{"id":13942,"date":"2019-12-21T17:06:34","date_gmt":"2019-12-21T11:36:34","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=13942"},"modified":"2021-07-02T15:35:39","modified_gmt":"2021-07-02T10:05:39","slug":"2nd-puc-statistics-previous-year-question-paper-march-2015","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-statistics-previous-year-question-paper-march-2015\/","title":{"rendered":"2nd PUC Statistics Previous Year Question Paper March 2015"},"content":{"rendered":"

Students can Download 2nd PUC Statistics Previous Year Question Paper March 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 March 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 \u00d7 1=10)<\/span><\/p>\n

Question 1.
\nDefine expectation of life.
\nAnswer:
\nIt is the average number of years a person aged x can expected to live under the prevailing mortalite conditions.<\/p>\n

Question 2.
\nIf the general price level goes up by 80% between 2000 and 2012. What is the index number for 2012 with base 2000?
\nAnswer:
\nIndex number for 2012 = 180<\/p>\n

Question 3.
\nName the index number which satisfies circular test.
\nAnswer:
\nKelly\u2019s method and simple geometric mean of price relatives method.<\/p>\n

Question 4.
\nWhich index is used for the measurement of seasonal variation?
\nAnswer:
\nSeasonal Indices method.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIf p = \\(\\frac { 1 }{ 4 }\\) for a Bernoulli distribution, write down the p.m.f.
\nAnswer:
\np.m.f is P(x) = \\(\\left(\\frac{1}{4}\\right)^{x}\\left(1-\\frac{1}{4}\\right)^{1-x}\\) : x = 0,1
\n= (0.25)x<\/sup> (0.75)1-x<\/sup>: x = 0 , 1<\/p>\n

Question 6.
\nWhat is the probability that a normal variate takes a value greater than its mean?
\nAnswer:
\n0.5<\/p>\n

Question 7.
\nDefine sampling distribution of a statistic.
\nAnswer:
\nIt is distribution of the values of a statistic for different samples by same size.<\/p>\n

Question 8.
\nWhat is power of a test?
\nAnswer:
\nIt is the probability rejecting to, when it is not true : (1- \u03b2) i Here \u03b2 = P (Type II error)<\/p>\n

Question 9.
\nIf p = 0.02 and n = 64 then find SE (p).
\nAnswer:
\n\"2nd
\nHere Qo<\/sub> = 1 – Po<\/sub> = 1 – 0.02 = 0.98<\/p>\n

Question 10.
\nDefine acceptance sampling.
\nAnswer:
\nIt is the process of inspecting a random sample of goods and deciding whether to accept the entire lot based on the results.<\/p>\n

Question 11.
\nWhat is meant by lead time?
\nAnswer:
\nIt is the time gap between the placing of order and arrival of goods at the Inverotory (T).<\/p>\n

Question 12.
\nWhen a T.P. balanced?
\nAnswer:
\nWhen \u03a3ai -Availabilities = \u03a3bj – Requirements then the T-P is said to be balanced.<\/p>\n

Section – B<\/span><\/p>\n

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

Question 13.
\nIn a locality 10,000 live births occnred. The number of infant deaths was 500. Calculate infant mortality rate.
\nAnswer:
\n\"2nd<\/p>\n

Question 14.
\nState two characteristics of index numbers.
\nAnswer:<\/p>\n

    \n
  1. Index numbers are specialised averages.<\/li>\n
  2. Index numbers are expressed in ratios but for better comparision expressed in percentages.<\/li>\n<\/ol>\n

    Question 15.
    \nQuantity index number of current year is 250. If the number of units produced in the current year is 120 then find the number of units produced in the base year.
    \nAnswer:
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 16.
    \nDiagrammatically represent \u2018Business Cycle\u2019 with stages
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 17.
    \nDefine \u2018Interpolation\u2019 and \u2018Extrapolation\u2019.
    \nAnswer:
    \n\u2018Interpolation\u2019 is the technique of estimating a value of the dependent variable (y) for any intermediate value of the independent variables (x) \u2018Extrapolation\u2019 is the technique of estimating the value of dependent variable (y) for any value of the independent variable (x) which is outside the range of the given series.<\/p>\n

