{"id":14186,"date":"2019-12-21T17:52:00","date_gmt":"2019-12-21T12:22:00","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=14186"},"modified":"2021-07-02T15:35:38","modified_gmt":"2021-07-02T10:05:38","slug":"2nd-puc-statistics-previous-year-question-paper-june-2015","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-statistics-previous-year-question-paper-june-2015\/","title":{"rendered":"2nd PUC Statistics Previous Year Question Paper June 2015"},"content":{"rendered":"

Students can Download 2nd PUC Statistics Previous Year Question Paper June 2015, 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 Previous Year Question Paper June 2015<\/h2>\n

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

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

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

Question 1.
\nDefine cohort.
\nAnswer:
\nCohort is a group of individuals who are born at the same time and who experience the same mortality conditions.<\/p>\n

Question 2.
\nWhat is the value of index number for the base year?
\nAnswer:
\n100<\/p>\n

Question 3.
\nWhich index number shows upward bias?
\nAnswer:
\nLaspeyre\u2019s index number.<\/p>\n

Question 4.
\nGive an example for random variation.
\nAnswer:
\nShut down of production of a factory due to fire accident.<\/p>\n

Question 5.
\nWrite down the relationship between mean and variance of a Bernoulli distribution.
\nAnswer:
\nMean > Variance i.e. p > pq.<\/p>\n

Question 6.
\nFor a chi-square variate with 10 d.f., find the variance.
\nAnswer:
\nIf n = 10, then variance = 2n = 2 \u00d7 10 = 20<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nWhat is parameter space?
\nAnswer:
\nThe set of all admissible values of the parameter is called parameter space.<\/p>\n

Question 8.
\nGiven \u03c32<\/sup> = 16 cm2 and n = 49, calculate standard error of sample mean.
\nAnswer:
\n\u03c32<\/sup> = 16;
\n\u2234 \u03c3 = \\(\\sqrt{16}\\) = 4, n = 49
\nS.E (x\u0304) = \\(\\frac{\\sigma}{\\sqrt{n}}=\\frac{4}{\\sqrt{49}}=\\frac{4}{7}\\) = 0.5714<\/p>\n

Question 9.
\nWhat is type I error?
\nAnswer:
\nType I error is taking a wrong decision to reject the null hypothesis, when it is actually true.<\/p>\n

Question 10.
\nStatistical quality control helps in detecting which type of variation?
\nAnswer:
\nAssignable cause of variation.<\/p>\n

Question 11.
\nWhat do you mean by a non degenerate solution in a transportation problem?
\nAnswer:
\nIf the number of non-zero allocation are equal to (m + n – 1) then, the solution is called non-degenerate.<\/p>\n

Question 12.
\nWrite down one advantage of inventory?
\nAnswer:
\nInventory facilitates bulk purchase of raw materials at discount rates.<\/p>\n

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

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

Question 13.
\nIn a locality out of 20,000 births 100 mothers died due to child birth complications. Calculate maternal mortality rate.
\nAnswer:
\nMaternal mortality rate:
\n\"2nd<\/p>\n

Question 14.
\nIf \u03a3 p0<\/sub>q = 1200 and \u03a3p1<\/sub>q = 1500, compute suitable index number.
\nAnswer:
\n\"2nd<\/p>\n

Question 15.
\nMention the steps involved in the construction of consumer price index number.
\nAnswer:<\/p>\n

    \n
  1. Object and the scope<\/li>\n
  2. Conducting family budget survey.<\/li>\n
  3. Obtaining the price quatation.<\/li>\n
  4. Methods of constraction:<\/li>\n<\/ol>\n

    \"2nd<\/p>\n

    Question 16.
    \nWrite down the normal equations for fitting linear trend.
    \nAnswer:
    \nLinear equation is: y = a + bx
    \nNormal equations are : na + b\u03a3x = \u03a3 y
    \na\u03a3x + b\u03a3x2<\/sup> = \u03a3 xy.<\/p>\n

    Question 17.
    \nWrite down the conditions for the application of binomial expansion method of interpolation and extrapolation.
    \nAnswer:
    \nThe conditions are:<\/p>\n

