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Question Number 58253 by pete last updated on 20/Apr/19

$$\mathrm{There}\mathrm{are}100,150\mathrm{and}250\mathrm{students}\mathrm{in}$$$$\mathrm{forms}\mathrm{one},\mathrm{two}\mathrm{and}\mathrm{three},\mathrm{respectively}$$$$\mathrm{in}\mathrm{a}\mathrm{school}.\mathrm{If}\mathrm{the}\mathrm{mean}\mathrm{ages}\mathrm{of}\mathrm{tbe}\mathrm{students}\mathrm{in}$$$$\mathrm{the}\mathrm{forms}\mathrm{are}15.6\mathrm{years},16.8\mathrm{years}\mathrm{and}$$$$18\mathrm{years}\mathrm{respectively},\mathrm{find}$$$$\mathrm{i}.\mathrm{the}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{students}\mathrm{in}\mathrm{the}\mathrm{forms}$$$$\mathrm{ii}.\mathrm{correct}\mathrm{to}\mathrm{one}\mathrm{decimal}\mathrm{place},\mathrm{the}\mathrm{mean}$$$$\mathrm{age}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{students}$$

Answered by tanmay last updated on 20/Apr/19

$$N=n_1+n_2+n_3=100+150+250=500$$$$mean=\stackrel{-}{x}=\frac{n_1{x}_1+n_2{x}_2+n_3{x}_3}{n_1+n_2+n_3}$$$$\stackrel{-}{x}=\frac{100\times 15.6+150\times 16.8+250\times 18}{500}$$$$=\frac{1560+2520+4500}{500}$$$$=\frac{8580}{500}=17.16$$

Commented by pete last updated on 20/Apr/19

$$\mathrm{thanks}\mathrm{sir}$$

Commented by tanmay last updated on 20/Apr/19

$$mostwelcome..$$

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