Страница 385 номер 1222, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(2,\ 3,\ 5,\ 3,\ 7\) и \(4,\ 7,\ 5,\ 6\)
\({\overline{X}}_{1} = \dfrac{2 + 3 \cdot 2 + 5 + 7}{5} = \dfrac{20}{5} = 4,\)
\(D_{1} = \dfrac{(2 - 4)^{2} + 2(3 - 4)^{2} + (5 - 4)^{2} + (7 - 4)^{2}}{5} =\)
\(= \dfrac{4 + 2 + 1 + 9}{5} = \dfrac{16}{5} = 3\dfrac{1}{5},\)
\({\overline{X}}_{2} = \dfrac{4 + 7 + 5 + 6}{4} = \dfrac{22}{4} = 5{,}5,\)
\(D_{2} = \dfrac{(4 - 5{,}5)^{2} + (7 - 5{,}5)^{2} + (5 - 5{,}5)^{2} + (6 - 5{,}5)^{2}}{4} =\)
\(= \dfrac{2{,}25 + 2{,}25 + 0{,}25 + 0{,}25}{4} = \dfrac{5}{4} = 1\dfrac{1}{4},\)
\(D_{1} > D_{2};\)

2) \(- 1,\ 3,\ 4\) и \(- 2,\ 0,\ 2,\ 4,\ 5\)
\({\overline{X}}_{1} = \dfrac{-1 + 3 + 4}{3} = 2,\)
\(D_{1} = \dfrac{(-1 - 2)^{2} + (3 - 2)^{2} + (4 - 2)^{2}}{3} = \dfrac{14}{3} = 4\dfrac{2}{3},\)
\({\overline{X}}_{2} = \dfrac{-2 + 0 + 2 + 4 + 5}{5} = \dfrac{9}{5} = 1{,}8,\)
\(D_{2} = \dfrac{(-2 - 1{,}8)^{2} + (0 - 1{,}8)^{2} + (2 - 1{,}8)^{2} + (4 - 1{,}8)^{2} + (5 - 1{,}8)^{2}}{5} =\)
\(= \dfrac{14{,}44 + 3{,}24 + 0{,}04 + 4{,}84 + 10{,}24}{5} = \dfrac{32{,}8}{5} = 6{,}56,\)
\(D_{1} < D_{2}.\)