{"id":13971,"date":"2019-12-21T16:07:27","date_gmt":"2019-12-21T10:37:27","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=13971"},"modified":"2021-07-02T15:35:40","modified_gmt":"2021-07-02T10:05:40","slug":"2nd-puc-statistics-model-question-paper-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-statistics-model-question-paper-2\/","title":{"rendered":"2nd PUC Statistics Model Question Paper 2 with Answers"},"content":{"rendered":"

Students can Download 2nd PUC Statistics Model Question Paper 2 with Answers, Karnataka 2nd 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 2nd PUC Statistics Model Question Paper 2 with Answers<\/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 may be used.<\/li>\n
  3. All working steps should be clearly shown.<\/li>\n<\/ol>\n

    Section – A<\/span><\/p>\n

    I. Answer any TEN the following questions. 10 x 1 = 10<\/span><\/p>\n

    Question 1.
    \nDefine Vital Statistics.
    \nAnswer:
    \nVital statistics is the science applied to the analysis and interpretation of numerical facts regarding vital events occurring in a human population.<\/p>\n

    Question 2.
    \nWhat is the value Index number?
    \nAnswer:
    \nValue index number indicates the effect of combination of price and quantity changes in between two time periods.<\/p>\n

    Question 3.
    \nWrite the formula of Dorbish Bowley’s price Index number.
    \nAnswer:
    \nDorbish-Bowley Price Index number is –
    \n\"2nd<\/p>\n

    Question 4.
    \nWhat do you mean by components of time series?
    \nAnswer:
    \nThe factors which are responsible for the fluctuations\/ variations in the time series are called components of time series.<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nIn which probability distribution mean and variance are equal?
    \nAnswer:
    \nIn poisson distribution mean and variance are equal i.e., mean (\u03bb) = variance (\u03bb).<\/p>\n

    Question 6.
    \nDefine S.N.V.
    \nAnswer:
    \nA normal variate with mean \u03bc = 0 and SD( \u03c3) = 1 is called S.N.V. It is denoted by Z.
    \ni.e., Z = \\(\\frac{x-\\mu}{\\sigma}\\) ~N(0, 1)<\/p>\n

    Question 7.
    \nWhat is the type I error?
    \nAnswer:
    \nType I error is taking a wrong decision to reject the null hypothesis (H0) when it is actua ly true.<\/p>\n

    Question 8.
    \nDefine Standard error.
    \nAnswer:
    \nThe standard deviation of the sampling distribution of sample statistic is called standard error.<\/p>\n

    Question 9.
    \nDefine critical region.
    \nAnswer:
    \nThe set of those values of the test statistic which lead to the rejection of the null Hypothesis is called critical region, also called as rejection region.<\/p>\n

    Question 10.
    \nIn a transportation when do you say that a solution is degenerate?
    \nAnswer:
    \nThe number of positive allocation in a basic feasible solution (B.F.S) is less than (m + n – 1), then the solution is called to be degenerate.<\/p>\n

    Question 11.
    \nWhat is mixed strategy in a 2 person – zero – sum game?
    \nAnswer:
    \nWhile playing a game, mixed strategy of a player is his pre-decision to chose his course of action according to certain pre-assigned probability.<\/p>\n

    \"KSEEB<\/p>\n

    Question 12.
    \nWhat is lead time?
    \nAnswer:
    \nLead time is the time gap between placing order and arrival of goods at the inventory.<\/p>\n

    Section – B<\/span><\/p>\n

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

    Question 13.
    \nThe woman population of child bearing age group in a certain locality is 100000. The number of live births in the year in the locality is 2300. Find G.F.R.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 14.
    \nDefine consumer price index number.
    \nAnswer:
    \nConsumer price index number is the index number of the cost met by a specified class of consumers in buying a basket of goods and services.<\/p>\n

    Question 15.
    \nMention any two uses of analysis of Time Series.
    \nAnswer:
    \nAnalysis of time series is useful to<\/p>\n

      \n
    1. Understand the past behaviour of the data.<\/li>\n
    2. Predicting the future trend.<\/li>\n<\/ol>\n

      Question 16.
      \nWrite the normal equation for fitting the linear trend.
      \nAnswer:
      \nna + b\u03a3x = \u03a3y
      \na\u03a3x + b\u03a3x2<\/sup> = \u03a3xy<\/p>\n

