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bdkbZμ6: vifdj.k% vFkZ ,oa fo'ks"krk,¡] vifdj.k osQ lkis{k ,oa fujis{k eki] jsat] prqFkZd fopyu ,oa 'kred foLrkj




                                                 Q − Q 1                                              uksV
                                                  3
                                                    2      Q − Q 1
                                                            3
                            prqFkZd fopyu xq.kkad =      =
                                                 Q + Q 1   Q + Q 1
                                                  3
                                                            3
                                                    2
            mnkgj.k (Illustration) 5: fuEu nks Jsf.k;ksa esa vifdj.k dh rqyuk prqFkZd ekiksa }kjk dhft,μ
              Å¡pkbZ (bap esa) %  58  56  62   61    63    64   65    59    62    65    55
              Hkkj (ikS.M esa) %  117  112  127  123  125  130  106   119   121  132   108
            gy (Solution):

            prqFkZd ekiksa }kjk vifdj.k dh rqyuk osQ fy, prqFkZd fopyu xq.kkad Kkr fd;k tk,xkμ
             Øe %     1     2      3     4      5     6     7      8     9     10     11

             Å¡pkbZ %  55   56    58     59    61     62    62    63     64    65     65
             Hkkj %  106    108   112   117    119   121    123   125   127    130   132
                            (A) Å¡pkbZ                              (B) Hkkj
                    Q  = Size of G F  N + I 1 J  item  Q  = Size of G F  N + I 1 J  item
                                     th
                                                                         th
                     1       H  4 K                     1        H  4 K
                      = Size of G F H 11 + I 1 J  or 3  item   = Size of G F H 11 + I 1 J  or 3  item
                                 4 K
                                                             4 K
                                                                     rd
                                         rd
                   Q  = 58                     Q  = 112
                     1                          1
                              3(N + 1)                           3(N + 1)
                                                                         th
                   Q  = Size of         item          Q  = Size of         item
                                      th
                     3           4                      3           4
                             3(11 + 1)                    3(11 + 1)
                       = Size of      = 9  item   = Size of        = 9  item
                                                                     th
                                        th
                                 4                           4
                   Q  = 64                            Q  = 127
                   3                                   3
                                     Q − Q                                 Q −  Q
                prqFkZd fopyu xq.kkad =   3  1         prqFkZd fopyu xq.kkad =   3  1
                                     Q + Q 1                               Q +  Q 1
                                      3
                                                                             3
                     64 − 58   6                            127 −  112  15
                   =        =     = .049                  =          =     = .063
                     64 + 58  122                           127 +  112  239
            vr% Hkkj esa Å¡pkbZ dh vis{kk vf/d vifdj.k gSA
            mnkgj.k (Illustration) 6: fuEu caVu ls vifdj.k osQ prqFkZd xq.kkad dk ifjdyu dhft,μ

              in dk osQUnzh; eku%  1    2     3     4     5     6     7     8     9    10
              vko`fÙk        %    2     9    11    14    20    24    20    16     5     2

            gy (Solution): fn;s x;s eè;&fcUnq osQ vk/kj ij oxkZUrj Kkr fd;s tk;saxsA eè; fcUnq dk vUrj 1 gSA
            vr% 0.5 eè; ewY; esa ?kVkus o tksM+us ij fupyh o Åijh lhek izkIr gks tk,xhA









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