    Question 18.
    \nFor a hypergeometric distribution the parameters are a = 7, b = 5 and n = 8. Find the variance.
    \nAnswer:
    \n\"2nd<\/p>\n

    19.
    \nIf z1<\/sub> and z2<\/sub> are two independent SNVs, then name the distribution of z1<\/sub>2<\/sup>, + z2<\/sub>2<\/sup> and find its mean.
    \nAnswer:
    \n(z1<\/sub>2<\/sup>, + z2<\/sub>2<\/sup>,) ~ \u03c72<\/sup>(2)<\/sub> d.f. ie; is a chi-square distribution with n = 2 degress of freedom mean = n = 2.<\/p>\n

    20.
    \nWrite two applications of t-test.
    \nAnswer:
    \nUnder small sample tests t-distribution is used<\/p>\n

      \n
    1. to test whether the population has the given mean,<\/li>\n
    2. to test whether two population means are equal (=\/>\/<)<\/li>\n<\/ol>\n

      21.
      \nMention two conditions for applicability of Chi-square test of goodness of fit.
      \nAnswer:<\/p>\n

        \n
      1. Total number of frequency N should be large.<\/li>\n
      2. Theoretical\/Expected frequency each Ei should be 5 or more (Ei<\/sub> \u2265 5) if any one is below it should be pooled with the adjacent frequencies (untill together 5 or more)<\/li>\n
      3. If any one of the parameter is estimated from the data for each such estimation one degrees of freedom should be lessened.<\/li>\n<\/ol>\n

        22.
        \nIn a floor mat manufacturing company, the average number of defects per square meter of floor mat is known to be 4. Find the upper control limit for the number of defects.
        \nAns.
        \nGiven: \u03bb1<\/sup> =4. The upper control limit for C – chart: UCL = \\(\\lambda^{1}+3 \\sqrt{\\lambda^{1}}=4+3 \\sqrt{4}=10\\)<\/p>\n

        23.
        \nUnder what conditions is E.O.Q. model with shortage applicable.
        \nAnswer:<\/p>\n

          \n
        • Uniform demand<\/li>\n
        • Production is Instantaneous<\/li>\n
        • Lead time is Zero<\/li>\n
        • Shortages are allowed.<\/li>\n<\/ul>\n

          24.
          \nFrom the following T.P. test whether the solution is non-degenerate:
          \n\"2nd
          \nAnswer:
          \nHere no. positive allocations = 4; m + n – 1 = No. of rows + No. of columns -1 = 3 + 3 -1 = 5
          \nHere no. of allocations < (m + n – 1) the solution is degenerate \/ not a non-degenerate.<\/p>\n

          \"KSEEB<\/p>\n

          Section – C<\/span><\/p>\n

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

          )Question 25.
          \nCompute the Gross reproduction rate from the following data :
          \n\"2nd
          \nAnswer:
          \n\"2nd
          \n\"2nd
          \n\"2nd
          \nW. S. F. R (15-19) = \\(\\frac { 19 }{ 1658 }\\) \u00d7 100 = 11.459
          \nSimilarly for remaing age groups W.S.F.R can be computed. Here i = 5
          \nG. R. R. = 5 \u00d7 200.272 = 1001.36 Female births \/ 1000 women of (15-49) years.<\/p>\n

          Question 26.
          \nBriefly explain TRT and FRT.
          \nAnswer:
          \nTRT: Time reversal test requires that the index number computed backward (P10<\/sub>) should be reciprocal of the index number computed time forward (P10<\/sub>)
          \ni.e., P10<\/sub> = \\(\\frac{1}{\\mathrm{P}_{01}}\\); Exccept the factor 100.<\/p>\n