      Question 17.
      \nWrite any two assumptions of interpolation and extrapolations.
      \nAnswer:<\/p>\n

        \n
      1. There are no sudden jumps in the values of dependent variable from one period to another.<\/li>\n
      2. There will be no conscutive missing values in the series.<\/li>\n<\/ol>\n

        Question 18.
        \nState under what conditions Binomial distribution tends to Poisson distribution.
        \nAnswer:
        \nBinomial distribution tends to poisson distribution under the following conditions:<\/p>\n

          \n
        1. When n is large i.e., n \u2192 \u221e<\/li>\n
        2. P is very small i.e., p \u2192 0 and<\/li>\n
        3. np = \u03bb is fixed.<\/li>\n<\/ol>\n

          Question 19.
          \nWhat are the value of \u03b21<\/sub>, and \u03b22<\/sub> in Normal distribution?
          \nAnswer:
          \nIn Normal distribution, \u03b21<\/sub> = 0 and \u03b22<\/sub> = 3.<\/p>\n

          \"KSEEB<\/p>\n

          Question 20.
          \nDefine level of significance and critical value.
          \nAnswer;<\/p>\n

            \n
          1. Probability of occurence of Type I error is called level of significance (\u03b1).<\/li>\n
          2. The value of test statistics which separates the critical region (i.e., rejection region) and acceptance region is called critical value (k).<\/li>\n<\/ol>\n

            Question 21.
            \nWhat are the conditions under which chi-square test of goodness of fit is applied?
            \nAnswer:<\/p>\n

              \n
            1. The total frequency N should be large.<\/li>\n
            2. The expected frequencies (E) should be 5 or more. If any E is less than 5, it should be pooled with adjacent frequency.<\/li>\n
            3. If any parameter is estimated, one degrees of freedom should be lessened.<\/li>\n<\/ol>\n

              Question 22.
              \nWhen no solution and unbounded solutions existed in L.P.P.?
              \nAnswer:
              \nNo solution exists in L.P.P. when the feasible solution does not exists.
              \nUnbounded solution exists in L.P.P. when the feasible region is not unbounded or the optimum value of ‘z’ may be infinity.<\/p>\n

              Question 23.
              \nWhat do you mean by replacement problems?
              \nAnswer:
              \nReplacement problem is the problem of deciding the age at which the old equipments are replaced.<\/p>\n

              Question 24.
              \nDefine Setup cost and Holding cost.
              \nAnswer:
              \nIt is the cost of setting up of machines for production or the cost of placing the order for the goods.
              \nIt is the cost of carrying or holding the goods in stock until the goods are sold o used.<\/p>\n

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

              III. Answer any EIGHT questions. ( 8 x 5 = 40 )<\/span><\/p>\n

              Question 25.
              \nWrite down the any two difference between CBR and GFR with respective formula.
              \nAnswer:<\/p>\n\n\n\n\n\n
              CBR<\/strong><\/td>\nGFR<\/strong><\/td>\n<\/tr>\n
              a. It takes into consideration of the whole population<\/td>\na. It takes into consideration of the women population of childbearing age.<\/td>\n<\/tr>\n
              b. It does not give importance to age and sex composition of the population.<\/td>\nb. It give importance to sex but not age composition of the population.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              \"2nd
              \n\"2nd<\/p>\n

              Question 26.
              \nExplain brielfy the steps involved in the construction of price Index number.
              \nAnswer:
              \nThe following are the steps\/stages\/principles\/Heads involved:-<\/p>\n

                \n
              1. Purpose and the scope<\/li>\n
              2. Selection of base period<\/li>\n
              3. Selection of commodities or itsm<\/li>\n
              4. Selection of price list<\/li>\n
              5. Selection an average Selection of weights<\/li>\n
              6. Selection of formula<\/li>\n<\/ol>\n

                \"KSEEB<\/p>\n

                Question 27.
                \nConstruct cost of living Index number using a suitable formula and interpret it.
                \nAnswer:
                \n\"2nd
                \nHere price Index = p and expenditure = w
                \n\u2234 Use family budget method.
                \n(CPI) I01<\/sub> = \\(\\frac{\\sum p w}{\\sum w}\\)
                \n\"2nd<\/p>\n

