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Question-152903

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Question Number 152903 by DELETED last updated on 03/Sep/21
Answered by DELETED last updated on 03/Sep/21
1.a). x^− = ((Σf_i .x_i )/(Σf_i )) =((2×14,5+9×18,5+12×22,5+9×26,5+5×30,5+3×34,5)/(2+9+12+9+5+3)) =((29+166,5+270+238,5+152,5+103,5)/(40)) =((195,5+508,5+256)/(40))=((960)/(40))=24 1.b). median=l_i +((((N/2) − <Σf_i ))/f_i ) ×C med=20,5 + (((20−11))/(12))×4 =20,5 + 3=23,5//
$$\left.\mathrm{1}.\mathrm{a}\right).\:\overset{−} {\mathrm{x}}\:=\:\frac{\Sigma\mathrm{f}_{\mathrm{i}} .\mathrm{x}_{\mathrm{i}} }{\Sigma\mathrm{f}_{\mathrm{i}} }\: \\ $$$$\:=\frac{\mathrm{2}×\mathrm{14},\mathrm{5}+\mathrm{9}×\mathrm{18},\mathrm{5}+\mathrm{12}×\mathrm{22},\mathrm{5}+\mathrm{9}×\mathrm{26},\mathrm{5}+\mathrm{5}×\mathrm{30},\mathrm{5}+\mathrm{3}×\mathrm{34},\mathrm{5}}{\mathrm{2}+\mathrm{9}+\mathrm{12}+\mathrm{9}+\mathrm{5}+\mathrm{3}} \\ $$$$=\frac{\mathrm{29}+\mathrm{166},\mathrm{5}+\mathrm{270}+\mathrm{238},\mathrm{5}+\mathrm{152},\mathrm{5}+\mathrm{103},\mathrm{5}}{\mathrm{40}} \\ $$$$=\frac{\mathrm{195},\mathrm{5}+\mathrm{508},\mathrm{5}+\mathrm{256}}{\mathrm{40}}=\frac{\mathrm{960}}{\mathrm{40}}=\mathrm{24} \\ $$$$\left.\mathrm{1}.\mathrm{b}\right).\:\mathrm{median}=\mathrm{l}_{\mathrm{i}} +\frac{\left(\frac{\mathrm{N}}{\mathrm{2}}\:−\:<\Sigma\mathrm{f}_{\mathrm{i}} \right)}{\mathrm{f}_{\mathrm{i}} }\:×\mathrm{C} \\ $$$$\:\:\:\mathrm{med}=\mathrm{20},\mathrm{5}\:+\:\frac{\left(\mathrm{20}−\mathrm{11}\right)}{\mathrm{12}}×\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{20},\mathrm{5}\:+\:\mathrm{3}=\mathrm{23},\mathrm{5}// \\ $$$$ \\ $$
Answered by DELETED last updated on 03/Sep/21
2). Modus=L_m +(δ_1 /(δ_1 +δ_2 )) × c =43,5+(((20−12))/((20−12)+(20−15))) ×5 =43,5+(8/(8+5)) ×5 =43,5+((40)/(13))=43,5+3,08=46,58//
$$\left.\mathrm{2}\right).\:\mathrm{Modus}=\mathrm{L}_{\mathrm{m}} +\frac{\delta_{\mathrm{1}} }{\delta_{\mathrm{1}} +\delta_{\mathrm{2}} }\:×\:\mathrm{c} \\ $$$$\:\:\:\:\:\:\:=\mathrm{43},\mathrm{5}+\frac{\left(\mathrm{20}−\mathrm{12}\right)}{\left(\mathrm{20}−\mathrm{12}\right)+\left(\mathrm{20}−\mathrm{15}\right)}\:×\mathrm{5} \\ $$$$\:\:\:=\mathrm{43},\mathrm{5}+\frac{\mathrm{8}}{\mathrm{8}+\mathrm{5}}\:×\mathrm{5} \\ $$$$\:\:\:=\mathrm{43},\mathrm{5}+\frac{\mathrm{40}}{\mathrm{13}}=\mathrm{43},\mathrm{5}+\mathrm{3},\mathrm{08}=\mathrm{46},\mathrm{58}// \\ $$
Answered by DELETED last updated on 03/Sep/21
