Страница 384 номер 1218, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(3,\ 8,\ 5,\ 6\)
\(\overline{X} = \dfrac{3 + 8 + 5 + 6}{4} = \dfrac{22}{4} = 5{,}5,\)
\(D = \dfrac{(3 - 5{,}5)^{2} + (8 - 5{,}5)^{2} + (5 - 5{,}5)^{2} + (6 - 5{,}5)^{2}}{4} =\)
\(= \dfrac{6{,}25 + 6{,}25 + 0{,}25 + 0{,}25}{4} = \dfrac{13}{4} = 3{,}25,\)
\(\sigma = \sqrt{D} = \sqrt{3{,}25} \approx 1{,}8;\)

2) \(4,\ 7,\ 3,\ 9\)
\(\overline{X} = \dfrac{4 + 7 + 3 + 9}{4} = \dfrac{23}{4} = 5{,}75,\)
\(D = \dfrac{(4 - 5{,}75)^{2} + (7 - 5{,}75)^{2} + (3 - 5{,}75)^{2} + (9 - 5{,}75)^{2}}{4} =\)
\(= \dfrac{3{,}0625 + 1{,}5625 + 7{,}5625 + 10{,}5625}{4} =\)
\(= \dfrac{22{,}75}{4} = 5{,}6875,\)
\(\sigma = \sqrt{D} = \sqrt{5{,}6875} \approx 2{,}38;\)

3) \(4,\ 1,\ 3,\ 2,\ 2\)
\(\overline{X} = \dfrac{4 + 1 + 3 + 2 \cdot 2}{5} = \dfrac{12}{5} = 2{,}4,\)
\(D = \dfrac{(4 - 2{,}4)^{2} + (1 - 2{,}4)^{2} + (3 - 2{,}4)^{2} + 2(2 - 2{,}4)^{2}}{5} =\)
\(= \dfrac{2{,}56 + 1{,}96 + 0{,}36 + 0{,}32}{5} = \dfrac{5{,}2}{5} = 1{,}04,\)
\(\sigma = \sqrt{D} = \sqrt{1{,}04} \approx 1{,}02;\)

4) \(3,\ 2,\ 1,\ 1,\ 5\)
\(\overline{X} = \dfrac{3 + 2 + 1 \cdot 2 + 5}{5} = \dfrac{12}{5} = 2{,}4,\)
\(D = \dfrac{(3 - 2{,}4)^{2} + (2 - 2{,}4)^{2} + 2(1 - 2{,}4)^{2} + (5 - 2{,}4)^{2}}{5} =\)
\(= \dfrac{0{,}36 + 0{,}16 + 3{,}92 + 6{,}76}{5} = \dfrac{11{,}2}{5} = 2{,}24,\)
\(\sigma = \sqrt{D} = \sqrt{2{,}24} \approx 1{,}5;\)

5) \(2,\ - 1,\ 3,\ - 2,\ 5\)
\(\overline{X} = \dfrac{2 - 1 + 3 - 2 + 5}{5} = \dfrac{7}{5} = 1{,}4,\)
\(D = \dfrac{(2 - 1{,}4)^{2} + (-1 - 1{,}4)^{2} + (3 - 1{,}4)^{2} + (-2 - 1{,}4)^{2} + (5 - 1{,}4)^{2}}{5} =\)
\(= \dfrac{0{,}36 + 5{,}76 + 2{,}56 + 11{,}56 + 12{,}96}{5} = \dfrac{33{,}2}{5} = 6{,}64,\)
\(\sigma = \sqrt{D} = \sqrt{6{,}64} \approx 2{,}58;\)

6) \(- 2,\ 4,\ - 3,\ - 1,\ 6\)
\(\overline{X} = \dfrac{-2 + 4 - 3 - 1 + 6}{5} = \dfrac{4}{5} = 0{,}8,\)
\(D = \dfrac{\begin{gathered}(-2 - 0{,}8)^{2} + (4 - 0{,}8)^{2} + (-3 - 0{,}8)^{2} + {} \\ {} + (-1 - 0{,}8)^{2} + (6 - 0{,}8)^{2}\end{gathered}}{5} =\)
\(= \dfrac{7{,}84 + 10{,}24 + 14{,}44 + 3{,}24 + 27{,}04}{5} = \dfrac{62{,}8}{5} = 12{,}56,\)
\(\sigma = \sqrt{D} = \sqrt{12{,}56} \approx 3{,}54.\)