Упр.1218 ГДЗ Алимов 10-11 класс (Алгебра)
- Найти дисперсию и среднее квадратичное отклонение выборки: 1) $$3, 8, 5, 6$$; 2) $$4, 7, 3, 9$$; 3) $$4, 1, 3, 2, 2$$; 4) $$3, 2, 1, 1, 5$$; 5) $$2, -1, 3, -2, 5$$; 6) $$-2, 4, -3, -1, 6$$.
1) $$\overline{x}=\frac{3+8+5+6}{4}=\frac{22}{4}=5{,}5$$
$$D=\frac{(3-5{,}5)^2+(8-5{,}5)^2+(5-5{,}5)^2+(6-5{,}5)^2}{4}$$
$$D=\frac{6{,}25+6{,}25+0{,}25+0{,}25}{4}=\frac{13}{4}=3{,}25$$
$$\sigma=\sqrt{D}=\sqrt{3{,}25}\approx 1{,}8$$
Ответ: $$D=3{,}25,\ \sigma\approx 1{,}8$$
2) $$\overline{x}=\frac{4+7+3+9}{4}=\frac{23}{4}=5{,}75$$
$$D=\frac{(4-5{,}75)^2+(7-5{,}75)^2+(3-5{,}75)^2+(9-5{,}75)^2}{4}$$
$$D=\frac{3{,}0625+1{,}5625+7{,}5625+10{,}5625}{4}=\frac{22{,}75}{4}=5{,}6875$$
$$\sigma=\sqrt{D}=\sqrt{5{,}6875}\approx 2{,}38$$
Ответ: $$D=5{,}6875,\ \sigma\approx 2{,}38$$
3) $$\overline{x}=\frac{4+1+3+2+2}{5}=\frac{12}{5}=2{,}4$$
$$D=\frac{(4-2{,}4)^2+(1-2{,}4)^2+(3-2{,}4)^2+(2-2{,}4)^2+(2-2{,}4)^2}{5}$$
$$D=\frac{2{,}56+1{,}96+0{,}36+0{,}16+0{,}16}{5}=\frac{5{,}2}{5}=1{,}04$$
$$\sigma=\sqrt{D}=\sqrt{1{,}04}\approx 1{,}02$$
Ответ: $$D=1{,}04,\ \sigma\approx 1{,}02$$
4) $$\overline{x}=\frac{3+2+1+1+5}{5}=\frac{12}{5}=2{,}4$$
$$D=\frac{(3-2{,}4)^2+(2-2{,}4)^2+(1-2{,}4)^2+(1-2{,}4)^2+(5-2{,}4)^2}{5}$$
$$D=\frac{0{,}36+0{,}16+1{,}96+1{,}96+6{,}76}{5}=\frac{11{,}2}{5}=2{,}24$$
$$\sigma=\sqrt{D}=\sqrt{2{,}24}\approx 1{,}5$$
Ответ: $$D=2{,}24,\ \sigma\approx 1{,}5$$
5) $$\overline{x}=\frac{2+(-1)+3+(-2)+5}{5}=\frac{7}{5}=1{,}4$$
$$D=\frac{(2-1{,}4)^2+(-1-1{,}4)^2+(3-1{,}4)^2+(-2-1{,}4)^2+(5-1{,}4)^2}{5}$$
$$D=\frac{0{,}36+5{,}76+2{,}56+11{,}56+12{,}96}{5}=\frac{33{,}2}{5}=6{,}64$$
$$\sigma=\sqrt{D}=\sqrt{6{,}64}\approx 2{,}58$$
Ответ: $$D=6{,}64,\ \sigma\approx 2{,}58$$
6) $$\overline{x}=\frac{-2+4-3-1+6}{5}=\frac{4}{5}=0{,}8$$
$$D=\frac{(-2-0{,}8)^2+(4-0{,}8)^2+(-3-0{,}8)^2+(-1-0{,}8)^2+(6-0{,}8)^2}{5}$$
$$D=\frac{7{,}84+10{,}24+14{,}44+3{,}24+27{,}04}{5}=\frac{62{,}8}{5}=12{,}56$$
$$\sigma=\sqrt{D}=\sqrt{12{,}56}\approx 3{,}54$$
Ответ: $$D=12{,}56,\ \sigma\approx 3{,}54$$
