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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
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 arithmetic mean and standard deviation when we combine the results of mathematics for all classes of form four. Assuming the number of students of 4a,4b and 4c are 30,45,25.
$$\mathrm{There}\:\mathrm{are}\:\mathrm{three}\:\mathrm{classes}\:\mathrm{of}\:\mathrm{form}\:\mathrm{four} \\ $$$$\mathrm{which}\:\mathrm{are}\:\mathrm{4a},\:\mathrm{4b}\:\mathrm{and}\:\mathrm{4c}. \\ $$$$\mathrm{in}\:\mathrm{mathematics}\:\mathrm{result}\:\mathrm{the}\:\mathrm{arithmetic}\:\mathrm{mean}\:\mathrm{of} \\ $$$$\mathrm{4a}\:\mathrm{4b}\:\mathrm{and}\:\mathrm{4c}\:\mathrm{are}\:\mathrm{70},\:\mathrm{60}\:\mathrm{and}\:\mathrm{80}\:\mathrm{and}\:\mathrm{their} \\ $$$$\mathrm{standard}\:\mathrm{deviation}\:\mathrm{are}\:\mathrm{3},\:\mathrm{3}.\mathrm{4},\:\mathrm{and}\:\mathrm{2}.\mathrm{5} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{arithmetic}\:\mathrm{mean}\:\mathrm{and}\:\mathrm{standard}\:\mathrm{deviation}\:\mathrm{when} \\ $$$$\mathrm{we}\:\mathrm{combine}\:\mathrm{the}\:\mathrm{results}\:\mathrm{of}\:\mathrm{mathematics}\:\mathrm{for}\: \\ $$$$\mathrm{all}\:\mathrm{classes}\:\mathrm{of}\:\mathrm{form}\:\mathrm{four}.\:\mathrm{Assuming}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{students}\:\mathrm{of}\: \\ $$$$\mathrm{4a},\mathrm{4b}\:\mathrm{and}\:\mathrm{4c}\:\mathrm{are}\:\mathrm{30},\mathrm{45},\mathrm{25}. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18
A.M=((n_1 x_1 +n_2 x_2 +n_3 x_3 )/(n_1 +n_2 +n_3 )) =((30×70+45×60+25×80)/(30+45+25)) =((2100+2700+2000)/(100))=68
$${A}.{M}=\frac{{n}_{\mathrm{1}} {x}_{\mathrm{1}} +{n}_{\mathrm{2}} {x}_{\mathrm{2}} +{n}_{\mathrm{3}} {x}_{\mathrm{3}} }{{n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{30}×\mathrm{70}+\mathrm{45}×\mathrm{60}+\mathrm{25}×\mathrm{80}}{\mathrm{30}+\mathrm{45}+\mathrm{25}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2100}+\mathrm{2700}+\mathrm{2000}}{\mathrm{100}}=\mathrm{68} \\ $$
Commented by mondodotto@gmail.com last updated on 22/Oct/18
sir we need to find arithmetic meam and standard deviation
$$\mathrm{sir}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:\mathrm{arithmetic}\:\mathrm{meam} \\ $$$$\mathrm{and}\:\mathrm{standard}\:\mathrm{deviation}\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18
pls check...
$${pls}\:{check}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18
N_1 =30 N_2 =45 N_3 =25 X_1 ^− =70 X_2 ^− =60 X_3 ^− =80 X_(123) ^− =((N_1 X_1 ^− +N_2 X_2 ^− +N_3 X_3 ^− )/(N_1 +N_2 +N_3 )) =((30×70+45×60+25×80)/(30+45+25)) =((6800)/(100))=68 d_1 =X_1 ^− −X_(123) ^− =70−68=2 d_2 =60−68=−8 d_3 =80−68=12 σ_1 =3 σ_2 =3.4 σ_3 =2.5 formula σ_(123) =(√((N_1 σ_1 ^2 +N_2 σ_2 ^2 +N_3 σ_3 ^2 +N_1 d_1 ^2 +N_2 d_2 ^2 +N_3 d_3 ^2 )/(N_1 +N_2 +N_3 ))) pls put value and calculate by your self...
$${N}_{\mathrm{1}} =\mathrm{30}\:\:{N}_{\mathrm{2}} =\mathrm{45}\:\:\:{N}_{\mathrm{3}} =\mathrm{25} \\ $$$$\overset{−} {{X}}_{\mathrm{1}} =\mathrm{70}\:\:\:\overset{−} {{X}}_{\mathrm{2}} =\mathrm{60}\:\:\:\overset{−} {{X}}_{\mathrm{3}} =\mathrm{80} \\ $$$$\overset{−} {{X}}_{\mathrm{123}} =\frac{{N}_{\mathrm{1}} \overset{−} {{X}}_{\mathrm{1}} +{N}_{\mathrm{2}} \overset{−} {{X}}_{\mathrm{2}} +{N}_{\mathrm{3}} \overset{−} {{X}}_{\mathrm{3}} }{{N}_{\mathrm{1}} +{N}_{\mathrm{2}} +{N}_{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{30}×\mathrm{70}+\mathrm{45}×\mathrm{60}+\mathrm{25}×\mathrm{80}}{\mathrm{30}+\mathrm{45}+\mathrm{25}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{6800}}{\mathrm{100}}=\mathrm{68} \\ $$$${d}_{\mathrm{1}} =\overset{−} {{X}}_{\mathrm{1}} −\overset{−} {{X}}_{\mathrm{123}} =\mathrm{70}−\mathrm{68}=\mathrm{2} \\ $$$${d}_{\mathrm{2}} =\mathrm{60}−\mathrm{68}=−\mathrm{8} \\ $$$${d}_{\mathrm{3}} =\mathrm{80}−\mathrm{68}=\mathrm{12} \\ $$$$\sigma_{\mathrm{1}} =\mathrm{3}\:\:\:\sigma_{\mathrm{2}} =\mathrm{3}.\mathrm{4}\:\:\:\sigma_{\mathrm{3}} =\mathrm{2}.\mathrm{5} \\ $$$${formula} \\ $$$$\sigma_{\mathrm{123}} \\ $$$$=\sqrt{\frac{{N}_{\mathrm{1}} \sigma_{\mathrm{1}} ^{\mathrm{2}} +{N}_{\mathrm{2}} \sigma_{\mathrm{2}} ^{\mathrm{2}} +{N}_{\mathrm{3}} \sigma_{\mathrm{3}} ^{\mathrm{2}} +{N}_{\mathrm{1}} {d}_{\mathrm{1}} ^{\mathrm{2}} +{N}_{\mathrm{2}} {d}_{\mathrm{2}} ^{\mathrm{2}} +{N}_{\mathrm{3}} {d}_{\mathrm{3}} ^{\mathrm{2}} }{{N}_{\mathrm{1}} +{N}_{\mathrm{2}} +{N}_{\mathrm{3}} }}\: \\ $$$${pls}\:{put}\:{value}\:{and}\:{calculate}\:{by}\:{your}\:{self}… \\ $$
Commented by mondodotto@gmail.com last updated on 22/Oct/18
oky thanx
$$\mathrm{oky}\:\mathrm{thanx} \\ $$

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