{"id":14253,"date":"2019-12-23T11:12:55","date_gmt":"2019-12-23T05:42:55","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=14253"},"modified":"2021-07-02T15:35:11","modified_gmt":"2021-07-02T10:05:11","slug":"2nd-puc-statistics-model-question-paper-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-statistics-model-question-paper-3\/","title":{"rendered":"2nd PUC Statistics Model Question Paper 3 with Answers"},"content":{"rendered":"

Students can Download 2nd PUC Statistics Model Question Paper 3 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 3 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 \u00d7 1 = 10 )<\/span><\/p>\n

    Question 1.
    \nMention a source of vital statistics.
    \nAnswer:
    \nRegistration method\/census method<\/p>\n

    Question 2.
    \nWhat is the index number for the base year.
    \nAnswer:
    \nIndex number for the base year = 100.<\/p>\n

    Question 3.
    \nMention a use of index number.
    \nAnswer:
    \nIndex numbers are used in fixation of bank rates, grant of dearness allowance to employees<\/p>\n

    Question 4.
    \nGive an example for \u2018seasonal variation\u2019 in time series analysis.
    \nAnswer:
    \nSales of cool drinks, ice creams are more in the summer season.<\/p>\n

    Question 5.
    \nMention a feature of Bernoulli distribution
    \nAnswer:
    \nBernoulli distribution has parameter – p or mean = p or variance = p (1 – p)<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nWrite the relationship between mean and variance of Poisson distribution.
    \nAnswer:
    \nMean = variance<\/p>\n

    Question 7.
    \nDefine point estimation.
    \nAnswer:
    \nIn point estimation a single statistic is used to provide an estimate of the population parameter OR .while estimating unknown parameter if a special value is proposed as an estimate is called point estimation.<\/p>\n

    Question 8.
    \nWrite down the expression for SE(p)
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 9.
    \nWhat is the degree of freedom used in test of goodness of fit when one of the parameter is estimated?
    \nAnswer:
    \nDegrees of freedom = (n – 1 – 1 ) = (n – 2)<\/p>\n

    Question 10.
    \nWhat is meant by multiple solution in a L.P.P?
    \nAnswer:
    \nIf an L.P.P. has many optimal solutions, it is said to have multiple solution.<\/p>\n

    Question 11.
    \nWhat is meant by pure strategy?
    \nAnswer:
    \nWhile playing a game, pure strategy of a player is his determined decision to adopt a specified course of action, irrespective of the course of action of the opponent.<\/p>\n

    Question 12.
    \nWhat are control charts in S.Q.C?
    \nAnswer:
    \nControl chart is a graphical device which is used to verify whether the production process is in control or not.<\/p>\n

    \"KSEEB<\/p>\n

    Section – B<\/span><\/p>\n

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

    Question 13.
    \nCalculate I.M.R. for the following data.
    \n\"2nd
    \nAnswer:
    \nInfant Mortality Rate is
    \n\"2nd<\/p>\n

    Question 14.
    \nWhat are T.R.T. and F.R.T. in index numbers?
    \nAnswer:
    \nT.R.T (Time Reversal Test) and F.R.T (Factor Reversal Test) are Reliability Tests used to choose an appropriate formula which satisfies these tests.
    \nHere T.R.T : P01<\/sub> \u00d7 P10<\/sub> = 1 ; and P01<\/sub> \u00d7 P10<\/sub> = \\(\\frac{\\Sigma P_{1} q_{1}}{\\Sigma P_{o} q_{0}}\\) (except 100)<\/p>\n

    Question 15.
    \nGiven P01<\/sub>(L)<\/sup> =125 and P01<\/sub>(P)<\/sup> = 122 find P01<\/sub>(DB)<\/sup> = 122.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 16.
    \nMention a merit and demerit of moving averages method of measuring trend in a time series.
    \nAnswer:
    \nMerit: The method is simple.
    \nDemerit: The method does not provide trend values at the beginning and at. the end of the time series.<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nExpand (y – l)6<\/sup>.
    \nAnswer:
    \n(y-l)6<\/sup> = y6<\/sup> – 6y5<\/sup> + 15y4<\/sup> – 20y3<\/sup> +15y2<\/sup> – 6y1<\/sup> + y0<\/sup> = 0<\/p>\n

