Distribution 6399

For the statistical file:

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

Determine the sample variance and the median, and from the empirical distribution function, determine P (2.1 <ξ <3.5).

Correct answer:

r =  1.1782
m =  3.35
P =  0.413

Step-by-step explanation:

$$\begin{gathered} 2.3;3.4;1.8;3.2;3.2;1.9;3.3;4.5;4.3;5.0;4.8;4.3;4.3;1.9 \\ n=14 \\ x=(2.3+3.4+1.8+3.2+3.2+1.9+3.3+4.5+4.3+5.0+4.8+4.3+4.3+1.9)/n=(2.3+3.4+1.8+3.2+3.2+1.9+3.3+4.5+4.3+5.0+4.8+4.3+4.3+1.9)/14=\frac{241}{70}=3\frac{31}{70}\doteq 3.4429 \\ r=((2.3-x{)}^2+(3.4-x{)}^2+(1.8-x{)}^2+(3.2-x{)}^2+(3.2-x{)}^2+(1.9-x{)}^2+(3.3-x{)}^2+(4.5-x{)}^2+(4.3-x{)}^2+(5.0-x{)}^2+(4.8-x{)}^2+(4.3-x{)}^2+(4.3-x{)}^2+(1.9-x{)}^2)/n=((2.3-3.4429{)}^2+(3.4-3.4429{)}^2+(1.8-3.4429{)}^2+(3.2-3.4429{)}^2+(3.2-3.4429{)}^2+(1.9-3.4429{)}^2+(3.3-3.4429{)}^2+(4.5-3.4429{)}^2+(4.3-3.4429{)}^2+(5.0-3.4429{)}^2+(4.8-3.4429{)}^2+(4.3-3.4429{)}^2+(4.3-3.4429{)}^2+(1.9-3.4429{)}^2)/14=1.1782 \end{gathered}$$

$$\begin{gathered} 1.8,1.9,1.9,2.3,3.2,3.2,3.3,3.4,4.3,4.3,4.3,4.5,4.8,5. \\ m=(3.3+3.4)/2=3.35 \end{gathered}$$

$$\begin{gathered} s=\sqrt{r}=\sqrt{1.1782}\doteq 1.0854 \\ {P}_1=0.108 \\ {P}_2=0.521 \\ P=P_2-P_1=0.521-0.108=0.413 \end{gathered}$$

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