      \n
    • The values of the independent variable (x) should have a common difference, i.e., in A.P.<\/li>\n
    • The value of ‘X’, for which the value of’Y’ is tobe interpolated must be one of the value of x.<\/li>\n<\/ul>\n

      Question 18.
      \nWrite down the Bernoull distribution with parameter p = 0.25.
      \nAnswer:
      \nThe Bernoulli distribution is:-with p = 0.25
      \nP(x) = (0.25)x<\/sup> (1-0.25)1-x<\/sup> ; x = 0, 1
      \n= (0.25)x<\/sup> (0.75)1-x<\/sup> x = 0,1
      \nOR
      \n\"2nd<\/p>\n

      \"KSEEB<\/p>\n

      Question 19.
      \nMention two features of a student\u2019s t-distribution.
      \nAnswer:<\/p>\n

        \n
      1. The parameter is \u2018n\u2019 called d.f.<\/li>\n
      2. Mean = E(x) = 0 and
        \nVar(x) = \\(\\frac{n}{n-2}\\) for n > 2<\/li>\n<\/ol>\n

        Question 20.
        \nWhat are point estimation and interval estimation?
        \nAnswer:
        \nEstimation of unknown parameter by proposing a specific value as an estimate is called point estimation.
        \nWhile estimating the unknown parameter by proposing an interval, which is likely to contain the parameter is called interval estimation.<\/p>\n

        Question 21.
        \nWrite down two utilities of standard error.
        \nAnswer:<\/p>\n

          \n
        1. It is used in interval estimation, to write down the confidence intervals.<\/li>\n
        2. It is used in testing of hypothesis totest whether the difference between the sample statistic and the population parameter is significant or not.<\/li>\n<\/ol>\n

          Question 22.
          \nWhat do you mean by process control and product control in statistical quality control?
          \nAnswer:
          \nControlling the quality of the product during the manufacturing process itself is the Process control. Controlling the quality of the finished products\/ manufactured products is called product control.<\/p>\n

          Question 23.
          \nMention two method of obtaining initial basic feasible solution for a transportation problem.
          \nAnswer:<\/p>\n

            \n
          1. Matrix-Minim method and<\/li>\n
          2. North-west corner rule.<\/li>\n<\/ol>\n

            Question 24.
            \nGiven R = 1000 units\/month, C3<\/sub> = Rs. 350 and C1<\/sub> = Rs. 0.20\/unit\/months, find Q\u00b0.
            \nAnswer:
            \n\"2nd<\/p>\n

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

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

            Question 25.
            \nFrom the following data calculate GFR, ASFR(20 -24) and ASFR(30 – 39)
            \n\"2nd
            \nAnswer:
            \n\"2nd<\/p>\n

            Question 26.
            \nDefine index number. Mention three limitations of index numbers.
            \nAnswer:
            \nIndex number is a statistical measure designed to show an average change in a variable or group of related variables over a period.
            \nThe limitations are:<\/p>\n

              \n
            • Many formulae are used and they give different answers for the index.<\/li>\n
            • As the customs and habits change from time to time, the use of commodities also vary.<\/li>\n
            • There is an ample scope for bias in the construction of index number.<\/li>\n<\/ul>\n

              \"KSEEB<\/p>\n

              Question 27.
              \nCompute cost of living index number.
              \n\"2nd
              \nAnswer:
              \nWeights are given, use family budget method: I01<\/sub> = \\(\\frac{\\sum p w}{\\sum w}\\) P = \\(\\frac{p_{1}}{p_{0}} \\times 100\\)\u00a0 ; w-weights
              \n\"2nd
              \nConsumer \/ cost of living index number:
              \n\"2nd
              \nThere is net increase of cost of living by 39.96% in the current year as compared to base year.<\/p>\n

              Question 28.
              \nCompute the trend values by finding five yearly moving averages for the following data.
              \n\"2nd
              \nAnswer:
              \n\"2nd
              \n\u0177 – trend values.
              \nOverall trend in the values of times series shows upward trend.<\/p>\n