    Question 18.
    \nIn a N.D. of Q1<\/sub>= 22 ems and Q3<\/sub> = 34 ems then find median.
    \nAnswer:
    \nIn a normal distribution the two quartiles are equidistant from the median
    \n\"2nd<\/p>\n

    Question 19.
    \nFind the area under the Normal curve to the right of z = 1.3.
    \nAnswer:
    \nArea under normal curve to the right of z = 1.3
    \n= Area from 0 to \u221e – Area from 0 to 1.3
    \n=0.5 – 0.4032 = 0.0968<\/p>\n

    Question 20.
    \nMention any two features of t-distribution.
    \nAnswer:
    \n(a) Range is ( -\u221e , \u221e )
    \n(b) Mean = 0 (mean = median = mode = 0).<\/p>\n

    Question 21.
    \nMention two application of tests.
    \nAnswer:
    \n(a) To test whether a population has given variance.
    \n(b) To test \u2018goodness of fit\u2019 of theoretical frequencies to a observed frequencies.<\/p>\n

    Question 22.
    \nMention two situations when replacement is carried out.
    \nAnswer:
    \nReplacement’s carried out when
    \n(a) Machine and their spare parts deteriorates with time.
    \n(b) The maintenance cost increases and efficiency reduces.<\/p>\n

    Question 23.
    \nWrite the formula for minimum average cost in E.O.Q. model with shortages, giving meaning of notations.
    \nAnswer:
    \n\"2nd
    \nHere
    \nC1<\/sub> = carrying\/ Holding\/ Maintenance cost
    \nC2<\/sub> = Shortage\/penalty cost per unit\/unit time
    \nC3<\/sub> = set up cost\/ capital cost .
    \nR = Demand Rate<\/p>\n

    Question 24.
    \nWhat is meant by defect and defectives in S.Q.C?
    \nAnswer:
    \nA \u2018defect\u2019 is a quality characteristic which does not conform to specifications.
    \nThe presence of one or more defects in a product is \u2018defective.<\/p>\n

    \"KSEEB<\/p>\n

    Question 25.
    \nFrom the following data, calculate Total Fertitlity Rate.
    \n\"2nd
    \nAnswer:
    \nTotal Fertility Rate.(T.F.R) =5 \u03a3 Quinquennial ASFR\u2019s
    \n\"2nd
    \n\"2nd
    \nTFR = 5\u03a3 Quinquennial ASFR\u2019 = 5 \u00d7 559.9 = 2799.5<\/p>\n

    Section – C<\/span><\/p>\n

    III. Answer any EIGHT questions. ( 8 \u00d7 5 = 40 )<\/span><\/p>\n

    Question 26.
    \nExplain steps while constructing consumer price index number.
    \nAnswer:
    \nThe following are main steps in the construction of cost of living index numbers.
    \n(a) Object and scope.
    \n(b) Conducting family budget survey.
    \n(c) Obtaining price quotations.
    \n(d) Method of construction.<\/p>\n

    (a) Object and scope:
    \nDecide the class of consumers for which index number is required. Such as whether index number is meant for govt, employees, bank employees, merchants, farmers etc. Also the geographical location as that of city, town, or a community of people who are having the similar consumption pattern.<\/p>\n

    (b) Conducting family budget survey:
    \nA sample survey regarding the average expenses of familities on various items of consumption is conducted. In the survey, the information regarding commodities consumed by the families, their quality and the respective budget is collected. The items included are under the heads of i) food (ii) clothing (ii) fuel and lighting and (iii) miscellaneous
    \nEach of these can be divided into smaller groups as food may be include wheat, rice, dhal etc which are consumed by the people for whom index is meant.<\/p>\n

