Pre štatistický

Pre štatistický súbor:

2,3; 3,4; 1,8; 3,2; 3,2; 1,9; 3,3; 4,5; 4,3; 5,0; 4,8; 4,3; 4,3; 1,9

určte výberový rozptyl a medián, a z empirickej distribučnej funkcie určte P(2,1 < ξ < 3,5).

Správna odpoveď:

r =  1,1782
m =  3,35
P =  0,413

Postup správneho riešenia:

$$\begin{gathered} 2,3;3,4;1,8;3,2;3,2;1,9;3,3;4,5;4,3;5,0;4,8;4,3;4,3;1,9 \\ n=14 \\ x=(2,3+3,4+1,8+3,2+3,2+1,9+3,3+4,5+4,3+5,0+4,8+4,3+4,3+1,9)/n=(2,3+3,4+1,8+3,2+3,2+1,9+3,3+4,5+4,3+5,0+4,8+4,3+4,3+1,9)/14=\frac{241}{70}=3\frac{31}{70}\doteq 3,4429 \\ r=((2,3-x{)}^2+(3,4-x{)}^2+(1,8-x{)}^2+(3,2-x{)}^2+(3,2-x{)}^2+(1,9-x{)}^2+(3,3-x{)}^2+(4,5-x{)}^2+(4,3-x{)}^2+(5,0-x{)}^2+(4,8-x{)}^2+(4,3-x{)}^2+(4,3-x{)}^2+(1,9-x{)}^2)/n=((2,3-3,4429{)}^2+(3,4-3,4429{)}^2+(1,8-3,4429{)}^2+(3,2-3,4429{)}^2+(3,2-3,4429{)}^2+(1,9-3,4429{)}^2+(3,3-3,4429{)}^2+(4,5-3,4429{)}^2+(4,3-3,4429{)}^2+(5,0-3,4429{)}^2+(4,8-3,4429{)}^2+(4,3-3,4429{)}^2+(4,3-3,4429{)}^2+(1,9-3,4429{)}^2)/14=1,1782 \end{gathered}$$

$$\begin{gathered} 1,8,1,9,1,9,2,3,3,2,3,2,3,3,3,4,4,3,4,3,4,3,4,5,4,8,5, \\ m=(3,3+3,4)/2=3,35 \end{gathered}$$

$$\begin{gathered} s=\sqrt{r}=\sqrt{1,1782}\doteq 1,0854 \\ {P}_1=0,108 \\ {P}_2=0,521 \\ P=P_2-P_1=0,521-0,108=0,413 \end{gathered}$$

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