Упр.1218 ГДЗ Алимов 10-11 класс (Алгебра)

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. $$\overline{x}=\frac{3+8+5+6}{4}=\frac{22}{4}=5{,}5$$

   $$D=\frac{(3-5{,}5)^2+(8-5{,}5)^2+(5-5{,}5)^2+(6-5{,}5)^2}{4}$$

   $$D=\frac{6{,}25+6{,}25+0{,}25+0{,}25}{4}=\frac{13}{4}=3{,}25$$

   $$\sigma=\sqrt{D}=\sqrt{3{,}25}\approx 1{,}8$$

2. $$\overline{x}=\frac{4+7+3+9}{4}=\frac{23}{4}=5{,}75$$

   $$D=\frac{(4-5{,}75)^2+(7-5{,}75)^2+(3-5{,}75)^2+(9-5{,}75)^2}{4}$$

   $$D=\frac{3{,}0625+1{,}5625+7{,}5625+10{,}5625}{4}=\frac{22{,}75}{4}=5{,}6875$$

   $$\sigma=\sqrt{D}=\sqrt{5{,}6875}\approx 2{,}38$$

3. $$\overline{x}=\frac{4+1+3+2+2}{5}=\frac{12}{5}=2{,}4$$

   $$D=\frac{(4-2{,}4)^2+(1-2{,}4)^2+(3-2{,}4)^2+(2-2{,}4)^2+(2-2{,}4)^2}{5}$$

   $$D=\frac{2{,}56+1{,}96+0{,}36+0{,}16+0{,}16}{5}=\frac{5{,}2}{5}=1{,}04$$

   $$\sigma=\sqrt{D}=\sqrt{1{,}04}\approx 1{,}02$$

4. $$\overline{x}=\frac{3+2+1+1+5}{5}=\frac{12}{5}=2{,}4$$

   $$D=\frac{(3-2{,}4)^2+(2-2{,}4)^2+(1-2{,}4)^2+(1-2{,}4)^2+(5-2{,}4)^2}{5}$$

   $$D=\frac{0{,}36+0{,}16+1{,}96+1{,}96+6{,}76}{5}=\frac{11{,}2}{5}=2{,}24$$

   $$\sigma=\sqrt{D}=\sqrt{2{,}24}\approx 1{,}5$$

5. $$\overline{x}=\frac{2+(-1)+3+(-2)+5}{5}=\frac{7}{5}=1{,}4$$

   $$D=\frac{(2-1{,}4)^2+(-1-1{,}4)^2+(3-1{,}4)^2+(-2-1{,}4)^2+(5-1{,}4)^2}{5}$$

   $$D=\frac{0{,}36+5{,}76+2{,}56+11{,}56+12{,}96}{5}=\frac{33{,}2}{5}=6{,}64$$

   $$\sigma=\sqrt{D}=\sqrt{6{,}64}\approx 2{,}58$$

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

1. $$D=3{,}25,\ \sigma\approx 1{,}8$$
2. $$D=5{,}6875,\ \sigma\approx 2{,}38$$
3. $$D=1{,}04,\ \sigma\approx 1{,}02$$
4. $$D=2{,}24,\ \sigma\approx 1{,}5$$
5. $$D=6{,}64,\ \sigma\approx 2{,}58$$
6. $$D=12{,}56,\ \sigma\approx 3{,}54$$