                Question 28.
                \nA Bernoulli variate with parameter p = 0.4, then find the probability for x = 1 and write down mean and variance.
                \nAnswer:
                \nGiven : p = 0.4, \u2234 1 – p = 1 – 0.4 = 0.6
                \nThen p(x = 1) = (0.4)1<\/sup> (1 – 0.4)1-1<\/sup> = 0.4 x 0.60<\/sup> = 0.4
                \nMean p = 0.4, variance = p( 1 – p) = 0.4 x 0.6 = 0.24.<\/p>\n

                Question 29.
                \nEstimate the sale of a business concern for the year 2012.
                \n\"2nd
                \nAnswer:
                \nLet X and Y be the year and sales.
                \n\"2nd
                \nThere are four known values of y. So, using binominal expansion;
                \n(y-l)4<\/sup> = 0 : y4<\/sub><-4y3<\/sub> + 6y2<\/sub>-4y1<\/sub> + yo<\/sub> = 0 ……….. (1)
                \n130-4 (76)+ 6(50)-4 y1<\/sub> + 26 = 0
                \n130-304 +300-4y1<\/sub> + 26 = 0
                \n152 – 4y1<\/sub> = 0 \u2234 -y1<\/sub>=\\(\\frac{-152}{4}\\)\u00a0 y1<\/sub> = 38
                \nRewrite equation (1) by increasing suffixes of by \u2018 1\u2019 we get y5<\/sub>
                \ny5<\/sub> – 4y4<\/sub> + 6y3<\/sub> – 4y2<\/sub> + y1<\/sub>=0
                \ny5<\/sub> – 4(130) + 6(76) – 4(50) + 38 = 0; Since y1<\/sub> = 38
                \ny5<\/sub> – 520 + 456 – 200 + 38 = 0
                \n\u2234 y5<\/sub> = 226.<\/p>\n

                Question 30.
                \nIn a college 40% are girls. In a random sample of 5 students, find the probability of getting,(i) 2 girls (ii) at least 3 girls.
                \nAnswer:
                \nLet x be the number of girls is a Binomial variate with the parameters n = 5,
                \np = 40% = 0.4. \u2234 q = 1 – p = 0.6.
                \nThe p.m.f. is –
                \nP(x) = ncx <\/sub>px <\/sup>qn – x<\/sup>; x = 0,1,2, …… n
                \n= 5cx<\/sub> (0.4)x<\/sup> (0.6)5-x<\/sup>; x = 0,1,2, …….. 5
                \n(i) p(2 are girls) = p(x = 2) = 5C2<\/sub>(0.4)2<\/sup>(0.6)5-2<\/sup>
                \n=10 x 0.16 x 0.216
                \n= 0.3456<\/p>\n

                \"KSEEB<\/p>\n

                (ii) p(atleast 3 girls) = p(x \u22653)
                \n=p(x = 3) + p(x = 4) + p(x = 5)
                \n= 5C3<\/sub> (0.4)3<\/sup>(0.6)5-3<\/sup> + 5C4<\/sub>(0.4)4<\/sup>(0.6)5-4<\/sup>+ 5C5<\/sub>(0.4)5<\/sup>(0.6)5-5 <\/sup> = 10 x 0.064 x 0.36 + 5 x 0.0256 x 0.6 + 1 x 0.01024 x 1
                \n= 0.2304 + 0.0768 + 0.01024
                \np(x \u2265 3) =0.3174.<\/p>\n

                Question 31.
                \nA Factory has produced 10,000 electric bulbs. Assuming that the expected life of these bulbs is Normally distributed with mean 120 days and S.D. 20 days. Find the number of bulbs which will burn (i) before 90 days (it) between 80 and 140 days.
                \nAnswer:
                \nLet ‘x’ be the life of bulbs is a Normal variate with mean \u00b5 = 120 and S.D. \u03c3 = 20 days, then,
                \n\"2nd<\/p>\n

                (i) p(bulb burn before 90 days)
                \n\"2nd
                \n= p(z < -1.5)
                \n= Area from (-1.5) to (-\u221e)
                \n= Area from 0 to (-\u221e) – Area from 0 to (-1.5)
                \np(x < 90) = 0.5 – 0.4342 = 0.0658
                \n\u2234 Number of bulbs bum before 90 days
                \n= p(x < 90) x N
                \n= 0.0658 x 10000 = 658
                \n\"2nd<\/p>\n

