{"id":13380,"date":"2019-12-19T11:13:37","date_gmt":"2019-12-19T05:43:37","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=13380"},"modified":"2021-07-02T15:35:52","modified_gmt":"2021-07-02T10:05:52","slug":"2nd-puc-statistics-question-bank-chapter-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-statistics-question-bank-chapter-4\/","title":{"rendered":"2nd PUC Statistics Question Bank Chapter 4 Interpolation and Extrapolation"},"content":{"rendered":"

You can Download Chapter 4 Interpolation and Extrapolation Questions and Answers, Notes, 2nd PUC Statistics Question Bank with Answers<\/a>\u00a0Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Statistics Question Bank Chapter 4 Interpolation and Extrapolation<\/h2>\n

2nd PUC Statistics Interpolation and Extrapolation One and Two Marks Questions and Answers<\/h3>\n

Question 1.
\nDefine Interpolation and Extrapolation
\nAnswer:
\nInterpolation is the technique of estimating the value dependent variable(Y) for any intermediate ) value of the independent variable(X). I
\nExtrapolation is the technique of estimating the value of dependent variable (Y) any value of independent variable (X) which is outside the given series.<\/p>\n

Question 2.
\nMention different methods of interpolation.
\nAnswer:<\/p>\n

    \n
  • Binomial expansion method and<\/li>\n
  • Newton’s advancing difference method<\/li>\n<\/ul>\n

    Question 3.
    \nWhat is interpolation?
    \nAnswer:
    \nInterpolation is the technique of estimating the value dependent variable( Y) for any intermediate value of the independent variable(X).<\/p>\n

    \"KSEEB<\/p>\n

    Question 4.
    \nWhat is meant by extrapolation?
    \nAnswer:
    \nExtrapolation is the technique of estimating the value of dependent variable (Y) any value of independent variable (X) which is outside the given series.<\/p>\n

    Question 5.
    \nWrite down the assumptions of interpolation and extrapolation.
    \nAnswer:<\/p>\n

      \n
    • The assumptions made in interpolation and extrapolations are:<\/li>\n
    • There are no sudden jumps in the values of dependent variable(Y) from one period to another(X).<\/li>\n
    • The rate of change of figures (Y) from one period to another(X) is uniform.<\/li>\n
    • There will be no consecutive missing values in the series.<\/li>\n<\/ul>\n

      Question 6.
      \nDifferentiate between interpolation and extrapolation.
      \nAnswer:
      \nThe procedure of estimating the missing value of y for a given value of x, where x is within the limits x0 and xn we use Interpolation. Here “Interpolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is within the limits or the range of the independent variable”.<\/p>\n

      But if the value of y is to be estimated for a value of x which is outside the limits x, and x, then procedure Extrapolation is used. Here “Extrapolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is outside the limits or the range of the independent variable”<\/p>\n

      Question 7.
      \nWrite down the conditions for application of Binomial expansion method of interpolation.
      \nAnswer:
      \nFollowing conditions are applied binomial interpolation method:<\/p>\n

        \n
      • The X-variable (independent variable) advances by equal intervals say 15, 20, 25, 30 or say 2, 4, 6, 8, 10 etc.<\/li>\n
      • The value of X for which the value of Y is to be estimated must be one of the values of X.<\/li>\n<\/ul>\n

        Question 8.
        \nWrite down the formula of Binomial expansion method of for 5 and 6 known values of y.
        \nAnswer:
        \n\u22065<\/sup>0<\/sub> or (y – 1)5<\/sup> = y5<\/sub> – 5y4<\/sub> + 10y3<\/sub> – 10y2<\/sub> + 5y1<\/sub> – y0<\/sub> = 0
        \n\u22066<\/sup>0<\/sub> or (y – 1)6<\/sup> = y6<\/sub> – 6y5<\/sub> + 15y4<\/sub> – 20y3<\/sub> + 15y2<\/sub> – 6y1<\/sub> + y0<\/sub> = 0<\/p>\n