          FRT: Factor reversal test requires that the product of price and quantity index numbers both computed time forward is equal to the total change in the value.
          \n\"2nd<\/p>\n

          Question 27.
          \ncompute the cost of living index number by using the following data
          \n\"2nd
          \nAnswer:
          \nWeights (W) are given , so use family budged method to compute cost of living imdex number C.P.I.\/IO = \\(\\frac{\\Sigma \\mathrm{PW}}{\\Sigma \\mathrm{W}}\\) ; Where, p = \\(\\frac{P_{1}}{P_{0}} \\times 100\\)\u00a0 the price relative and W – Weghts
          \n\"2nd
          \nI01<\/sub> = \\(\\frac { 3560 }{ 27 }\\) = 131.85 \u2245 132
          \nThere is net in crease of cost of living by 32% (132-100) in the current year as compared to base year.<\/p>\n

          Question 28.
          \nCompute trend values by using 4-yearly moving averages method:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n
          Year<\/td>\nSales (000)<\/td>\n<\/tr>\n
          2000<\/td>\n70<\/td>\n<\/tr>\n
          2001<\/td>\n85<\/td>\n<\/tr>\n
          2002<\/td>\n75<\/td>\n<\/tr>\n
          2003<\/td>\n70<\/td>\n<\/tr>\n
          2004<\/td>\n70<\/td>\n<\/tr>\n
          2005<\/td>\n65<\/td>\n<\/tr>\n
          2006<\/td>\n60<\/td>\n<\/tr>\n
          2007<\/td>\n55<\/td>\n<\/tr>\n
          2008<\/td>\n60<\/td>\n<\/tr>\n
          2009<\/td>\n75<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Answer:
          \nLet x and y be the year and sales.
          \n\"2nd
          \nFrom the trend values the sales shows down word trend.<\/p>\n

          \"KSEEB<\/p>\n

          Question 29.
          \nInterpolate and extrapolate the production for the year 2006 and 2010 for the following data:
          \n\"2nd
          \nAnswer:
          \nLet x and y be the year and production.
          \n\"2nd
          \nThe number of known values of y; n = 4 and so, the normal expansion (y – 1)4<\/sup>= 0.
          \ny4<\/sub> – 4y3<\/sub> + 6y2<\/sub> – 4y1<\/sub> + y0<\/sub> = 0
          \n26 – 4(15) + 6 (10) – 4y1<\/sub> + 5 = 0
          \n26 – 60 + 60 – 4y1<\/sub> + 5 = 0
          \n31 – 4y1<\/sub> = 0; \u2234 y1<\/sub>= \\(\\frac { 31 }{ 4 }\\) = 7.75 .
          \nSo, the production for the year 20.6 = 7.75
          \nFor the entropolation for y5<\/sub> consider the equation 1, increase the suffixes of y by 1, we get
          \ny5<\/sub> – 4y4<\/sub> + 6y3<\/sub> – 4y2<\/sub> + y1<\/sub>= 0
          \ny5<\/sub> – 4(26)+ 6 (15) – 4(10) + 7.75 = 0
          \ny5 <\/sub>– 104 + 90-40 + 7.75 = 0
          \ny5<\/sub> – 46.25 = 0; \u2234 y5<\/sub> = 46.25
          \n\u2234 The production for the year 2010 = 46.25 tons.<\/p>\n

          Question 30.
          \nIn a certain school 40% of the students have opted for first language Kannada. Assuming 20 teachers take a sample of 4 students each, how many teachers will report that 2 or 3 students opted for first language Kannada?
          \nAnswer:
          \nLet x be the numbers students opted first language kannada is a Binomnal variate with the parameters.
          \nn = 4; P = 40% = 0.4; q = 1 – p = 0.6 and N = 20.
          \nThe p. m. f. is P(x) = n<\/sup>Cx<\/sub>Px<\/sup> qn-x<\/sup> = 1 – p = 0.6 and N = 20
          \nP(x) = 4<\/sup>Cx<\/sub>(0.4)x<\/sup> (0.6)4 – x<\/sup> ; x = 0,1,2,3,4
          \nNumber of teachers will report 2 or more students opted for first language kannada
          \n= P(x = 2) + P(x = 3) = 4<\/sup>C2<\/sub>(0.4)2<\/sup> (0.6)4 – 2<\/sup> + 4<\/sup>C3<\/sub>(0.4)3<\/sup>(0.6)4-3<\/sup>
          \n= 0.3456 + 0.1536 = 0.4992 \u00d7 20 = 9.98<\/p>\n

