Mogę prosić o rozwiązanie

Zad. 1
[tex]Zestaw: 1, 8, 3, 2, 7, 2, 3, 5, 2\\\left[\begin{array}{ccccccc}\text{Dane liczbowe}&1&2&3&5&7&8\\\text{Liczba wskazan}&1&3&2&1&1&1\end{array}\right] \\\\n=9\\X=\frac{1*1+3*2+2*3+1*5+1*7+1*8}{9}=\frac{1+6+6+5+7+8}9=\frac{33}9=\frac{11}3[/tex]
[tex]\sigma=\sqrt{\frac{(1-\frac{11}3)^2+3*(2-\frac{11}3)^2+2*(3-\frac{11}3)^2+(5-\frac{11}3)^2+(7-\frac{11}3)^2+(8-\frac{11}3)^2}{9}}\\\sigma=\sqrt{\frac{(-\frac83)^2+3*(-\frac53)^2+2*(-\frac23)^2+(\frac43)^2+(\frac{10}3)^2+(\frac{13}3)^2}{9}}\\\sigma=\sqrt{\frac{\frac{64}9+3*\frac{25}9+2*\frac49+\frac{16}9+\frac{100}9+\frac{169}9}{9}}\\\sigma=\sqrt{\frac{\frac{349}{9}+\frac{75}9+\frac89}{9}}\\\sigma=\sqrt{\frac{\frac{432}9}{9}}\\\sigma=\sqrt{\frac{432}9*\frac19}\\[/tex]
[tex]\sigma=\sqrt{\frac{432}{81}}\\\sigma=\frac{12\sqrt3}9\\\sigma=\frac{4\sqrt3}3[/tex]
Zad. 2
[tex]\left[\begin{array}{cccccccccc}\text{Dane liczbowe}&2&4&6&8\\\text{Liczba wskazan}&12&17&8&13\end{array}\right] \\n=12+17+8+13=50\\X=\frac{12*2+17*4+8*6+13*8}{50}=\frac{24+68+48+104}{50}=\frac{244}{50}=\frac{122}{25}=4.88[/tex]
[tex]\sigma=\sqrt{\frac{12*(2-4.88)^2+17*(4-4.88)^2+8*(6-4.88)^2+13*(8-4.88)^2}{50}}\\\sigma=\sqrt{\frac{12*(-2.88)^2+17*(-0.88)^2+8*(1.12)^2+13*(3.12)^2}{50}}\\\sigma=\sqrt{\frac{99.5328+13.1648+10.0352+126.5472}{50}}\\\sigma=\sqrt{\frac{249.28}{50}}\\\sigma=\sqrt{4.9856}\\\sigma=2.23284571...=2.23[/tex]