        Write the formula to find the value of ‘x’ in finding the missing value of ‘y’ using Newton’s method of interpolation.
        \n\"2nd
        \nWrite down the Newton\u2019s formula for interpolation.
        \n\"2nd
        \nExample 1.
        \nEstimate the production for the year 2010 and 2012 from the following data.
        \n\"2nd
        \nSolution:
        \nLet X and Y be year and production.
        \n\"2nd
        \nSince the known values of ‘Y’ are four, the fourth order leading differences will be zero, i.e. (y- l)4<\/sup> = 0 or \u03944<\/sup> = 0
        \n(y – 1 )4<\/sup> = y4<\/sub> – 4y3<\/sub> + 6y2<\/sub> – 4y1<\/sub> + y0<\/sub> = 0 (i) and the second equation can be obtained by increasing the suffixes of each term of ‘Y’ by one, keeping the. coefficients same;
        \nwe get
        \ny5<\/sub> – 4y4<\/sub> + 6y3<\/sub> – 4y2<\/sub> + y1<\/sub> = 0 ……(ii)
        \nFrom equation (i) 52 – 4y3<\/sub> + 6 (28) – 4(18) + 8 = 0
        \n52 – 4y3<\/sub> + 168 -72 + 8 = 0
        \nSo, by simplifying, – 4y3<\/sub> = – 156
        \n\u2234 y3<\/sub> = 39 ie. Production for the year 2010 is 139 (tons).
        \nFrom (ii) y5<\/sub> – 4y4<\/sub> + 6y3<\/sub> – 4y2<\/sub> + y1<\/sub> = 0; Here y3<\/sub> = 39
        \ny5<\/sub> – 4 x 52 + 6 x (39) – 4 x 28 + 18 = 0
        \ny5<\/sub> – 208 + 234 – 112 + 18 = 0
        \n\u2234 y5<\/sub> = 68 i.e. the production for the year 2012 is 68(tons)<\/p>\n

        \"KSEEB<\/p>\n

        Example 2.
        \nCost of living indices of a Banglore for some years are given below. Interpolate the missing index number for and
        \n\"2nd
        \nSolution:
        \nSince the known values are 5, the fifth leading differences will be zero, i.e. \u03945<\/sup> = 0
        \n\u03945<\/sup> = (y – 1 )5<\/sup> = y5<\/sub> – 5y4<\/sub> +10y3<\/sub> – 10y2<\/sub> + 5y1<\/sub> – y0<\/sub> = 0 ………….(i)
        \nAnd the second equation can be obtained by, increasing the suffixes of each term of’y’ by one, keeping the coefficients same;
        \nie. (y- 1)5<\/sup> = y6<\/sub> – 5y5<\/sub> + 10y4<\/sub> – 10y3 <\/sub>+ 5y2<\/sub> – y, = 0 ………. (ii) determine the value of Y2<\/sub> from equation (i)
        \n162 – 5(142) +10 (128) – 10y2 + 5(112) – 100 = 0
        \nby simplifying, -10y2<\/sub> = -1192
        \ny2<\/sub> = 119.2 Hence, the missing Index number for 1990 is 119.2.
        \nFrom (ii) y6<\/sub> – 5(162) + 10(142) – 10(128) + 5(119.2) – 112 = 0 Here y2<\/sub> =119.2
        \ny6<\/sub> – 810 + 1420 – 1280 + 596- 112 = 0.
        \n\u2234 y6<\/sub> = 186
        \nHence the cost of living index number for the year 2010 is 186.<\/p>\n