                (ii) p(bulb bum in between 80 and 140 days)
                \n= p(80<x< 140)
                \n\"2nd
                \n= p(-2 < z < 1) .
                \n= Area from (-2) to 1
                \n= Area from 0 to (-2) + Area from 0 to 1
                \np(80 < x < 140) = 0.4772 + 0.3412 = 0.8185
                \n\u2234 Number of bulbs bum in between 80 and 140 days
                \n= p(80 < x < 140) x N
                \n= 0.8185 x 10,000
                \n= 8185.
                \n\"2nd<\/p>\n

                Question 32.
                \nA random sample of 100 jack fruits weigh on an average 10 lbs. With variance of 41 lbs2<\/sup>. Test at 1% level of significance that the expected average weight of jackfruits of that varieties is 11 lbs.
                \nAnswer:
                \nGiven : n = 100, X\u0304 = 10 lbs. S2<\/sup> = 41 lbs2<\/sup>, \u2234 S = 6.4 lbs, \u03bc = 11 lbs.
                \nH0 <\/sub>: Average weight of Jack fruits is 11 lbs.
                \n(i.e., H0<\/sub> : \u03bc = 11lbs)
                \nH1<\/sub>: Average weight of Jack fruits differ from 11 lbs.
                \n(i.e., H1<\/sub> : \u03bc \u2260 11lbs) { Two Tail test }
                \nThe test statistics is –
                \n\"2nd
                \nZCal<\/sub> = -1.56
                \nAt \u03b1 – 1% the two Tail critical values are -k1<\/sub> = -2.58 and k2<\/sub> = 2.58. .
                \nHere zcal<\/sub> lies between k1<\/sub> and k2<\/sub>.
                \n\u2234 H0<\/sub> is accepted.
                \nConclusion: The average weight of Jack fruit is 11 lbs.
                \n\"2nd<\/p>\n

                Question 33.
                \nIn a college there were 50 students in II pu commerce. Among them 15 students . passed in preparatory examination. Whereas in annual examination 42 students passed. However among those who passed in the preparatory examination only 12 students passed in the annual examination. Test at 1% level of significance whether passing in the preparatory examination and passing in the annual examination are independent.
                \nAns.
                \nH0<\/sub> : Pass in Preparatory and pass in annual examination are independent
                \nH1<\/sub> : Pass in preparatory and passes in annual examination, are not independent.
                \nThe 2 x 2 contingency table is:
                \n\"2nd
                \nHere one of the class frequency at b = 3, is less than 5, so, apply Yate’s correction.
                \nThe test statistic is –
                \n\"2nd
                \n\u03c72<\/sup> = 00071
                \nAt a = 1% the upper Tail critical value for 1 d.f. k = 6.63 Here
                \n\u03c72<\/sup>cal<\/sub> < K .
                \nH0<\/sub> is accepted
                \nConclusion Pass in preparatory and pass in annual examination are independent.<\/p>\n

                \"KSEEB<\/p>\n

                Question 34.
                \nSolve the following game using principle of dominance.
                \n\"2nd
                \nAnswer:
                \nBy principle of dominance, in the pay off matrix, each pay off in the first row exceeds the corresponding pay-off in the third row.
                \nA1<\/sub>\u00a0dominates A3<\/sub>,
                \n\"2nd
                \nHere B3<\/sub> dominates B1<\/sub>, B2 and B4<\/sub> and so delete B1<\/sub>, B2<\/sub> and B4<\/sub>.
                \n\"2nd
                \nA1<\/sub> dominates A2<\/sub>, delete A2<\/sub>.
                \n\"2nd
                \nThus saddle point occurs at (1, 3)
                \nThe solutions –<\/p>\n

                  \n
                1. Strategy for A is A1<\/sub><\/li>\n
                2. Strategy for B is B3<\/sub><\/li>\n
                3. Value of the game v = 10<\/li>\n<\/ol>\n

                  Question 35.
                  \nAn engineering company has a machine whose purchase price is Rs. 7000. The expected maintenance cost \u00a3nd resale values in different years are given below:
                  \n\"2nd
                  \nWhen should the machine be replaced?
                  \nAnswer:
                  \nHere p = 7000, let Ci<\/sub> and Sn<\/sub> be the maintenance cost and resale value, then,
                  \nTotal annual cost: T = (p – Sn<\/sub>) + \u03a3Ci<\/sub> and
                  \nAnnual average cost: \\(A(n)=\\frac{T}{n}\\) ; n = years
                  \n\"2nd
                  \nHere, annual average cost A(n) is minimum for n = 5 years. Therefore the optimal replacement age is n = 5 years. The is machine should be replaced for every 5 years.
                  \nMinimum average cost: A(5) = Rs. 3020.<\/p>\n