          Question 31.
          \nIn a text book, on an average 0.3 mistakes per page is found. If there are 500 pages in that text book, in how many pages will there be at the most two mistakes?
          \nAnswer:
          \nLet x be the numbers of print mistakes pef page is a poisson variate with the parameter \u03bb = 0.3, then the p.m.f. is :
          \n\"2nd
          \nNumber of pages having at the most 2 mistakes
          \n= P(x \u2265 2)N = 1 – P(x < 2)N = 1 – [P(x = 0) + P(x = 1)
          \n\"2nd
          \n= 1 – e-0.3<\/sup>(1 + 0.3) = [1 – 0.7048 \u00d7 1.3] = 1 – 0.9630
          \n= 0.037 \u00d7 500 = 18.5 symbol undi 19 pages<\/p>\n

          Question 32.
          \nA random sample of 400 tins of Vanaspathi has mean weight 4.96 kg and S.D. 0.4 kg. Test at 1% L. O. S. that the average weight of tins of Vanaspathi is less than 5 kg.
          \nAnswer:
          \nGiven : n = 400; x\u0304 = 4.96; S = 0.4, \u03b1 = 1% \u00b5 = 5
          \nH0<\/sub> : Mean weight of tens is 5 kg (Ho<\/sub>: p = 5kg) –
          \nH1<\/sub> : Mean weight of tens is less than 5kg
          \n(H1<\/sub> : \u00b5 < 5 kg) { Lower tail test; -K}
          \n\"2nd
          \nunder H0<\/sub> the test statistic is :-
          \n\"2nd
          \nZcal<\/sub> = -2
          \nAt \u03b1 = 1% the lower tail critical value – K = -2.33
          \nHere Zcal<\/sub> lies in A.R .
          \n\u2234 H0<\/sub> is Accepted
          \nConclusion: Mean weight of tins is 5kg.<\/p>\n

          Question 33.
          \nFollowing is the data regarding I.Q. of five students before and after performing yoga.
          \n\"2nd
          \nIs training effective? (Use 5% L.O.S.)
          \nAnswer:
          \nH0<\/sub> : Mean I.Q. be for and after Yoga is same (H0<\/sub>: \u00b51<\/sub> = \u00b52<\/sub>)
          \nH1<\/sub> : Mean I.Q. has increased after Yoga (H, : \u00b51<\/sub> < \u00b52<\/sub>) {Upper tail test + K}
          \nUnder H0<\/sub>, the paired t-test statistic is
          \n\"2nd
          \nlet x and y be I. Q. before and after yoga.
          \n\"2nd
          \n\"2nd
          \nAt \u221d = 5% for (q-1) = 5-1 = 4.d.f.
          \nThe lower tail critical value -K = -2.13
          \nHere tcal lies in R. R.
          \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted
          \nCondusion : I. Q. has improved after yoga
          \n\"2nd<\/p>\n

          \"KSEEB<\/p>\n

          Question 34.
          \nFor the following data find out the control limits for X\u0304 – chart. (Given A2<\/sub> = 0.577)
          \n\"2nd
          \nAnswer:
          \nStandands not known; for X\u0304 chart we need
          \n\"2nd
          \nThe control limits for X\u033f – chart K = 6
          \nC.L = X\u033f = 49.5
          \nL.C.L= X\u033f – A2<\/sub>R\u0304
          \n= 49.5 – 0.577 \u00d7 10.16
          \n= 43.63<\/p>\n

