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vFkZ'kkL=k esa lkaf[;dh; fof/;k¡
uksV 3. v[kf.Mr Js.kh esaμblosQ ifjdyu dh nks jhfr;k¡ gSaμ
(a) U;wure oxZ dh fupyh lhek dks U;wure ewY; vkSj vf/dre oxZ dh Åijh lhek dks vf/
dre ewY; ekuk tkrk gSA
(b) mPpre oxkZUrj osQ eè; ewY; dks vf/dre ,oa U;wure oxkZUrj osQ eè; ewY; dks U;wure
ewY; eku ysrs gSaA
R = L – S
fVIi.khμ(1) [kqys flj okys oxkZUrjksa esa foLrkj Kkr ugha fd;k tk ldrkA
(2) lekos'kh oxkZUrj dks igys viothZ oxkZUrjksa esa cny ysrs gSaA
foLrkj xq.kkadμfoLrkj vifdj.k dk ,d fujis{k eki gS tks rqyukRed vè;;u osQ fy, vuqi;qDr gS] vr%
rqyuk djus osQ fy, foLrkj dk lkis{k eki Kkr fd;k tkrk gSA ;fn foLrkj dks pje inksa osQ ;ksx ls foHkkftr
dj fn;k tk, rks mls foLrkj xq.kkad ;k ijkl xq.kkad dgrs gSaA
L − S
foLrkj xq.kkad (CR) =
L + S
mnkgj.k (Illustration) 1: fuEu lead Js.kh dk foLrkj vkSj mldk xq.kkad Kkr dhft,μ
20, 35, 25, 30, 16, 14, 13, 28, 38, 40, 10
gy (Solution):
S = 10, L = 40
foLrkj = L – S = 40 – 10 = 30
L − S 40 − 10 30
foLrkj xq.kkad = = = 0.6
L + S 40 + 10 50
mnkgj.k (Illustration) 2: fuEu caVu esa foLrkj o mlosQ xq.kkad dk ifjdyu dhft,μ
oxZ% 0–5 5–10 10–15 15–20 20–25 25–30
vko`fÙk% 4 8 5 15 10 11
gy (Solution):
I. L = 30, S = 0 II. L = 27.5 S = 2.5
foLrkj L – S = 30 – 0 = 30 foLrkj = L – S = 27.5 – 2.5 = 25
L − S 30 − 0 L − S 27.5 − 2.5 25
CR = = = 1 CR = = =
L + S 30 + 0 L + S 27.5 + 2.5 30
CR = 0.83
mnkgj.k (Illustration) 3: fuEu caVu ls foLrkj xq.kkad Kkr dhft,μ
oxZ% 1—5 6—10 11—15 16—20 21—25 26—30
vko`fÙk% 8 7 6 15 11 9
gy (Solution) : lekos'kh oxkZUrjksa dks igys viothZ oxkZUrjksa esa cny fy;k tkrk gSA mDr leadksa esa U;wure
lhek 0.5 o vfèkdre lhek 30.5 gSA
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