                  Question 36.
                  \nGlass panes are manufactured by a firm. Broken corner, excessive thickness, excessive colour tint are considered defect. The number of defects in randomly selected panes are noted as below :
                  \n\"2nd
                  \nDraw chart for defects and analyse the data.
                  \nAnswer:
                  \nHere, number of defects are given, so use c-chart, standards are not given.
                  \nLet C and K be the number of defects and sample number.
                  \n\\(\\bar{c}=\\frac{\\Sigma c}{K}\\)
                  \n= \\(\\frac{23}{8}=2.875\\)
                  \nThe control limits for c-chart are :-
                  \nC.L. = C\u0304 =2.875
                  \nL.C.L. = C\u0304 – 3\u221aC\u0304 = 2.875 – 3 >\/2.875 . ;
                  \n= -2.211 \u2245 0 (Taken as 0)
                  \nV.C.L. = C\u0304 + \u221aC\u0304
                  \n= 2.875 + 3\u221a2.875
                  \n= 7.961
                  \n\"2nd
                  \nInterpretation:
                  \nAll the points are within the control limits. And so, the process is in control.<\/p>\n

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

                  IV. Answer the following Questions<\/span><\/p>\n

                  Question 37.
                  \nCalculate Fisher’s and Dorbish-Bowley’s Index numbers and test whether they satisfies T.R.T and F.R.T.
                  \n\"2nd
                  \nAnswer:
                  \n\"2nd
                  \n\"2nd
                  \n\u2234 Fisher\u2019s Index number satisfies T.R.T.
                  \n\"2nd
                  \nDorbish – Bowely Index number does not satify T.R.T.
                  \n\"2nd
                  \nDorbish – Bowely’s :
                  \n\"2nd
                  \n\u2234 Dorbish – Bowely Index number does not satisfy T.R.T.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 38.
                  \n(a) In a college, out of 60 students randomly selected 40 passed in II PU examination in first class. At 5% level of significance can we conclude that on an average 50% of the students in the college pass in first class.
                  \nAnswer:
                  \nGiven: n = 60, x = 40, \u2234 p = \\(\\frac{x}{n}=\\frac{40}{60}\\) = 0.67
                  \np0<\/sub> \u2234 q = 1 – p0<\/sub>
                  \nH0<\/sub> : The proportion of first class pass students \u00a1s 0.5
                  \n(i.e., H0<\/sub> : p0<\/sub> = 0.5)
                  \nH1 : The proportion of first class pass students differs from 0.5; {Two Tail test)
                  \n(i.e., H1<\/sub> : p0<\/sub> \u2260 0.5) .
                  \nThe test statistic is –
                  \n\"2nd
                  \nz cal<\/sub> = 2.63
                  \nAt \u03b1 = 5% the two Tail crtitical values are -k1<\/sub> = 1.96 and k2<\/sub> = 1.96.
                  \nHere zcal<\/sub> is outside the interval -k1<\/sub> and k2<\/sub> \/zcal<\/sub> is in rejection region.
                  \n\u2234 H0<\/sub> is rejected and H1<\/sub>is accepted.
                  \nConclusion : The proportion of first class pass students differs from 0.5<\/p>\n

                  (b) Calculate standardised death rates. Which population seems to be healthier ?
                  \n\"2nd
                  \nAnswer:
                  \nLet A and B be ASDR’s of population A and B and P be the standard population.
                  \n\"2nd
                  \nSDR(A)<\/sub> = \\(\\frac{1110}{100}\\) =11.10
                  \nSDR(A)<\/sub> < SDR(B)<\/sub>
                  \nPopulation A is healthier than population B.<\/p>\n

