            \nAnswer:
            \n\"1st
            \ne.g.Kannada = \\(\\frac { 75 }{ 517 }\\) \u00d7 100 = 15 %
            \nPercentage bar diagram showing the marks of students A and B.
            \n\"1st<\/p>\n

            Question 29.
            \nCalculate geometric mean from the following data
            \n\"1st
            \nAnswer:
            \n\"1st<\/p>\n

            \"KSEEB<\/p>\n

            Question 30.
            \nCalculate the Spearman’s rank correlation co-efficient for the following data marks of 8 students in statistics (x) and mathematics (y).
            \n\"1st
            \nAnswer:
            \nLet Rx<\/sub> and Ry<\/sub> be the ranks of X and Y
            \n\"1st
            \nSpearman\u2019s rank correlation coefficients:
            \n\"1st
            \nThere exists a high degree of positive correlation between x and y.<\/p>\n

            Question 31.
            \nCalculate yule’s association from the following data.
            \n\"1st
            \nAnswer:
            \nThe given data can be written under 2 \u00d7 2 contingency table as below.
            \n\"1st
            \nThere exists a low degree of negative association between marriage and failure.<\/p>\n

            \"KSEEB<\/p>\n

            Question 32.
            \nThe number of births taking place annually in a village are as follows:
            \n\"1st
            \nFind the missing value for the year 2008.
            \nAnswer:
            \n\"1st
            \nSince there are 5 known values of y, so expand the binomial expansion (y – 1)5<\/sup> = 0,
            \ny5<\/sub> – 5<\/sup>y4<\/sub> + 10<\/sup>y3<\/sub> – 10<\/sub>y2<\/sup> + 5<\/sup>y1<\/sub> – y0<\/sub> = 0
            \n248 – 5 \u00d7 255 + 10 \u00d7 260 – 10<\/sup>y2<\/sub> + 5 \u00d7 280 – 300 = 0.
            \n248 – 1275 + 2600 – 10<\/sup>y2<\/sub> + 1400 – 300 = 0.
            \ny2<\/sub> = 263.7<\/p>\n

            Question 33.
            \nState and prove addition theorem of probability for two non-mutually exclusive events.
            \nAnswer:
            \nStatement: Let A and B be any two events (Subsets of sample spaces) with respective probability P(A) and P(B). Then, the probability of occurence of atleast one of these events is:
            \nP(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
            \nwhere P(A \u2229 B) is the simultaneous occurence of A and B.<\/p>\n

            Proof: A random experiment results in \u2018n\u2019 exhaustive outcomes out of which movt comes are favourable to event \u2018A\u2019, m2 outcomes are favourable to event \u2018B\u2019 and \u2018p\u2019 outcomes are common to both A and B.
            \nConsider the venn diagram:
            \n\"1st
            \n\"1st
            \nThe event of occurrence of at least one out of A and B is (A \u222a B) has (m1<\/sub> + m2<\/sub> – p) favourable outcomes.
            \n\"1st
            \nSubstituting the values from (1)
            \nP(A \u222a B) = P(A) + P(B) – P(A \u2229B)
            \nQ.E.D.<\/p>\n

            Question 34.
            \nA bag contains S white and 4 green balls. Two balls are drawn from the bag at random. Find the probability that they are:
            \n1. of the same colour.
            \n2. of different colours.
            \nAnswer:
            \nn = Two balls can be drawn from 5 white and 4 green balls in 9C2<\/sub> = 36 ways.
            \n1. m = 2 balls of the same colour can be drawn in (2 W or 2 G) = (5C2<\/sub> \u00d7 4C2<\/sub>) = (10 + 6) = 16 ways.
            \nLet P(A) = P( 2 balls of the same colour) = \\(\\frac { m }{ n }\\) = \\(\\frac { 16 }{ 36 }\\) =\\(\\frac { 4 }{ 9 }\\)<\/p>\n

            2. P(A1<\/sup>) = P(2 balls of different colours) = 1 P(A) = 1 – \\(\\frac { 4 }{ 9 }\\) = \\(\\frac { 5 }{ 9 }\\)<\/p>\n

            \"KSEEB<\/p>\n

            Question 35.
            \nCalculate the Karl Pearson’s co-efficient of correlation from the following data.
            \n\"1st
            \nAnswer:
            \n\"1st
            \nKarl Pearson co-efficient of correlation
            \n\"1st
            \nThere exists a low degree of positive correlation between x and y.<\/p>\n

            Question 36.
            \nCalculate D5<\/sub> and P28<\/sub> from the following data.
            \n\"1st
            \nAnswer:
            \n\"1st
            \n\"1st
            \n\"1st
            \n\"1st<\/p>\n

            \"KSEEB<\/p>\n

            Section-D<\/span><\/p>\n

            IV. Answer any two of the following questions: (2 \u00d7 10 = 20)<\/span><\/p>\n

            Question 37.
            \nCalculate Mean, Median and Mode from the following data.
            \n\"1st
            \nAnswers:
            \nLet x and f be the marks and number of students.
            \n\"1st
            \n\"1st
            \nMode = Size of the highest frequency
            \n= Highest frequency is 17. Its size (x) is = 30
            \n\u2234 z = 30<\/p>\n