          U.C.L =X\u033f + A2<\/sub>R\u0304
          \n= 49.5 + 0.577 \u00d7 10.16
          \n= 55.36<\/p>\n

          Question 35.
          \nGraphically solve the given L.P.P.
          \nMaximize Z = 12x + 40y
          \nS.t.
          \n3.v + 3y \u2264 12
          \n6x + 3y \u2264 18 and x, y \u2265 0.
          \nAnswer:
          \nConsider the constraints as equalities and find co-ordinates
          \n3x + 3y = 12;
          \nPut x = 0; We get y = 4; (0,4)
          \nPut y = 0; We get x = 4; (4,0)
          \n6x + 3y = 18;
          \nPut x = 0; We get y = 6; (0,6)
          \nPuty = We get n = 3; (3,0)
          \nand x = 0; y = o
          \n\"2nd
          \nFrom the graph the feasible region exists at the comer points 0(0,0), A (0,4), B(2,2), C(3,0)
          \nDetermination of optimun value of objective function.
          \n\"2nd
          \nFrom the above table. Objective function \u2018Z\u2019 maximizes at the comer point A (0,4) The suggested optimun solution for the L.P.P. is x = 0, y = 4 and z A = 160
          \nOR
          \n(For Blind Students Only)
          \nWrite down the steps in the graphical method of solving LPP.
          \nAnswer:
          \nFinding a solution to a L.P.P. by graphical method.<\/p>\n

            \n
          • Consider the constraints as equalities.<\/li>\n
          • Find the co-ordinates for each constraints.<\/li>\n
          • Draw the graph and represent by straight line.<\/li>\n
          • Identify the feasible region, which satisfies the non-negativity restrictions and constraints . simultaneously.<\/li>\n
          • Locate the comer points of the feasible region.<\/li>\n
          • Find the value of the objective function each of the corner points of the feasible region.<\/li>\n
          • get the optimom value and suggest the solution for the L.P.P.<\/li>\n<\/ul>\n

            Question 36.
            \nThe following are the maintenance and depreciation costs per year of a vehicle whose purchase price is Rs. 50,000. –
            \n\"2nd
            \nWhen should the vehicle be replaced?
            \nAnswer:
            \nGiven Depreciation cast: (P-sn) and maintenance cost Ci and P = 50,000.
            \nThe Vehicle is suggested to replace when its Annual average maintenance cost is A(n) is least\/ minimum over a period.
            \n\"2nd
            \nHere T = (P-Sn) + \u03a3Ci
            \nFrom the above table the Annual average maintenance cast A(n) is least for the year n = 6 and so, the vehicle should replace after the use of 6 years, and A(6) = Rs. 17083.33<\/p>\n

            \"KSEEB<\/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 two towns compute standardized death rates and comment:
            \n\"2nd
            \nAnswer:
            \n\"2nd
            \nbe the standardised death rates of town A and B and Ps. The standard population, A and B be their respective Age specific death rates.
            \n\"2nd
            \n\"2nd
            \nASDR (0-20) = \\(\\frac { 128 }{ 8000 }\\) \u00d7 1000 = 18
            \nSDR (A) = \\(\\frac { 19,05,000 }{ 1,00,000 }\\)
            \n\"2nd
            \nASDR (0 – 20) = \\(\\frac { 72 }{ 4000 }\\) \u00d7 1000 = 18
            \nSDR (B) = \\(\\frac { 16,75,000 }{ 1,00,000 }\\) = 16.75
            \nHere S.D.R. (B) < S.D.R. (A) and so, town B is healthier than town A.<\/p>\n

            Question 38.
            \nCompute Marshall-Edgeworth\u2019s and Dorbish-Bowley\u2019s price index numbers from the following data:
            \n\"2nd
            \nAnswer:
            \n\"2nd
            \n\"2nd
            \n\"2nd<\/p>\n