                  Question 39.
                  \n(a) From the following data, test whether mean life of brand B bulbs are longer than that of brand A bulbs.
                  \n\"2nd
                  \nAnswer:
                  \nHere n1<\/sub> and n2<\/sub> are less than 30, So, use small sample test.
                  \nGiven: n1<\/sub> = 12, n2<\/sub> = 15
                  \nX\u03041<\/sub> = 1240 X\u03042<\/sub> = 1370
                  \nS1<\/sub>2 <\/sup>= 1000 ,S2<\/sub>2 <\/sup>= 2000
                  \nH0<\/sub> : Mean life of Brand A and Brand B bulbs are same.
                  \n(j.e., H0<\/sub> . \u00b51<\/sub> = \u00b52<\/sub>)
                  \nMean life of Brand B bulbs are-longer than Brand A bulbs.
                  \n(i.e., H1<\/sub>. \u00b51<\/sub> < \u00b52<\/sub>) {Lower\/Left Tail test}
                  \nThe test statistic is –
                  \n\"2nd
                  \nAt \u03b1 = 5% the left Tail crtitical value for (n1<\/sub> + n2<\/sub> – 2) = (12 + 15 – 2) = 25 d.f. (from t-table values is -k = -1.71
                  \n\"2nd
                  \nHere tcal<\/sub> = -8.189 is in rejection region\/ .tcal<\/sub>< -k.
                  \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
                  \nConclusion: Mean life of Brand B bulbs are longer than Brand A bulbs,<\/p>\n

                  (b) Calculate 5 yearly moving averages for the following data :
                  \n\"2nd
                  \nAnswer:
                  \nCompetitive of 5 yearly moving average m = 5
                  \n\"2nd<\/p>\n

                  Question 40.
                  \nThe following table shows the number of defective blades in, each packet of 5 each.
                  \n\"2nd
                  \nTest whether Binomial distribution is a good fit to this observed data.
                  \nAnswer:
                  \nThe null hypothesis is
                  \nH0 <\/sub>: Binomial distribution is a good fit to the observed data.
                  \nH1<\/sub> : Binomial distribution is not a good fit to the observed data.
                  \nTo test H0<\/sub>, we fit Binomial distribution to the data.
                  \nThe parameter \u2018p\u2019 is estimated by finding mean.
                  \n\"2nd
                  \nHere let x and f be the no. of defectives and number of packets.
                  \n5p = \\frac{150}{50} = 3
                  \n\u2234 p = \\frac{3}{5} = 0.6 \u2234q 1 – p = 0.4
                  \nThe p.m.f is
                  \np(x) = ncx <\/sub>px <\/sup>qn – x<\/sup>; x = 0,1,2, …… n
                  \n= 5cx <\/sub>0.6x <\/sup>0.45 – x<\/sup>; x = 0,1,2, …… n
                  \nTheoretical frequency : Tx<\/sub> = p(x) x N
                  \nT0<\/sub> = (x=0) x 50 = 5C0 <\/sub>(0.6)0 <\/sup>(0.4)5-0<\/sup> x 50
                  \nT0<\/sub>\u00a0= 0.512
                  \nUsing recurrence relation of theoretical frequency
                  \n\"2nd
                  \nThus the fitted Binomial distribution is (Approx)
                  \n\"2nd
                  \nThen the test statistic is
                  \n\"2nd
                  \nHere one parameter \u2018p\u2019 is estimated from the data.
                  \n\"2nd
                  \nHere first two and last one theoretical frequencies are less than 5. Therefore they are pooled with the adjacent frequencies such that they are 5 or more.
                  \n\u03c72<\/sup>cal<\/sub> = 3.2095
                  \nAt \u03b1 = 5% the upper tail critical value for (n – 2) = (3 – 2) = 1 d.f. is k = 3.84
                  \nHere \u03c72<\/sup>cal<\/sub> < k.
                  \n\u2234 H0<\/sub> is accepted.
                  \n\"2nd
                  \nConclusion: Binomial distribution is good fit to the observed data.<\/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.
                  \nIf 98% of electric bulbs manufactured by a company are known to be non-defective. What is the probability that a sample of 150 electric bulbs taken from the meduction process from the company, what would contain (i) Exactly on detective electric bulbs (ii) More than twe detective electric bulbs?
                  \nAnswer:
                  \nLet x denote number of detective electric bulbs is a binomial variate with the parameters n = 50 q = 98% are non-defective P = 1 – 9 = 2% = 0.02 are detectives.
                  \nHere n is very large and P is very small under these condition it tends to poisson distribution with the parameter \u03bb = np = 150 x 002 ,
                  \n\u03bb = 3
                  \nThen the p.m.f of poission distribution is
                  \n\"2nd
                  \n(i) P (exactly one defective bulb)
                  \nP(x = 1) = \\(\\frac{e^{-3} \\cdot 3^{1}}{11}\\) = e-3<\/sup>.3= 0.0498 x 3 = 0.1494
                  \n(ii) P (More than two detective bulbs)
                  \n= P (x > 2)
                  \n= 1 – P(x \u2264 2)
                  \n= 1 – [P(x = 0)+P(x = 1 )+P(x = 2)]
                  \n\"2nd
                  \n= 1 – e-3<\/sup> [1 + 3 + 4.5] = 1 – 0.0498 x 8.5
                  \nP(x > 2) = 1.04233 = 0.5767.<\/p>\n

