Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1218

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\[1)\ 3,\ 8,\ 5,\ 6;\]

\[\overline{X} = \frac{3 + 8 + 5 + 6}{4} = \frac{22}{4} = 5,5.\]

\[= \frac{{2,5}^{2} + {2,5}^{2} + {0,5}^{2} + {0,5}^{2}}{4} =\]

\[= \frac{6,25 + 6,25 + 0,25 + 0,25}{4} =\]

\[= \frac{13}{4} = 3,25.\]

\[\sigma = \sqrt{D} = \sqrt{3,25} \approx 1,8.\]

\[2)\ 4,\ 7,\ 3,\ 9;\]

\[\overline{X} = \frac{4 + 7 + 3 + 9}{4} = \frac{23}{4} = 5,75.\]

\[= \frac{22,74}{4} \approx 5,68.\]

\[\sigma = \sqrt{D} = \sqrt{5,68} \approx 2,38.\]

\[3)\ 4,\ 1,\ 3,\ 2,\ 2;\]

\[\overline{X} = \frac{4 + 1 + 3 + 2 + 2}{5} = \frac{12}{5} =\]

\[= 2,4.\]

\[= \frac{5,2}{5} = 1,04.\]

\[\sigma = \sqrt{D} = \sqrt{1,04} \approx 1,02.\]

\[4)\ 3,\ 2,\ 1,\ 1,\ 5;\]

\[\overline{X} = \frac{3 + 2 + 1 + 1 + 5}{5} = \frac{12}{5} =\]

\[= 2,4.\]

\[= \frac{11,2}{5} = 2,24.\]

\[\sigma = \sqrt{D} = \sqrt{2,24} \approx 1,49.\]

\[5)\ 2,\ - 1,\ 3,\ - 2,\ 5;\]

\[\overline{X} = \frac{2 - 1 + 3 - 2 + 5}{5} = \frac{7}{5} = 1,4.\]

\[= \frac{33,2}{5} = 6,64.\]

\[\sigma = \sqrt{D} = \sqrt{6,64} \approx 2,57.\]

\[6)\ - 2,\ 4,\ - 3,\ - 1,\ 6;\]

\[\overline{X} = \frac{- 2 + 4 - 3 - 1 + 6}{5} = \frac{4}{5} =\]

\[= 0,8.\]

\[= \frac{62,8}{5} = 12,56.\]

\[\sigma = \sqrt{D} = \sqrt{12,56} \approx 3,54.\]