              Question 29.
              \nUsing Newton\u2019s forward difference method to find the value of \u2018y\u2019 when x – 25.
              \n\"2nd
              \nAnswer:
              \nThe number of known values ofy, n = 5, so prepare leading difference table up to A4.
              \n\"2nd
              \n\"2nd
              \n\"2nd<\/p>\n

              Question 30.
              \nThe probability that a bomb hits the target is \\(\\frac { 1 }{ 4 }\\) Five bombs are aimed at the target. Find the probability that:
              \n(i) 3 bombs hit the target
              \n(ii) at the most two bombs hit the target.
              \nAnswer:
              \nLet x denote the number of bombs hit the target is a Binomial variate with the parameters
              \nn = 5, P = \\(\\frac { 1 }{ 4 }\\) = 0.25 and q= 1 – p= 1 – 0.25 = 0:75.
              \nThe p.m.f is:-
              \np(x) = n<\/sup>cx<\/sub> px<\/sup> qn-x<\/sup>: x = 0,1,2 n
              \np(x)= 5<\/sup>cx<\/sub> (0.25)x<\/sup> (0.75)5-x<\/sup> x = 0,1,2……..5
              \n(i) p(3 bombs hit the target) =p(x = 3) = 5<\/sup>c3<\/sub> (0.25)3<\/sup> (0.75)5-3<\/sup> = 10 \u00d7 0.015625 \u00d7 0.5625
              \n= 0.08789.<\/p>\n

              (ii) p(at most two bombs hit the target) = p(x \u2264 2)
              \n= p(x = 0) + p[x = l) + p(x = 2)
              \n= 5<\/sup>c0<\/sub> (0.25)\u00b0 x (0.75)5-0<\/sup> + 5<\/sup>c1<\/sub> (0.25)1<\/sup> (0.75)5-1<\/sup> + 5<\/sup>c2<\/sub> (0.25)2<\/sub> (0.75)5-2<\/sup>
              \n= 1 \u00d7 1 \u00d7 0.2373 + 5 \u00d7 0.25 \u00d7 0.3164 + 10 \u00d7 0.0625 \u00d7 0.4218
              \n= 0.2373 + 0.3955 + 0.2637 = 0.8965.<\/p>\n

              Question 31.
              \nIn a Hyper-geometric distribution if a = 4, b = 6 and n = 4, find mean and variance.
              \nAnswer:
              \nFor Hyper-geometric distribution: a = 4, b = 6, n = 4
              \n\"2nd<\/p>\n

              \"KSEEB<\/p>\n

              Question 32.
              \nFrom the following data test whether mean daily wages of workers of Factory-A and Factory-B are equal. (Use 1% I.o.s.)
              \n\"2nd
              \nAnswer: Given:
              \nn1<\/sub>=200, n2<\/sub> = 450, x\u03041<\/sub>=195, x\u03042<\/sub> – 200, S1<\/sub> = 20, S2<\/sub> = 30, \u03b1 = 1%
              \nH0<\/sub> : Mean daily wages of workers of factory A and B are equal (H0<\/sub>: \u03bc1<\/sub> = \u03bc2<\/sub>)
              \nH1<\/sub> : Mean daily Wages of workers of factory A and B are not equal (H1<\/sub> : \u03bc1<\/sub> \u2260 \u03bc2<\/sub>)
              \n\"2nd
              \nUnder H0<\/sub>, the test statistic is:-
              \n\"2nd
              \nZcal = – 2.449.
              \nAt \u03b1 = 1% the two tail critical values are \u00b1K = \u00b12.58 Here Zcal lies in A.R (accepetance region), there fore H0<\/sub> is accepted.
              \nConclusion: Mean daily wages of workers of factory A and B workers are equal (i.e., \u00b51<\/sub> = \u00b52<\/sub>)<\/p>\n