    (c) Obtaining price quotations:
    \nWhile constucting the cost of living index number, retail prices of commodities are to be collected. The price lists are obtained from different agencies from different places. Then, they are averaged and these averages are used in the construction of index numbers. Here price lists of current period as well as base period should be collected.<\/p>\n

    (d) Methods of construction:
    \nThere are two methods of construction.
    \n(i) Aggregative expenditure method (A.E.M)
    \n\"2nd
    \n(ii) Family Budget Method (F.B.M).
    \n\"2nd
    \nPrice<\/p>\n

    Question 27.
    \nCalculate cost of living index number by family budget method.
    \n\"2nd
    \nAnswer:
    \nConsumer price index number by family budget method = \\(\\frac{\\sum p w}{\\sum w}\\) ; p = \\(\\frac{p_{1}}{p_{0}} \\times 100\\) ; w = Weights.
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 28.
    \nObtain trend using 4 yearly moving averages method from the following data. Also plot original and trend values on a graph.
    \n\"2nd
    \nAnswer:
    \nHere m = 4
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 29.
    \nUse Newtons advancing difference method to find the value of y when x = 30.
    \n\"2nd
    \nAnswer:
    \nSince the value to be estimate does not one of the data (x) series. So, apply Newton’s method. The number of known value of 4 are 5, prepare the leading difference upto \u03944<\/sup>.
    \n\"2nd<\/p>\n

    Question 30.
    \nThe number of print mistakes per page in a text book follows P.D. with mean = 2. If there are 560 pages in a book, in how may pages would you expect mistakes?
    \nAnswer:
    \nLet x be the number of mistakes per page in a text book is a Poisson variate with the parameter \u03bb = 2 Then
    \nThe p.m.f is –
    \nP(x) = \\(\\frac{e^{-i} \\lambda}{x !}\\) ;x = 0,1,2 ………….
    \nThe number of pages expect mistakes = P (there are mistakes) \u00d7 N
    \n= P(x \u2265 1) \u00d7 N .
    \n= 1 – p(x < l) \u00d7 5600
    \n= {1 – P(x = 0)} \u00d7 5600
    \n= |1 – 0.1353| \u00d7 5600 = 4842 pages<\/p>\n

    \"KSEEB<\/p>\n

    Question 31.
    \nFrom the following data obtain theoretical frequencies, assuming p= 1\/2 , for a.B.D.
    \nimag\u00a0 64
    \nAnswer:
    \nLet X is a binomial variate with (P = q) the parameters p = \\(\\frac{1}{2}\\) ;q = \\(\\frac{1}{2}\\) and n = l
    \nThen the p.m.f. is:
    \nP(x) = ncx <\/sub>px <\/sup>qn – x<\/sup>; x = 0,1,2, …… n
    \n\"2nd
    \nTheoretical frequency Tx<\/sub> = P(x) \u00d7 N
    \nT(X = 0) = T0<\/sub> = P(x=0) \u00d7 128
    \n\"2nd
    \n= 1 \u00d7 1 \u00d7 0.0078125 \u00d7 128
    \n\u2234T0<\/sub> = 1
    \n\"2nd
    \n\"2nd
    \nThe fitted observed and theoritical frequency distribution is:
    \n\"2nd
    \n\u2234 Here the distribution is symmetric, since P = q = \\(\\frac{1}{2}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 32.
    \nIn a sample of 600 students of a college 400 are found to use dot pens. In another college from a sample of 900 students, 450 were found to use dot pens. Test whether the two colleges differ significantly with respect to the habit of using dot pens.
    \nAnswer:
    \nGiven n1<\/sub> = 600, x1<\/sub> = 400, \u2234P1<\/sub> = \\(\\frac{x_{1}}{n_{1}}-\\frac{400}{600}=0.67\\) .
    \nn2<\/sub> = 900, x2<\/sub> = 450 \u2234P2<\/sub> = \\(\\frac{x_{2}}{n_{2}}-\\frac{450}{900}\\) = 0.5
    \nH0<\/sub> : The proportion of students using dot pens in the two colleges are same
    \n(i.e. H2<\/sub> : P01<\/sub> = P02<\/sub> = P0<\/sub>)
    \nThe proportion of students using dot pens in the two colleges are not same,
    \n(i.e. H1<\/sub> : P01<\/sub> \u2260 P02<\/sub> = P0<\/sub>) (two tail test)
    \nThe test statistic is –
    \n\"2nd
    \n\"2nd
    \nZcal<\/sub> =6.5097
    \nAt \u03b1 = 5% the two tail critical values are -k1<\/sub>:, =1.96 and k2<\/sub> =1.96
    \nHere Zcal<\/sub> is in rejection region \/ Zcal<\/sub> > K2<\/sub> .
    \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
    \nConclusion : The proportion of students using dot pens in the two colleges are not same.<\/p>\n