        Example 3.
        \nEstimate the number of children in the ages 16 and 22 years from the following data:
        \n\"2nd
        \nSolution:
        \nLet X and Y be age and no. of children.
        \n\"2nd
        \nvalues of y are given, then 5th leading difference will be zero.
        \n\u03945<\/sup>0<\/sub> = (y – 1 )5<\/sup> = y5<\/sub> – 5y4<\/sub> + 10y3<\/sub> – 10y2<\/sub> + 5y1<\/sub> – y0<\/sub> = 0 …. (i) and the second equation can be obtained as, increasing the suffixes of each term of ‘Y’ by one, keeping the coefficients same; we get :
        \n\u03945<\/sup> = (y – 1)5<\/sup> = y6<\/sub> – 5y5<\/sub>+ 10y4<\/sub> – 10y3<\/sub> + 5y2<\/sub> – y1<\/sub> =0 …….(ii)
        \nFrom equation (i)
        \ny5<\/sub> – 5y4<\/sub> + 10y3<\/sub> – 10y2 <\/sub>+ 5y1<\/sub> – y0<\/sub> =0
        \nie\u201e 98 – 5(88) + 10y3 – 10(65) + 5(55) – 50 = 0
        \n98 – 440 + 10y3<\/sub> – 650 + 275 – 50 = 0
        \nSo, by simplifying, 10y3<\/sub> – 767 = 0
        \n\u2234 y3<\/sub> = 76.7 = 77 children.
        \nFrom (ii)
        \ni-e, y6<\/sub> – 5y5<\/sub> + 10y4<\/sub> – 10y3<\/sub> + 5y2<\/sub> – y1<\/sub> = 0;
        \nPut y3<\/sub>= 76.7 we get, y6<\/sub> – 5(98) + 10(88) – 10(76.7) + 5(65) – 55 = 0
        \ni.e, y6<\/sub> – 490 + 880 – 767 + 325 – 55 = 0
        \n\u2234 y6<\/sub> = 107 children.<\/p>\n

        \"KSEEB<\/p>\n

        Example 4.
        \nFind out the missing values in the following the data.
        \n\"2nd
        \nSolution:
        \n\"2nd
        \nSince 4 known values of y are given, then 4th leading difference may be zero.
        \n\u03944<\/sup>0<\/sub> = (y – 1 )4<\/sup> = y4<\/sub> – 4y3<\/sub> + 6y2<\/sub> – 4y1<\/sub> + y0<\/sub> = 0 (i) and the second equation can be obtained
        \nby increasing the suffixes of each term of ‘Y’ by one, keeping the coefficients same; we get:
        \n\u03944<\/sup>1<\/sub> = (y – l)4<\/sup> = y5<\/sub> – 4y4<\/sub> + 6y3<\/sub> – 4y2<\/sub> + y1<\/sub> = 0 …………. (ii)
        \nFrom equation (i) y4<\/sub> – 4y3<\/sub> + 6y2<\/sub> – 4y1<\/sub> + y0<\/sub> = 0
        \nie., 38 – 4(33) + 6y2 – 4(20) + 13 = 0
        \nSo, by simplifying, 38 – 132 + 6y2 – 80 + 13 = 0 ;
        \n6y2<\/sub> – 161
        \nWe get y2<\/sub> =26.83
        \nFrom (ii) y5<\/sub> – 4y4<\/sub> + 6y3<\/sub> – 4y2<\/sub> + y = 0 = 0
        \ny5<\/sub> – 4(38) + 6(33) -4(26.83) + 20 = 0
        \ny5<\/sub> – 152 + 198 – 107.32 + 20 = 0
        \n\u2234 y = 41.32.<\/p>\n

        Example 5.
        \nEstimate probable life expectation of life of an average Indian at the ages 25 and 40.
        \n\"2nd
        \nSolution:
        \nSince the known values are 5, the estimation is based on the expansion of \u03945<\/sup> = (y – 1)5 <\/sup>= 0
        \n\u2234 \u03945<\/sup>= (y – l)5<\/sup> = y5<\/sub> – 5y4<\/sub> + 10y3<\/sub> – 10y2<\/sub> + 5y1<\/sub> – y0<\/sub> = 0
        \nWe have to determine the value of y3<\/sub>;
        \n20.1 – 5(23.1) + 10y3<\/sub> – 10(29.1) + 5(32.2) – 35.4 = 0
        \nSo, by simplifying, 10y3<\/sub> – 260.8 = 0 .
        \n\u2234 y3<\/sub> = 26.08 years.
        \nHence, the probable expectation of life at the age 25 is 26.08 years.
        \nNow expand (y – l)5 = 0 with change of subscript, keeping coefficients as it is.
        \ny6<\/sub> – 5y5<\/sub> + 10y4<\/sub> – 10y3<\/sub> + 5y2<\/sub> – y1<\/sub> = 0
        \ny6<\/sub> – 5(20.1) + 10(23.1) – 10(26.08) + 5(29.1)-32.2 = 0
        \ny6<\/sub>– 17 = 0, y6<\/sub> = 17 years.<\/p>\n