              Question 33.
              \nFrom the following data, test whether literacy and smoking are independent at 5% level of significances.
              \n\"2nd
              \nAnswer:
              \nChi-Square test for Independence of attribute
              \nH0<\/sub>: Literacy and smoking are independent.
              \nH1<\/sub> : Literacy and smoking are dependent
              \nThe given data can be written in a 2 \u00d7 2 contingency table as below:
              \n\"2nd
              \n\"2nd
              \nunder H0<\/sub> the \u03c72<\/sup> – test statistic is:-
              \n\"2nd
              \nAt \u03b1 = 5% for 1 .d.f the upper tail critical value K2<\/sub> = 3.83. Here \u03c72<\/sup>cal<\/sub>\u00a0lies in A.R.
              \n\u2234 H0<\/sub> is accepted
              \nconclusion: Literacy and smoking are independent.<\/p>\n

              Question 34.
              \nThe following data relates to the number of knitting defects per unit length of cloth manufactured by a textile mill.
              \n\"2nd
              \nDraw control chart with \u03bb1<\/sup> – 3.
              \nOR
              \n(For Visually Challenged Students Only)
              \nBriefly explain single sample plan.
              \nAnswer:
              \nStandand is known: given \u03bb1<\/sup> = 3
              \nControl limits for number of defects is C-chart
              \nCentral line: C.L = \u03bb1<\/sup>\u00a0= 3
              \nLower control limit: L.C.L = \\(\\lambda^{\\prime}-3 \\sqrt{\\lambda^{\\prime}}=3-3 \\sqrt{3}\\)= -2.196 = 0
              \nUpper control limit: U.C.L =\\(\\lambda^{\\prime}+3 \\sqrt{\\lambda^{\\prime}}=3+3 \\sqrt{3}\\)= 8.2
              \n\"2nd
              \nConclusion: All the sample points lies with in the control limits, so the production process is under control.
              \nOR
              \nFor visually challanged students only<\/p>\n

              \"KSEEB<\/p>\n

              Question 35.
              \nA tailor gets a profit of Rs. 100 from a shirt and Rs. 170 from a pant. In a week of 56 hours, he uses 20 hours for stitching and 36 hours for cutting. He requires one hour to stitch a shirt and two hours to stitch a pant. For cutting he requires two hours for a shirt and three hours for a pant. Formulate L.P.P.
              \nAnswer:
              \nLet x and y the number of shirts and points
              \nThe objective function \/ L.P.P. is:-
              \nMaximize z= 100 x + 170 y
              \nSubject to constraints:
              \nStiching: x + 2y \u2264 20
              \nCutting: 2x + 3y \u2264 36 and x,y \u2265 0<\/p>\n

              Question 36.
              \nThe cost of a scooter is Rs. 36,000. Its resale value and maintenance cost at different years is given below:
              \n\"2nd
              \nDetermine the optimal year for replacement of the scooter.
              \nAnswer:
              \nP = 36,000 let sn<\/sub> be the resale value and Ci<\/sub> be the maintenance cost. The Annual average cost:
              \nA(n) = \\(\\frac { T }{ n }\\) . Here T-total maintenance cost = (P-Sn<\/sub>) + \u03a3Ci<\/sub>; n-the year of service.
              \n\"2nd
              \nFrom the above table A(n) is least for the year n = 5 so, the scooter is replaced after use of 5 years, (is the optimal year for replacement of the scooter)<\/p>\n

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

              IV. Answer any TWO of the following questions : 2 x 10 = 20<\/span><\/p>\n

              Question 37.
              \nFor the following data compute standardized death rates and comment.
              \n\"2nd
              \nAnswer:
              \n\"2nd
              \n\"2nd
              \nFor locality A:
              \nASDR(0- 20) = \\(\\frac { 72 }{ 4000 }\\) \u00d7 1000 = 18;
              \nFor locality B:
              \nA. S.D.R(B) = \\(\\frac { 128 }{ 8000 }\\) \u00d7 1000 = 16 similarly for other age groups can be calculated
              \nStandardised death rates for the two localities are:
              \n\"2nd
              \nComment: Here SDR(B) < SDR(A) and so, locality B is heather than locality A.<\/p>\n

              \"KSEEB<\/p>\n

              Question 38.
              \nCompute Fisher’s price index number. Show that it satisfies time reversal test and factor reversal tests.
              \n\"2nd
              \nAnswer:
              \n\"2nd
              \n\"2nd<\/p>\n