3). Q_2 =43,5+ (((40/2−16)/(12)))×5 =43,5+((4×5)/(12)) =43,5+(5/3)=43,5+1,67 =45,17// Mod=44,5+(((12−8))/((12−8)+(12−6))) ×5 =44,5+(4/(4+6)) × 5 =44,5+2=46,5//
$$\left.\mathrm{3}\right).\:\mathrm{Q}_{\mathrm{2}} =\mathrm{43},\mathrm{5}+\:\left(\frac{\mathrm{40}/\mathrm{2}−\mathrm{16}}{\mathrm{12}}\right)×\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{43},\mathrm{5}+\frac{\mathrm{4}×\mathrm{5}}{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{43},\mathrm{5}+\frac{\mathrm{5}}{\mathrm{3}}=\mathrm{43},\mathrm{5}+\mathrm{1},\mathrm{67} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{45},\mathrm{17}// \\ $$$$\:\:\:\:\:\mathrm{Mod}=\mathrm{44},\mathrm{5}+\frac{\left(\mathrm{12}−\mathrm{8}\right)}{\left(\mathrm{12}−\mathrm{8}\right)+\left(\mathrm{12}−\mathrm{6}\right)}\:×\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{44},\mathrm{5}+\frac{\mathrm{4}}{\mathrm{4}+\mathrm{6}}\:×\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{44},\mathrm{5}+\mathrm{2}=\mathrm{46},\mathrm{5}// \\ $$
Answered by DELETED last updated on 03/Sep/21
1a) x^− =((Σx_i .f_i )/(Σf_i )) =((15×2+19×9+23×12+27×9+31×5+35×2)/(40)) =((960)/(40))=24//
$$\left.\mathrm{1a}\right)\:\overset{−} {\mathrm{x}}=\frac{\Sigma\mathrm{x}_{\mathrm{i}} .\mathrm{f}_{\mathrm{i}} }{\Sigma\mathrm{f}_{\mathrm{i}} }\: \\ $$$$=\frac{\mathrm{15}×\mathrm{2}+\mathrm{19}×\mathrm{9}+\mathrm{23}×\mathrm{12}+\mathrm{27}×\mathrm{9}+\mathrm{31}×\mathrm{5}+\mathrm{35}×\mathrm{2}}{\mathrm{40}} \\ $$$$=\frac{\mathrm{960}}{\mathrm{40}}=\mathrm{24}// \\ $$
Answered by DELETED last updated on 03/Sep/21
Soal no 2 Rumus modus: M_o =L_i +((δ_1 /(δ_1 +δ_2 )))×C =(44−0.5)+(((20−12)/((20−12)+(20−15))))×5 =43,5+((8/(8+5)))×5 =43,5+((40)/(13))=43,5+3,08 =46,58//
$$\mathrm{Soal}\:\mathrm{no}\:\mathrm{2} \\ $$$$\mathrm{Rumus}\:\mathrm{modus}: \\ $$$$\mathrm{M}_{\mathrm{o}} =\mathrm{L}_{\mathrm{i}} +\left(\frac{\delta_{\mathrm{1}} }{\delta_{\mathrm{1}} +\delta_{\mathrm{2}} }\right)×\mathrm{C} \\ $$$$\:\:\:\:\:\:\:=\left(\mathrm{44}−\mathrm{0}.\mathrm{5}\right)+\left(\frac{\mathrm{20}−\mathrm{12}}{\left(\mathrm{20}−\mathrm{12}\right)+\left(\mathrm{20}−\mathrm{15}\right)}\right)×\mathrm{5} \\ $$$$\:\:\:\:=\mathrm{43},\mathrm{5}+\left(\frac{\mathrm{8}}{\mathrm{8}+\mathrm{5}}\right)×\mathrm{5} \\ $$$$\:\:\:\:=\mathrm{43},\mathrm{5}+\frac{\mathrm{40}}{\mathrm{13}}=\mathrm{43},\mathrm{5}+\mathrm{3},\mathrm{08} \\ $$$$\:\:\:\:=\mathrm{46},\mathrm{58}// \\ $$

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