    \"2nd<\/p>\n

    Question 33.
    \nA manufacturer of tyres claims that average life is more than 15000 kms under normal condition. A random sample of 15 tyres was tested and mean and S.D. was found to be 18000 kms and 4000 kms respectively. Test whether the manufacturer’s claim is justified.
    \nAnswer:
    \nGiven: \u03bc= 15000 kms, nj = 15, X\u0304 = 18000, s = 4000, \u03b1 = l%
    \nH0<\/sub> : Average life of tyres is 15000 kms (i.e. H0<\/sub> : \u03bc = 15000)
    \nH1<\/sub> : Average life of tyres is more than 15000 kms
    \n(H1<\/sub> : \u03bc >15000) {Right\/upper tail test)
    \nThe t-test statistic is ( Here n < 30 use t -test for mean)
    \n\"2nd
    \nAt \u03b1 = 1 % for (n – 1) = (15 ) = 14 d.f. the upper tail critical value k = 2.62
    \nHere t>cal<\/sub> > k\/tcal<\/sub> is in rejection region.
    \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
    \nConclusion : Average life of tyres is more than 15000 kms.
    \n\"2nd<\/p>\n

    Question 34.
    \nSolve the L.P.P graphically
    \nminimise Z = 3x + 24
    \nsubject to 5x + 4y \u2264 50
    \nx + 2y \u226510
    \nand x, y \u2265 0
    \nAnswer:
    \nConsider the constraints and written as equalities and are represented by straight lines, by identifying two points on each of these lines.
    \n5x + 4y = 50 put x = 0 we get y = 12.5 ………….\u00a0 (0,12.5)
    \nput y = 0, we get x = 10 ………………. (10,0)
    \nand x + 2y = 10 put x = 0 we get y = 5 (0,5)
    \nput y = 0 , we get x = 10 (10,0)
    \nThe equations are graphically represented as below
    \n\"2nd
    \nThe corner points of the feasible region are
    \nA(0,5), B(0,12.5) and C(10,0)
    \nConsider the comer points
    \n\"2nd
    \nSince \u2018Z\u2019 is to minimize which occurs at A(0,5)
    \nThe optimum solution is x = 0, y = 5
    \nThe optimum value of z = 10<\/p>\n

    \"KSEEB<\/p>\n

    Question 35.
    \nA firm is considering of replacing a certain equipment, whose purchasing price is Rs. 5,000, when should we do so?
    \n\"2nd
    \nAnswer:
    \nHere P = Rs. 5,000
    \nLet Ci<\/sub> and Sn<\/sub> be running cost and resale value (Rs.)
    \nAverage annual cost A(n) = \\(\\frac{T}{n}\\)
    \nHer-s T = (p – Sn<\/sub>) + \u03a3Ci<\/sub>
    \n\"2nd
    \nHere the average annual cost A(n) is minimum for n = 4
    \n\u2234 The optimal replacement age is n = 4 years i.e. the machine should replace (after running) for 4th year.<\/p>\n