            Question 38.
            \nCalculate Bowley’s co-efficient of Skewness of the following data
            \n\"1st
            \nAnswer:
            \nLet CI and f be the marks and number of students.
            \n\"1st
            \n\"1st
            \n\"1st
            \nThere exists a low degree skewness.<\/p>\n

            \"KSEEB<\/p>\n

            Question 39.
            \nFrom the following data find two regression equations.
            \n\"1st
            \nAnswer:
            \n\"1st
            \n\"1st
            \nRegression equation of x on y:
            \n(x – x\u0304) = bxy (y – y\u0304)
            \nx – 6 = -0. 95 (y -7)
            \n= -0.95 y + 6.65
            \nx = -0.95 y + 6.65 + 6
            \nx = -0.95y + 12.65<\/p>\n

            Regression equation of y on x:
            \n(y – y\u0304) = byx\u00a0 (x – x\u0304)
            \ny – 6 = -0.2474 (x – 6)
            \n= -0.2474 x + 1.4844
            \ny = – 0.2474 x + 1.4844 + 6
            \ny = – 0.2474 x + 7.4844.<\/p>\n

            Question 40.
            \n(a) The probability of two students A and B solving a problem are \\(\\frac { 1 }{ 2 }\\) and \\(\\frac { 3 }{ 4 }\\) respectively. If both of them independently try, what is the probability that the problem gets solved.
            \n(b) Find K and mean [E(x)] of the following distribution.
            \n\"1st
            \nAnswer:
            \n(a) Let the probability of student A and B be solving a problem be P (A) = \\(\\frac { 1 }{ 2 }\\) and P (B) = \\(\\frac { 3 }{ 4 }\\) Probability that the problem is solved is :-
            \nProbability that at least one will solve is
            \nP(A \u222a B) = P (A) + P (B) – P(A \u2229 B)
            \n= P(A) + P(B) – P(A) P(B)
            \n\\(=\\frac{1}{2}+\\frac{3}{4}-\\frac{1}{2} \\times \\frac{3}{4}\\)
            \nP(A \u222a B) = 0.5 + 0.75 – 0.375
            \n= 0.875<\/p>\n

            (b) We know that \u03a3P(x) = 1
            \n\\(\\frac{3}{8}+\\frac{1}{4}+k+\\frac{3}{16}+\\frac{1}{16}=1\\)
            \nk = 1 – \\(\\left(\\frac{3}{8}+\\frac{1}{4}+\\frac{3}{16}+\\frac{1}{16}\\right)\\)
            \n= 1 – 0.875
            \nk = 0.125
            \nThe probability distribution is
            \n\"1st
            \nMean = E(x) = \u03a3x P(x)
            \n\\(=0 \\times \\frac{3}{8}+1 \\times \\frac{1}{4}+2 \\times 0.125+3 \\times \\frac{3}{16}+4 \\times \\frac{1}{16}\\)
            \nE(x) = 1.3125<\/p>\n

            \"KSEEB<\/p>\n

            Section-E<\/span><\/p>\n

            V. Answer any two of the following questions : (2 x 5 = 10)<\/span><\/p>\n

            Question 41.
            \nThe following are the heights in centimeters of 40 students:
            \n\"1st
            \nFrom a frequency distribution by exclusive Cl taking lowest class interval as (145 -150) cms
            \nAnswer:
            \nFrequency distribution of height (in cms)
            \n\"1st<\/p>\n

            Question 42.
            \nDraw a histogram and find the value of mode.
            \n\"1st
            \nAnswer:
            \nConvert inclusive class internal into exclusive class internal.
            \n\"1st
            \nHistogram showing the weight of students (in kg)
            \n\"1st<\/p>\n

            \"KSEEB<\/p>\n

            Question 43.
            \nThe first four central moments of frequency distribution are 0,4,-2 and 2.4. comment on the skewness and Kurtosis.
            \nAnswer:
            \nGiven: \u00b51<\/sub> = 0, \u00b52<\/sub> = 4, \u00b53<\/sub> = -2, \u00b54<\/sub> = 2.4,
            \n\"1st
            \nThe distribution is low degree positive.
            \n\"1st
            \nHere \u03b22<\/sub>< 3. The distribution platykurtic.<\/p>\n

            Question 44.
            \nA bag contains 4 white and 5 red marbles. Two marbles are drawn one after the 1 other without changing. Find the probability of getting both as white marbles.
            \nAnswer:
            \nn = (4W + 5 R) = 9 ways to draw one ball from 9 balls
            \nLet probability of drawing white and red marbles be P(W) and P(R)
            \nP(drawing 2 white marbles one after the other)
            \n= P (2W) = P (white marble in I draw) and P (drawing white marble in II draw)
            \nP(WI \u2229 WII) = P (WI) P (WII\/ WI)
            \nP(A \u2229 B) = P(A) P(B\/A)
            \n\\(=\\frac{4}{9} \\times \\frac{3}{8}=\\frac{12}{72}=\\frac{1}{6}\\)
            \nHere, first drawn ball is not replaced.<\/p>\n","protected":false},"excerpt":{"rendered":"

            Students can Download 1st PUC Statistics Previous Year Question Paper March 2019 (North), Karnataka 1st PUC Statistics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Statistics Previous Year Question Paper March 2019 (North) Time: 3 Hrs 15 Min …<\/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\/17464"}],"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=17464"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17464\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17464"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17464"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17464"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}