{"id":12312,"date":"2019-12-16T16:29:07","date_gmt":"2019-12-16T10:59:07","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=12312"},"modified":"2021-07-02T15:36:36","modified_gmt":"2021-07-02T10:06:36","slug":"1st-puc-statistics-question-bank-chapter-8","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-statistics-question-bank-chapter-8\/","title":{"rendered":"1st PUC Statistics Question Bank Chapter 8 Interpolation and Extrapolation"},"content":{"rendered":"

You can Download Chapter 8 Interpolation and Extrapolation Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Statistics Question Bank Chapter 8 Interpolation and Extrapolation<\/h2>\n

1st PUC Interpolation and Extrapolation Question and Answers<\/h3>\n

Question 1.
\nWhat is Interpolation?
\nAnswer:
\n\u2018Interpolation 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\u2019<\/p>\n

Question 2.
\nWhat is extrapolation?
\nAnswer:
\nExtrapolation 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\u2019<\/p>\n

Question 3.
\nMention the situations where the technique of interpolation is used.
\nAnswer:
\nThe procedure of estimating the missing value of y for a given value of x, where x is within the limits x0<\/sub> and xn<\/sub> we use the technique Interpolation.<\/p>\n

Question 4.
\nWhat are the assumptions made in interpolation?
\nAnswer:
\nIn making use of the techniques of interpolation the following assumptions are made<\/p>\n

    \n
  • There are no sudden jumps in the values of independent variable from one period to another.<\/li>\n
  • The rate of change of figures from one period to another is uniform.<\/li>\n<\/ul>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nUsing Binomial expansion method expand (y – 1)5 = 0.
    \nAnswer:
    \nThe equation (y – 1 )5<\/sup> = 0 is:
    \ny5<\/sub> – 5y-4<\/sub> + 10y3<\/sub> -10y2<\/sub> + 5y1<\/sub> -y0<\/sub> = 0<\/p>\n

    Question 6.
    \nIf in a Binomial expansion y4<\/sub> – 4y3<\/sub> + 6y2<\/sub> – 4y1<\/sub> + y0<\/sub> = 0, if y3<\/sub>=157, y2<\/sub>=124, y1<\/sub> = 107 and\u00a0 y0<\/sub>=100, then find y4<\/sub>
    \nAnswer:
    \nSubstituting the known values in the equation, we obtain,
    \ny4 <\/sub>– 4(157) + 6(124) – 4(107)+ 100 = 0
    \ny4<\/sub> – 1056 + 844= 0
    \ny4 <\/sub>– 212= 0
    \n\u2234 y4<\/sub> = 212<\/p>\n

    Question 7.
    \nDistinguish 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<\/sub> and xn<\/sub> we use Interpolation. Here \u201cInterpolation 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\u201d<\/p>\n

    But if the value of y is to be estimated for a value of x which is outside the limits x0<\/sub> and xn<\/sub> then procedure Extrapolation is used. \u201cExtrapolation 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\u201d<\/p>\n

    Question 8.
    \nInterpolate the export of handlooms during 2008 from the following data
    \n\"1st
    \nAnswer:
    \nLet X and Y be years and Exports<\/p>\n\n\n\n\n\n\n\n\n\n\n
    (years)X<\/td>\n(Export in crores Rs. ) Y<\/td>\n<\/tr>\n
    1998 (x0<\/sub>)<\/td>\n10 (y0<\/sub>)<\/td>\n<\/tr>\n
    2000 (x1<\/sub>)<\/td>\n13 (y1<\/sub>)<\/td>\n<\/tr>\n
    2002 (x2<\/sub>)<\/td>\n15 (y2<\/sub>)<\/td>\n<\/tr>\n
    2004 (x3<\/sub>)<\/td>\n23 (y3<\/sub>) 18<\/td>\n<\/tr>\n
    2006 (x4<\/sub>)<\/td>\n26 (y4<\/sub>) 23<\/td>\n<\/tr>\n
    2008 (x5<\/sub>)<\/td>\n\u00a0– (y5<\/sub>)<\/td>\n<\/tr>\n
    2010 (x6<\/sub>)<\/td>\n32 (y6<\/sub>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Using binomial expansion method: Here we have to estimate y5<\/sub> Since there are 6 known values of y, the formula of estimation is based on the expansion of (y-1)6<\/sub>= 0
    \nWe get the equation,
    \ny6 <\/sub>– 6y5 <\/sub>+ 15y4 <\/sub>– 20y3 <\/sub>+ 15y2<\/sub> – 6y1<\/sub> + y0<\/sub>= 0
    \nSubstituting the known values in the equation, we get,
    \ni.e.32 – 6y5<\/sub>+ 15 x 23 -20 x 18 + 15 x15 -6 x 13 + 10 = 0
    \ni.e.32 – 6y5<\/sub>+ 345 – 360 + 225 – 78 + 10 = 0 ;
    \n-6y5<\/sub> = -174
    \n\u21d2 y5<\/sub> = 29
    \nHence the probable export for the year 2008 is 29 (crores Rs.)<\/p>\n