            Question 39.
            \nPopulation figures for a-city are given below:
            \n\"2nd
            \nFit a curve of the type Y = abx<\/sup> and estimate the population for the year 2014.
            \nAnswer:
            \nThe exponential curve is : Y = abx<\/sup>
            \nThe normal equations are : n loga + logb. \u03a3x = \u03a3loga (1)
            \n\u03a3 x + logb \u03a3x2 = logy …… (2)
            \nLet x and y be the years and the population.
            \n\"2nd
            \nSince \u03a3 x = 0;
            \nFrom (1)
            \n\"2nd
            \n\u2234 a = A.L (loga) = A.L (2.1960) = 157.036
            \nFrom (2)
            \n\"2nd
            \n\u2234 b = A.L (logb) = A.L (0.0399) = 1,0962
            \nThe fitted exponential trend equation is:
            \ny = abx<\/sup>
            \n\u0177 = 157.036(1.0962)x<\/sup>
            \nEstimation of the population for the year 2014; x = 4;
            \n\u0177(2014) = 157.036 (1.962)4<\/sup> = 226.7905 (000\u2019s)<\/p>\n

            Question 40.
            \nFive unbiased coins are tossed 128 times and the frequencies of number heads are noted below. Fit a Binomial distribution. Test at 5% L. O. S. that it is a good fit.
            \n\"2nd
            \nAnswer:
            \nLet x be the number of heads obtained is a Binomnal variate with the parameters n = 5, and P = \\(\\frac { 1 }{ 2 }\\) = 0.5. Sinpe the coin is unbiased \u2234 q = 1 – P = 1 – 0.5 = 0.5 N = 120 (Given as 128 is a misprint). Then the P.m.f. is :
            \nP(x) = n<\/sup>Cx<\/sub>Px<\/sup>qn-x<\/sup>, x = 0, 1, 2 …n
            \n= 5<\/sup>Cx<\/sub>(0.5)x<\/sup>(0.5)5-x<\/sup>, n = 0, 1,2 …n<\/p>\n

            Theoretical \/ Expected frequence: Tx = P(x) M.
            \nSo, T0<\/sub>\/E0<\/sub> = P(x = 0) \u00d7 100
            \n= 5<\/sup>Co<\/sub>(0.5)\u00b0 (0.5)5-0<\/sup> \u00d7 100 = 3.125<\/p>\n

            Using recurrence relation:
            \n\"2nd
            \nT1<\/sub> = \\(\\frac{5+1-1}{1}\\left(\\frac{0.5}{0.5}\\right)\\) T1-1<\/sub> = 5 \u00d7 3.125 = 15.625
            \nT0<\/sub> = \\(\\frac{5+1-2}{2}\\) \u00d7 1 \u00d7 15.625=31.25
            \nT3<\/sub> = \\(\\frac{5+1-3}{3}\\) \u00d7 l \u00d7 31.25 = 31.25
            \nT4<\/sub> = \\(\\frac{5+1-4}{4}\\) \u00d7 1 \u00d7 31.25 = 15.625 4
            \nT5<\/sub> = \\(\\frac{5+1-5}{5}\\) \u00d7 1 \u00d7 15.625 = 3.125
            \nThe fitted observed and theoretical frequency distribution is
            \n\"2nd
            \nChi-Square test:
            \nH0<\/sub>: B.D is a good fit (ie.,Oi<\/sub> = Ei<\/sub>)
            \nH1<\/sub>: B.D is not a good fit (i.e., Oi<\/sub> \u2260 Ei<\/sub>) {Upper tail test K2<\/sub>}
            \nUnder H0<\/sub>, the \u03c72<\/sup>– test statistic is :
            \n\"2nd
            \n\"2nd
            \n\u03c72<\/sup>cal<\/sub> = 15.6163
            \n\"2nd
            \nAt \u03b1 = 5% L.O.S for (n – 1) = 4 – 1 = 3 d.f the upper tail critical value K2<\/sub> = 7.81
            \nHere \u03c72<\/sup>cal <\/sub>lies in RR
            \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
            \nConclusion: B.D. is not a good fit.<\/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 weights of 1000 students are normally distributed with mean 55 kgs and S.D. 3 kgs. Find the number of students with weight between 57 kgs and 60 kgs.
            \nAnswer:
            \nLet x be the weights is a normal variate with mean: \u03bc = 55 kgs and S.D: \u03c3 = 3kgs. N = 1000
            \nThen S.N.V
            \n\"2nd
            \nNo of students with weight between 57kg and 60kg
            \n\"2nd
            \n= p(57 < \u00d7< 60)
            \n\\(=\\mathrm{P}\\left(\\frac{27-55}{3}<\\mathrm{Z}<\\frac{60-55}{3}\\right)\\)
            \n= P(0.67 < Z < 1.67)
            \n= Area from 0.67 to 1.67
            \n= Area from 0.67 to \u221e – Area from 1.67
            \n= 0.2514 – 0.0475
            \n= 0.2039 \u00d7 1000
            \n= 203.9 \u2245 204<\/p>\n