                  Question 42.
                  \nA normal variate has variance 8 – twenty sample observations of the variate have variance 9. Test at 1% level of significance whether the population variance is 8.
                  \nAnswer:
                  \nGiven ; \u03c32<\/sup> = 8, n = 20, s2<\/sup> = 9. \u03b1 = 1%
                  \nH0<\/sub>: Population variance is 9. (H0<\/sub>: \u03c32<\/sup> = 8)
                  \nH1<\/sub> : Population variance differs from 9. (H, : \u03c32<\/sup>\u2260 8); (two tailed test)
                  \nUnder H0<\/sub>, the \u03c72<\/sup> – test statistic is : –
                  \n\"2nd
                  \nHere \u03c72<\/sup>cal<\/sub> lies in acceptance region
                  \n.’. H0<\/sub> is accepted
                  \n\"2nd
                  \nConclusion: \u03c32<\/sup> = 8<\/p>\n

                  Question 43.
                  \nThe marks obtained by 5 students before and after attending coaching classes were given below. Test whether coaching was useful (use \u03b1 = 1%).
                  \n\"2nd
                  \nAnswer:
                  \nGiven : n1<\/sub>= n2<\/sub> = n = 5.
                  \nLet x and y be the marks before and after attending coaching class and let d = x – y.
                  \nH0<\/sub>: Mean marks before after coaching are same.
                  \n(i.e., H0<\/sub>. \u00b51<\/sub> = \u00b52<\/sub>)
                  \nH, : Mean marks after attending coaching are more (i e., 11, :p,<ft2)
                  \n(i.e., H0<\/sub>. \u00b51<\/sub> < \u00b52<\/sub>)
                  \nThe test statistic is –
                  \n\"2nd
                  \n\"2nd
                  \nAt \u03b1 = 1% the lower Tail critical value for (n – 1) = (5 – 1) = 4. d.f; -k = \u20143.75
                  \nHere tcal<\/sub> < k
                  \nH0<\/sub> is rejected and H1<\/sub> is accepted.
                  \nConclusion: Coaching is not useful.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 44.
                  \nFind the initial feasible solution to the transportation problem given below by North west corner rule.
                  \n\"2nd
                  \nAnswer:
                  \nSince total supply (70 + 30 + 50) = total demand (65 + 42 + 43). The problem is said to be balanced T.P.
                  \nBy N.W.C.R
                  \n\"2nd
                  \nFirst allocation is made at North West corner cell at (1,1) as : X11<\/sub> = min (70, 65) = 65
                  \nHere P is satisfied and so, allocate at the cell (1, 2) as : X12<\/sub> = min (70, 65, 42) = (5, 42) = 5
                  \nHere A is satisfied, next allocation is made at the cell (2, 2) as :
                  \nX22<\/sub>= min (30,42 – 5)
                  \n= min (30, 37) = 30
                  \nB is satisfied allocate at (3, 2) as
                  \nX32<\/sub>
                  \n= min (50, 42 – 35)
                  \n= min (50, 7) = 7
                  \nHere Q is satisfied Next allolate at (3, 3)
                  \nas : X33<\/sub> = min (50 – 7, 43) = min (43, 43) = 43
                  \nSo, all availabilities (ai<\/sub>) and requirements (bj<\/sub>) are satisfied
                  \nThe total transportation cost is
                  \n\"2nd
                  \nThe total transportation cost = 830 is the intitial besic feasible solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

                  Students can Download 2nd PUC Statistics Model Question Paper 2 with Answers, Karnataka 2nd 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 2nd PUC Statistics Model Question Paper 2 with Answers Time: 3 Hrs 15 Min Max. Marks: …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13971"}],"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=13971"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13971\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=13971"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=13971"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=13971"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}