    Question 36.
    \nGiven n = 4 and R. : 14, 8, 11, 9, 12, 8. find the control limits and construct R-chart.
    \nAnswer:
    \nHere n = 4, k = 6 and standards are not given \/ known.
    \n\\(\\bar{R}=\\frac{\\sum R}{k}=\\frac{62}{6}=10.33\\)
    \nThe control limits for R-chart are
    \nC.L = R\u0304 = 10.33
    \nL.C.L = D3<\/sub>R\u0304 = 0 \u00d7 10.00 = 0
    \nU.C.L = D4<\/sub>R\u0304 = 2.282 \u00d7 10.33 = 23.37
    \nThe R chart is
    \n\"2nd
    \nInterpretation: All points are within control limits, the process in control.<\/p>\n

    Section – D<\/span><\/p>\n

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

    Question 37.
    \n(a) Calculate St.D.R. (B) taking town A\u2019s population as standard.
    \n\"2nd
    \nAnswer:
    \nSince Town A’s population is taken as standard so compute SDR(B)<\/sub> = \\(\\frac{\\sum_{s} P_{s}}{\\sum P_{s}}\\)
    \nB =ADDR’s of B ; Ps<\/sub>=standard popn. of A
    \n\"2nd
    \n\"2nd<\/p>\n

    (b) Calculate CBR, GSR and ASFR (30 – 40) from the following data:
    \n\"2nd
    \nAnswer:
    \n\"2nd
    \n\"2nd
    \n(Note: In the above questions class intervals are not in order)<\/p>\n

    Question 38.
    \nConstruct Dorbish-Bowley and Marshall-Edgeworth’s Quantity index numbers. Comment on the result.
    \n\"2nd
    \nAnswer:
    \nDorbish-Bowley Quantity Index number is
    \n\"2nd
    \nThe quantity level in the year 2005 is 79% of the quantity level in the year 2004.<\/p>\n

    \"KSEEB<\/p>\n

    Question 39.
    \nMonthly income of employees follows Normal distribution with mean Rs. 4,000 and S.D. Rs. 100. Find the probability that a randomly selected employee has income,
    \n(a) > Rs. 3800 (b) > Rs. 4150.
    \nAnswer:
    \nLet x be the monthly income is a Normal variate with mean \u00b5 = 4000 and \u03c3 = 100 then S.N.V is
    \n\"2nd
    \n(a)
    \n\"2nd
    \n= P (z > -2)
    \n= Area from (-2) to \u221e = 0.9772
    \nOR = Area from 0 to (-2) + Area from 0 to \u221e
    \n= 0.4772 + 0.5 = 0.9772
    \n\"2nd<\/p>\n

    (b)
    \n\"2nd
    \n= P (z > 1.5)
    \n= Area from 1.5 to \u221e = 0.0668
    \nOR = Area from 0 to \u221e – Area from 0 to (1.5)
    \n= 0.5-0.4342 = 0.0658
    \n\"2nd<\/p>\n

    Question 40.
    \n(a) Test at 1% level of significance whether there is any significant difference between observed and expected frequencies.
    \n\"2nd
    \nH0<\/sub> : There is no significant difference between observed and expected frequencies.
    \nH1<\/sub> : There is a significant difference between observed and expected frequencies. (Right tail test )
    \nThe test statistic is:-
    \n\"2nd
    \n\u03c72<\/sup>cal <\/sub>= 3.67
    \nAt \u03b1 = 1 % the upper tail critical value for (n – 1) = (4 – 1 )= 3 d.f., k = 11.34
    \nHere \u03c72<\/sup>cal<\/sub> > k
    \n\u2234 H0<\/sub> is accepted.
    \nConclusion: There is no significant difference between observed and expected frequencies.<\/p>\n

