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bdkbZ—7% ekè; fopyu ,oa izeki fopyu




            gy (Solution): lekUrj ekè; ,oa izeki fopyu dk ifjdyu                                      uksV

                                     izR;{k jhfr X = 26                     ewY; oxZ jhfr
               Øekad      in ewY;       ls fopyu         fopyuksa osQ oxZ  in&ewY;ksa osQ oxZ
                                                                                 2
                                                               2
                            (X)         (d = X – X )          (d )              (X )
                 1.         26              0                  0                676
                 2.         24             – 2                 4                576
                 3.         29             + 3                 9                841
                 4.         22             – 4                16                484
                 5.         30             + 4                16                900
                 6.         19             – 7                49                361
                 7.         24             – 2                 4                576
                 8.         28             + 2                 4                784
                 9.         28             + 2                 4                784
                10.         30             + 4                16                900

                                                              2
                                                                               2
                ;ksx     ∑X = 260         ∑d = 0           ∑d  = 122         ∑X  = 6882
                                      ∑ X   260
            izR;{k jhfrμ lekUrj ekè; X =   =     = 26
                                       N    10
                                       Σ     X) 2   Σ(X −  d 2  122
                       izeki fopyu σ =           =       =      =  12.2
                                           N         N      10
                                  σ = 3.49
            ewY; oxZ jhfrμizeki fopyu
                                               ∑ F I
                                                                260
                                       ∑ X 2  −G J  2    6882  F I   2
                                                 X
                                                               G J  =  688.2 −
                                  σ =         H K    =        −  H K             676
                                         N      N         10     10
                                  σ =  12.2  = 3.49
            izeki fopyu xq.kkad (Coefficient of Standard Deviation)μizeki fopyu vifdj.k dk ,d fujis{k eki
            (Absolute measure) gSA nks Jsf.k;ksa osQ vifdj.k dh rqyuk djus osQ fy, bldk lkis{k eki (Relative
            measure) fudkyk tkrk gS] ftlosQ fy, izeki fopyu dks lekUrj ekè; ls Hkkx dj fn;k tkrk gSA bls
            vifdj.k dk izeki xq.kkad (Standard Coefficient of Dispersion) vFkok izeki fopyu xq.kkad (Coefficient

            of Standard Deviation) dgrs gSaA vr% lw=kkuqlkjμ
                                                           σ
                                 izeki fopyu xq.kkad (C of SD) =
                                                           X
            3. y?kq jhfr;k¡ (Short-cut Methods)μy?kq jhfr }kjk izeki fopyu Kkr djus dh fuEu izfØ;k gSμ
                 (a) fdlh Hkh ewY; dks vFkok fn;s gq, ewY;ksa esa ls fdlh ,d dks dfYir ekè; (A) eku fy;k tkrk gSA

                 (b) dfYir ekè; ls lHkh ewY;ksa osQ fopyu (dx = X – A) Kkr djosQ mldk ;ksx (∑dx) Kkr dj ysrs
                    gSaA
                                                         2
                 (c) lHkh fopyuksa osQ oxZ djosQ mu oxks± dk ;ksx (∑d x) Kkr dj ysrs gSaA



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