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bdkbZμ5 % ekè;] ekfè;dk rFkk cgqyd osQ vuqiz;ksx




            fd;k tkrk gSA                                                                             uksV
            fof/&(i)μ loZizFke izR;sd fopyu ;k in&fopyu eas 1 tksM+dj (dx +1) vFkok (d'x+1) Kkr dj fy;k tkrk
            gSA
            (ii) (dx + 1) ;k (d'x+1) esa mudh vko`fÙk;ksa dh xq.kk djosQ xq.kuiQyksa dk ;ksx Σ[f(dx+1] vFkok Σ[f(d'x+1)]Kkr
            dj fy;k tkrk gSA
            (iii) rRi'pkr~ fuEu lehdj.kksa dk iz;ksx fd;k tkrk gSμ
            Σfdx = Σ[f(dx+1)]–Σf    y?kq jhfr dk iz;ksx djus ij

            Σfd'x = Σ[f(d'x+1)]–Σf  in&fopyu jhfr dk iz;ksx djus ij
            (iv) ;fn mi;ZqDr lehdj.k osQ nksuksa i{k cjkcj gSa] rks le> ysuk pkfg, fd x.ku&fØ;k 'kq¼ gS vU;Fkk dksbZ
            =kqfV gks xbZ gSA






                    lekos'kh oxkZUrj fdls dgrs gSa\

            mnkgj.k (Illustration) 6: fuEu lkj.kh ls ekè; fudkfy,
              oxZ:    μ10     10μ20    20μ30     30μ40 40μ50     50μ60  60μ70 70μ80
              vko`fÙk :  22     38       54        75     72      64      31     10

            gy (Solution) : pw¡dh iwjh lead&ekyk esa oxZ&foLrkj (10) leku gSA vr% vfUre oxZ dh mQijh lhek 70
            + 10 = 80 ekuh tk;sxh vFkkZr~ ;g oxZ 70μ80 gksxk vkSj igyk oxZ 0μ10 gksxkA

             X            M.V.      f          d'x = A–35/10   fd'x       (d'x+1)      f(d'x+1)
             0–10     5          22       – 3            – 66     – 2       – 44
             10–20    15         38       – 2            – 76     – 1       – 38
             20–30    25         54       – 1            – 54       0           0
             30–40    35         75         0                0    +1        + 75
             40–50    45         72      + 1             + 72     +2        + 144
             50–60    55         64      + 2             +128     +3        + 192
             60–70    65         31      + 3             + 93     +4        + 124

             70–80    75         10       + 4            + 40     +5        +  50
                                 366                      137                 503

            [N = Σf = 366, A = 35, Σfd'x = 137]                         [Σf (d'x +1)= 503]
                   Σfd' x       137
                                            +
              =X  + A   ×  = i  35+  ×  10 =  35 3 74 = .  38 74
                                                     .
                     N          366
            pkfyZ;j tk¡p lw=k dk iz;ksx djs ijμ
            Σfd'x = Σ[f(d'x + 1)] – Σf or 137 = 503 – 366
            ∴ 137 = 137 vr% x.kuk&fØ;k esa dksbZ v'kqf¼ ugha gSA

            (7) vKkr ewY; ;k vko`fÙk dk fu/kZj.k (Location of Missing Size of Frequency)μ lekUrj ekè; dh
            ,d egRoiw.kZ fo'ks"krk ;g gS fd ;fn fdlh Js.kh osQ bu rhu ekuksa  X,N  vkSj ΣX esa dksbZ ls nks eku Kkr


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