    \"KSEEB<\/p>\n

    Question 9.
    \nInterpolate the missing figure.
    \n\"1st
    \nAnswer:
    \nLet X and Y be years and sales.<\/p>\n\n\n\n\n\n\n\n\n
    (years)X<\/td>\n(sales ) Y<\/td>\n<\/tr>\n
    2001 (x0<\/sub>)<\/td>\n100 (y0<\/sub>)<\/td>\n<\/tr>\n
    2002 (x1<\/sub>)<\/td>\n120 (y1<\/sub>)<\/td>\n<\/tr>\n
    2003 (x2<\/sub>)<\/td>\n(y2<\/sub>)<\/td>\n<\/tr>\n
    2004 (x3<\/sub>)<\/td>\n180 (y3<\/sub>)<\/td>\n<\/tr>\n
    2005 (x4<\/sub>)<\/td>\n210 (y4<\/sub>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Here we have to estimate y2 Since there are 4 known values of y, the formula of estimation is based on the expansion of (y – 1)4<\/sup>= 0
    \nWe get the equation,
    \nY4 <\/sub>-4 y3 <\/sub>+ 6y2 <\/sub>– 4y1<\/sub> + y0<\/sub>=o
    \nSubstituting the known values in the equation, we obtain,
    \ni.e.210 – 4 x 180 + 6y2<\/sub> – 4 x 120 +100 = 0
    \ni.e.210 – 720 + 6y2 <\/sub>– 480 +100 = 0;
    \n6y5<\/sub>= 890
    \n\u21d2 y2<\/sub> = 148.33
    \nHence the probable sales for the 2003 is 148.33<\/p>\n

    Question 10.
    \nInterpolate the index for 2008 from the following data.
    \n\"1st
    \nAnswer:
    \nUsing Binomial expansion method – let X and Y be year and Index number.<\/p>\n\n\n\n\n\n\n\n\n
    (Years) x<\/td>\n(Index No.) y<\/td>\n<\/tr>\n
    2006 (x0<\/sub>)<\/td>\n278 (y0<\/sub>)<\/td>\n<\/tr>\n
    2007 (x1<\/sub>)<\/td>\n281 (y1<\/sub>)<\/td>\n<\/tr>\n
    2008 (x2<\/sub>)<\/td>\n– (y2<\/sub>)<\/td>\n<\/tr>\n
    2009 (x3<\/sub>)<\/td>\n313 (y3<\/sub>)<\/td>\n<\/tr>\n
    2010 (x4<\/sub>)<\/td>\n322 (y4<\/sub>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    We have to estimate y2<\/sub> Since there are 4 known values of y, the formula of estimation is \u2022 based on the expansion of (y – 1)4<\/sup>=0
    \nWe get the equation,
    \ny4 <\/sub>– 4y3 <\/sub>+ 6y2 <\/sub>– 4y1<\/sub> yo<\/sub> = 0
    \nSubstituting the known values in the equation, we get,
    \ni.e. 322 – 4(313) + 6y2 <\/sub>– 4(281) + 278 = 0
    \ni.e. 322 – 1252 + 6y2 <\/sub>– 1124 + 278 = 0
    \n6y2<\/sub>= 1776
    \n\u21d2 y2<\/sub> = 296
    \nHence the probable value of the index number of 2008 is 296.<\/p>\n

    Question 11.
    \nFrom the following data interpolate the production of cement in 2007.
    \n\"1st
    \nAnswer:
    \nLet X and Y be the year and production<\/p>\n\n\n\n\n\n\n\n\n\n
    X<\/td>\nY<\/td>\n<\/tr>\n
    2005 (x0<\/sub>)<\/td>\n44 (y0<\/sub>)<\/td>\n<\/tr>\n
    2006 (x1<\/sub>)<\/td>\n90 (y1<\/sub>)<\/td>\n<\/tr>\n
    2007 (x2<\/sub>)<\/td>\n(y2<\/sub>)<\/td>\n<\/tr>\n
    2008 (x3<\/sub>)<\/td>\n160 (y3<\/sub>)<\/td>\n<\/tr>\n
    2009 (x4<\/sub>)<\/td>\n210 (y4<\/sub>)<\/td>\n<\/tr>\n
    2010 (x5<\/sub>)<\/td>\n290 (y5<\/sub>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Using Binomial expansion method We have to estimate y2<\/sub> Since there are 5 known values of y, the formula of estimation is, the expansion of (y-1)5<\/sup> = 0 We get the equation,
    \nY5<\/sub> – 5y4 <\/sub>+ 10y3<\/sub> – 10y2 <\/sub>+ 5y1<\/sub> – y0<\/sub>= 0.
    \nSubstituting the known values in the equation we obtain,
    \n290 – 5(210)+ 10 x 160 – 10y2<\/sub>+ 5(90) – 44 =0
    \n290 – 1050+ 1600 – 10y2<\/sub> + 450 – 44 =0
    \n-10y2 <\/sub>– 1246 =0
    \n– 10y2<\/sub> =1246
    \n\u2234 y2<\/sub> = \\(\\frac { 1246 }{ 10 }\\)
    \n\u21d2 y2<\/sub> = 127 is 124.6.<\/p>\n

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

    You can Download Chapter 8 Interpolation and Extrapolation Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Statistics Question Bank Chapter 8 Interpolation and Extrapolation 1st PUC Interpolation and Extrapolation Question and Answers …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12312"}],"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=12312"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12312\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=12312"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=12312"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=12312"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}