              Question 39.
              \nFit an exponential trend line of the form y = ab*. Also estimate the value for 2008.
              \n\"2nd
              \nAnswer:
              \n\"2nd
              \nHere \\(x=\\frac{x-\\bar{x}}{c}\\); x\u0304 = 2003; c = 1
              \nThe normal equations: n log a + log b. \u03a3x = \u03a3logy
              \nsince \u03a3x = 0; loga = \\(\\frac{\\Sigma \\log y}{n}=\\frac{7.853}{7}\\) = 1.1218
              \n\u2234 a = AL(loga) = AL(1.1218)
              \na = 13.2373
              \nlog a. \u03a3x + log b \u03a3x2<\/sup> = \u03a3x. logy<\/p>\n

              log b = \\(\\frac{\\Sigma x \\log y}{\\Sigma x^{2}}\\) ; logb= \\(\\frac{4.2071}{28}\\) =0.1502
              \n\u2234 b = AL (log b) = AL (0.1502) = 1.4132 ; The fitted exponential trend line is: y = abx<\/sup> ;
              \n\u0177 = 13.2373(1.4132)x<\/sup>.
              \nEstimation for the year 2008; x = 5; \u0177 = (2008) = 13.2373(1.4132)5<\/sup> = 74.6135.<\/p>\n

              Question 40.
              \nConsider the following data:
              \n\"2nd
              \nFit a Poisson distribution to the data and test the goodness of fit. (Use 5% L.O.S.)
              \nAnswer:
              \nThe parameter of Poisson variate \u03bb is:
              \nFrom the distribution: Mean = x\u0304 = \u03bb = \\(\\frac{\\Sigma f x}{N}\\)
              \n\"2nd
              \n\u03bb = 0.99
              \n\"2nd
              \nT0<\/sub> = 0.3716 \u00d7 100 = 37.16 = 37
              \n\"2nd
              \nT1<\/sub> = \\(\\frac{0.99}{1}\\)T0<\/sub> =0.99 \u00d7 37.16 = 36.7884 = 37
              \nT2<\/sub> = \\(\\frac{0.99}{2}\\) T1<\/sub> = 0.495 \u00d7 36.7884 = 18.2103 = 18
              \nT3<\/sub> = \\(\\frac{0.99}{3}\\)T2<\/sub> = 0.33 \u00d7 18.2103 = 6.0094 = 6
              \nT = \\(\\frac{0.99}{4}\\)T3<\/sub> = 0.2475 \u00d7 6.0094 = 1.4873 = 2
              \nT 5<\/sub>, = \\(\\frac{0.99}{5}\\)T4<\/sub> 0.198 \u00d7 1.4873 = 0.2945=0
              \nT 6<\/sub> = \\(\\frac{0.99}{6}\\) T5<\/sub> = 0.165 \u00d7 0.2945 = 0.0486= 0
              \nT7 or more <\/sub> =N – T6<\/sub>= 100 – 100 = 0.
              \nThe fitted observed and theoretial frequency distribution is:
              \n\"2nd
              \nH0<\/sub>: Poisson distribution is good fit {i.e., Oi<\/sub> = Ei<\/sub>}
              \nH1<\/sub>: Poisson distribution is not good fit {i.e., 0i<\/sub> \u2260 Ei<\/sub>}
              \nCHI – SQUARE TEST: Let 0\u2260 and E. be the observed (f) frequency and theoretical frequency (Tx)
              \nThen the \u03c72<\/sup>-test statistic is:
              \n\"2nd
              \n\u03bb is estimated from the data so (n – 2) d.f.
              \n\"2nd
              \n\u2234 \u03c72<\/sup>cal<\/sub>= 11.0979; n = 4
              \nAt \u03b1 = 5% for (n – 2) = 4 – 2 = 2 d.f the upper tail critical value K2<\/sub> = 9.49. Here \u03c72<\/sup>cal<\/sub> lies in
              \nR.R. \u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
              \nOR
              \nConclusion: 0i<\/sub> \u2260 Ei<\/sub>; i.e., P.D. is not a good fit.
              \n\"2nd<\/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 weekly wages of workers are normally distributed with mean Rs. 3,000 and S.D. Rs. 500. Find the probability of workers whose weekly wages will be.
              \n(i) More than Rs. 3,400
              \n(ii) Between Rs. 2,500 and Rs. 3,500.
              \nAnswer:
              \nLet A: be the weekly wages is a normal variate with the parameters mean : \u03bc = 3,000 and
              \nS. D: \u03c3 = 500.
              \n\"2nd
              \n= Area from (-1) to 1
              \n= Area from (-1) to \u221e – Area from 1 to \u221e
              \n= 0.8413 – 0.1587 = 0.6826.<\/p>\n