    (b) The following data gives time taken to travel from Bangalore to Mysore on 6 random days. Test at 5% the hypothesis that the mean travel time is 160 minutes.
    \n\"2nd
    \nAnswer:
    \nGiven n = 6 , \u03b1 = 5% , \u03bc = 160
    \nH0<\/sub> : Mean travel time is 160 min (i.e. H0<\/sub> : \u03bc= 160)
    \nH1<\/sub> : mean travel time differs from 160 min (H1<\/sub> : \u03bc \u2260 160) (Two Tail test)
    \nThe test statistic is:-
    \n\"2nd
    \nLet x be the time in min.
    \n\"2nd
    \n\"2nd
    \ns = 6.5
    \nX\u0304 = 150 + \\(\\frac{62}{6}\\)
    \nX\u0304 = 0.1135
    \n\"2nd
    \n\"2nd
    \ntcal <\/sub>= 0.1135
    \nAt \u03b1 =5% for (n – 1) = (6 – 1) = 5 d.f. the two Tail critical values are
    \n-k1<\/sub> =-257 and k2<\/sub>= 2.57
    \nHere tcal<\/sub> < -k1<\/sub> \/ tcal<\/sub> is in rejection region
    \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
    \nConclusion: Mean travel time differs from 160 min.<\/p>\n

    Section – E<\/span><\/p>\n

    V. Answer any TWO of the following questions: ( 2 \u00d7 5 = 10 )<\/span><\/p>\n

    Question 41.
    \nFor a Hypergeometric distribution, the parameters are a = 7, b = 5, and n = 8. What is the range? Find the S.D.
    \nAnswer:
    \nFor a hypergeometric distribution:
    \nRange = min (a, n)
    \n= min (7, 8)
    \nRange =0,1,2,3,4, 5, 6, 7.
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 42.
    \nFit a Poisson distribution and obtain the theoretical frequencies :
    \n\"2nd
    \nAnswer:
    \nIf X is a poisson variate the parameter \u03bb is obtained as :
    \n\"2nd
    \n\"2nd
    \nTheoretical frequency : Tx<\/sub> = P(X)N.
    \nT0<\/sub> = P(x = 0) \u00d7 150
    \n\"2nd
    \nT0<\/sub> = 0.1496 \u00d7 150 = 22.44
    \n\"2nd
    \nT6<\/sub> or more = N – [T0<\/sub> + T1<\/sub> + T2<\/sub> + T3<\/sub> + T4<\/sub> + T5<\/sub>] .
    \n= 150 – [22.44 + 42.64 + 40.50 + 25.65 + 12.18 + 4.63]
    \nT0<\/sub> or more = 150- 148.01 = 1.99.
    \nThe distribution of observed and theoretical frequency is : (Approximated)
    \n\"2nd<\/p>\n

    Question 43.
    \nOn an average a box contains 2 defective items. Find the probability that a randomly selected box has
    \n(a) 3 defective items (b) at the most 2 defective items
    \nAnswer:
    \nLet X be the average number of defective items is a Poisson variate with the parameter \u03bb = 2
    \nThe p.m.f. is :
    \n\"2nd<\/p>\n

    Question 44.
    \nA company has to supply 12000 units of a product per year to its customers. The shortage cost is assumed to be infinite. The holding cost is Rs. 0.2 per unit per month and set up cost is Rs. 350. Determine
    \n(a) Number of orders per year.
    \n(b) Minimum total variable cost per month.
    \nAnswer:
    \nR = 12,000; C1<\/sub> = Rs. 0.2 per unit per month
    \n= Rs.0.2 \u00d7 12 = 2.4 per year
    \nC3<\/sub> = Rs. 350 (shortage cost is not given use EOQ model I)
    \n(a) Number of orders per year: \\(n^{0}=\\frac{1}{t^{0}}\\)<\/p>\n

    \"2nd
    \n(b) Minimum total variable cost per month
    \n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

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