There-are-three-classes-of-form-four-which-are-4a-4b-and-4c-in-mathematics-result-the-arithmetic-mean-of-4a-4b-and-4c-are-70-60-and-80-and-their-standard-deviation-are-3-3-4-and-2-5-Find-the-arit

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Question Number 46191 by mondodotto@gmail.com last updated on 22/Oct/18

$$\mathrm{There}\mathrm{are}\mathrm{three}\mathrm{classes}\mathrm{of}\mathrm{form}\mathrm{four}$$$$\mathrm{which}\mathrm{are}4\mathrm{a},4\mathrm{b}\mathrm{and}4\mathrm{c}.$$$$\mathrm{in}\mathrm{mathematics}\mathrm{result}\mathrm{the}\mathrm{arithmetic}\mathrm{mean}\mathrm{of}$$$$4\mathrm{a}4\mathrm{b}\mathrm{and}4\mathrm{c}\mathrm{are}70,60\mathrm{and}80\mathrm{and}\mathrm{their}$$$$\mathrm{standard}\mathrm{deviation}\mathrm{are}3,3.4,\mathrm{and}2.5$$$$\mathrm{Find}\mathrm{the}\mathrm{arithmetic}\mathrm{mean}\mathrm{and}\mathrm{standard}\mathrm{deviation}\mathrm{when}$$$$\mathrm{we}\mathrm{combine}\mathrm{the}\mathrm{results}\mathrm{of}\mathrm{mathematics}\mathrm{for}$$$$\mathrm{all}\mathrm{classes}\mathrm{of}\mathrm{form}\mathrm{four}.\mathrm{Assuming}\mathrm{the}$$$$\mathrm{number}\mathrm{of}\mathrm{students}\mathrm{of}$$$$4\mathrm{a},4\mathrm{b}\mathrm{and}4\mathrm{c}\mathrm{are}30,45,25.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18

$$A.M=\frac{n_1{x}_1+n_2{x}_2+n_3{x}_3}{n_1+n_2+n_3}$$$$=\frac{30\times 70+45\times 60+25\times 80}{30+45+25}$$$$=\frac{2100+2700+2000}{100}=68$$

Commented by mondodotto@gmail.com last updated on 22/Oct/18

$$\mathrm{sir}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{arithmetic}\mathrm{meam}$$$$\mathrm{and}\mathrm{standard}\mathrm{deviation}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18

$$plscheck\dots$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18

$$N_1=30N_2=45N_3=25$$$${\stackrel{-}{X}}_1=70{\stackrel{-}{X}}_2=60{\stackrel{-}{X}}_3=80$$$${\stackrel{-}{X}}_{123}=\frac{N_1{\stackrel{-}{X}}_1+N_2{\stackrel{-}{X}}_2+N_3{\stackrel{-}{X}}_3}{N_1+N_2+N_3}$$$$=\frac{30\times 70+45\times 60+25\times 80}{30+45+25}$$$$=\frac{6800}{100}=68$$$$d_1={\stackrel{-}{X}}_1-{\stackrel{-}{X}}_{123}=70-68=2$$$$d_2=60-68=-8$$$$d_3=80-68=12$$$${\sigma }_1=3{\sigma }_2=3.4{\sigma }_3=2.5$$$$formula$$$${\sigma }_{123}$$$$=\sqrt{\frac{N_1{\sigma }_1^2+N_2{\sigma }_2^2+N_3{\sigma }_3^2+N_1{d}_1^2+N_2{d}_2^2+N_3{d}_3^2}{N_1+N_2+N_3}}$$$$plsputvalueandcalculatebyyourself\dots$$

Commented by mondodotto@gmail.com last updated on 22/Oct/18

$$\mathrm{oky}\mathrm{thanx}$$

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