              Question 42.
              \nIn a random sample of 1000 persons from a large population, 470 are females. Can it be concluded that males and females are in equal proportion in the population? (use a = 0.05).
              \nAnswer:
              \nGiven: n = 1000; x = 470; .v P = \\(\\frac { x }{ n }\\) = \\(\\frac { 470 }{ 100 }\\) 0.47 and P0<\/sub>\/P – population proportion = 0.5;
              \nQ0<\/sub> = 1 – p0<\/sub> = 0.5 [male and female are in equal in proportion]
              \nH0<\/sub>: Male and female are in equal proportion [P0<\/sub> = 0.5]
              \nH1<\/sub>: Male and female are not in equal proportion [p0<\/sub> \u2260 0.5]
              \n\"2nd
              \nunder H0<\/sub>, the test statistic is:-
              \n\"2nd
              \nAt \u03b1 = 0.5 the two tail critical values are \u00b11.96 Here zcal<\/sub> lies in A.R. \u2234 H0<\/sub> is accepted.
              \nConclusion: P0<\/sub> = 0.5 [male and female are in equal proportion]<\/p>\n

              Question 43.
              \nIt is required to test whether those who practice yoga have average blood sugar less than 120. A sample consisting of 17 persons who practice yoga is observed. If their mean blood sugar is 108 and S.D. is 8, what would you conclude? (Use 5% level of significance).
              \nAnswer:
              \nGiven: \u00b5 = 120; n = 17, x\u0304 = 108; s = 8 \u03b1 = 5% (n < 30 small samplet-test).
              \nH0<\/sub> : The average blood sugar is 120 (\u00b5 = 120)
              \nH1<\/sub> : Average blood sugar is less than 120 (\u00b5 < 120)
              \n\"2nd
              \nUnder H0<\/sub>, the t-test statistic is:-
              \n\"2nd
              \nAt \u03b1 = 5% for (n – 1) = 17 – 1 = 16 d.f. the lower tail critical value – K = – 1.75. Here tcal<\/sub> lies in R.R.
              \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
              \nConclusion: Average blood sugar is less than 120.<\/p>\n

              \"KSEEB<\/p>\n

              Question 44.
              \nFind the solution of the game by the principle of dominance for the following pay-off matrix of A.
              \nAnswer:
              \n\"2nd
              \nAnswer:
              \n\"2nd
              \nBy principle of dominance, all the payoffs of A4<\/sub> are more or equal to all the payoffs of A1<\/sub> and A3<\/sub>. SO, A4<\/sub> dominates A1<\/sub> and A3<\/sub>, delete A1<\/sub>, A3<\/sub>.
              \n\"2nd
              \nAll the payoffs of B2<\/sub> are less than all the payoffs of B1<\/sub> and B3<\/sub>. So B2<\/sub> dominates B1<\/sub> and B3<\/sub>, delete B1<\/sub>, B3<\/sub>.
              \n\"2nd
              \nA4<\/sub> dominates A2<\/sub>, delete A2<\/sub>
              \n\"2nd
              \nB2<\/sub> dominates B4<\/sub>, delete B4<\/sub>.
              \n\"2nd
              \nSaddle point occurs at (4,2) \/ A4<\/sub>, B2<\/sub>.
              \nSuggested optimal strategies are A4<\/sub>, B2<\/sub> value of the game \u03bd = 2 is the solution the game.<\/p>\n

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

              Students can Download 2nd PUC Statistics Previous Year Question Paper June 2015, 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 Previous Year Question Paper June 2015 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\/14186"}],"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=14186"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/14186\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=14186"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=14186"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=14186"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}