        \"KSEEB<\/p>\n

        Newton’s Method Problems:<\/p>\n

        Example 1.
        \nThe following data given below estimate the number of persons living in the age of 25 years.
        \n\"2nd
        \nSolution:
        \nSince, the value to be estimate does not one of the data(X) So, apply Newton’s method to estimate the number of persons living in the age 15 years.
        \nSince there are 4 known values of y, Prepare the leading difference until \u03943<\/sup>
        \n\"2nd
        \nThe first differences in each column are called leading differences.
        \n\"2nd
        \nThe number of persons at the age 15 years = 132.375 Example 2.<\/p>\n

        \"KSEEB<\/p>\n

        Example 2.
        \nBelow are given the wages earned by workers per day in a certain factory. Calculate the number of emloyees earning up to Rs. 450 per day.<\/p>\n\n\n\n\n\n\n\n\n\n
        Daily wages<\/td>\nNo.of Employees<\/td>\n<\/tr>\n
        Up to Rs.300<\/td>\n50<\/td>\n<\/tr>\n
        Up to Rs.400<\/td>\n150<\/td>\n<\/tr>\n
        Up to Rs.500<\/td>\n300<\/td>\n<\/tr>\n
        Up to Rs.600<\/td>\n500<\/td>\n<\/tr>\n
        Up to Rs.700<\/td>\n700<\/td>\n<\/tr>\n
        Up to Rs.800<\/td>\n800<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Solution:
        \nApplying Newton’s method to ascertain number of workers earning upto Rs. 450 per day. Since there are 6 known values of y, so prepare the leading difference table as below:
        \n\"2nd
        \n\"2nd
        \nyup to 450<\/sub> = =50 + 150 + 18.75 + 0 – 1.1718 = 217.58
        \nThus the number of workers earning up to Rs. 450 is 218.<\/p>\n

        Example 3.
        \nFrom the following table find out the number of workers earning salary less than Rs.25(000’s)
        \n\"2nd
        \nSolution:
        \nConvert the data into less than frequency table and find the difference table.
        \nSince there are 6 known values of y, Prepare the leading difference until 5th leading table as below:
        \n\"2nd
        \nSince, the value to be estimate is not in class limits of the data(X) so, apply Newton\u2019s method to estimate the number of workers who are drawing Less than Rs,25(000’s) Salary.
        \n\"2nd
        \nY(Less than 25) = 125 + 100 – 6.25 – 7 – 3.3594 – 0.2461
        \nThe number of workers who are drawing Less than Rs. 25(000’s) = 208.14 = 208.<\/p>\n

        \"KSEEB<\/p>\n

        Example 4.
        \nFrom the following data interpolate the value for X = 8.
        \n\"2nd<\/p>\n

        Solution:
        \nSince, the value to be estimate is not one of the value of X series so, apply Newton’s method to estimate the value at X = 8. Since there are 5 known values of y, Prepare the leading difference until 4th leading table as below:
        \n\"2nd
        \nThe first differences in each column are called Leading Differences.
        \n\"2nd
        \nY8<\/sub> = 19 +7.5 + 0.875 + 0.0625 – 0.1953;
        \n\u2234 Y8<\/sub> = 27.24.
        \nThe value of Y at X = 8 is 27.24.<\/p>\n","protected":false},"excerpt":{"rendered":"

        You can Download Chapter 4 Interpolation and Extrapolation Questions and Answers, Notes, 2nd PUC Statistics Question Bank with Answers\u00a0Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Statistics Question Bank Chapter 4 Interpolation and Extrapolation 2nd PUC Statistics Interpolation and Extrapolation One and Two …<\/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\/13380"}],"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=13380"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13380\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=13380"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=13380"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=13380"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}