            Question 42.
            \nA random sample of 100 people from a city in the year 2011 revealed that 65 were cricket match viewers. Another random sample of 100 people from same city in the year 2013 revealed that 75 were cricket match viewers. Examine whether there is a significant increase in proportion of cricket match viewership at 1% level of significance.
            \nAnswer:
            \nn1<\/sub> = 100; x1<\/sub>, = 65; n2<\/sub> 100; x2<\/sub> = 75
            \nH0<\/sub>: Proportion of cricket viewers in two years are same (i.e., P01<\/sub> = P02<\/sub>)
            \nH1<\/sub>: Proportion of cricket viewers has increased in two years (i.e.,P01<\/sub> < P02<\/sub>) {Lower tail test}
            \nUnder H0<\/sub>, the test statistic is:
            \n\"2nd
            \n\"2nd
            \n\"2nd
            \nAt 1% the lower tail critical value -K = -2.33
            \nHere Zcal<\/sub> lies in A.R. \u2234 H0<\/sub> is Accepted.
            \nConclusion: P01<\/sub> = P02<\/sub><\/p>\n

            Question 43.
            \nFrom the following data, test whether \u2018education\u2019 and \u2018employment\u2019 are independent at 5% level of significance.
            \n\"2nd
            \nAnswer:
            \nH0<\/sub>: Education and employment are independent
            \nH1<\/sub>: Education and employment are dependent.
            \nThe given data can be written in a 2 \u00d7 2 contingency table as below.
            \n\"2nd
            \nUnder H0<\/sub> the \u03c72<\/sup> -test statistics is
            \n\"2nd
            \nAt \u03b1 = 5% for l.df the upper tail critical value K2<\/sub> = 3.84. Here \u03c72<\/sup> -cal lies in R.R.
            \n\u2234 H0<\/sub> is rejected and H1<\/sub> is Accepted
            \nConclusion: Education and employment are dependent<\/p>\n

            \"KSEEB<\/p>\n

            Question 44.
            \nObtain the value of the game and mention the best strategies of the players.
            \n\"2nd
            \nAnswer:
            \n\"2nd
            \nUsing maximum-minimax principle –<\/p>\n

              \n
            • Minimum Payoff in each row is\/are circled.<\/li>\n
            • Pay off in each column are\/is boxed.<\/li>\n
            • From the above process the Payoff 3 at (2, 3) is circled as well as boxed, is the saddle point<\/li>\n
            • The suggested optimal strategies for the players are A2<\/sub>, B3<\/sub><\/li>\n
            • The value of the game V = 3<\/li>\n<\/ul>\n

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

              Students can Download 2nd PUC Statistics Previous Year Question Paper March 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 March 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\/13942"}],"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=13942"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13942\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=13942"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=13942"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=13942"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}