Solution:
Given the data below:
[tex]14,12,18,10,8,6,15,9,16,20[/tex]1) The range is the spread of data from the lowest to the highest value in the distribution.
In other words,
[tex]range=highest\text{ value - lowest value}[/tex]In this case,
[tex]\begin{gathered} highest\text{ value = 20} \\ lowest\text{ value = 6} \\ thus, \\ range\text{ = 20-6=14} \\ range\approx\text{ 14.0\lparen 1 decimal place\rparen} \end{gathered}[/tex]Thus, the range is 14.0 (1 decimal place)
2) Population variance:
The population variance is expressed as
[tex]\begin{gathered} \sum_{i=1}^n\frac{\left(x_i-\bar{x}\right)^2}{n} \\ where \\ \bar{x}\text{ is the mean} \\ n\text{ is the numberof observation} \end{gathered}[/tex]Thus, we have
[tex]\begin{gathered} \frac{(14-12.8)^2+(12-12.8)^2+(18-12.8)^2+(10-12.8)^2+(8-12.8)^2+(6-12.8)^2+(15-12.8)^2+(9-12.8)^2+(16-12.8)^2+(20-12.8)^2}{10} \\ =\frac{187.6}{10} \\ =18.76 \\ population\text{ variance}\approx18.8\text{ \lparen1 decimal place\rparen} \end{gathered}[/tex]Thus, the population variance is 18.8 (1 decimal place)
3) Standard deviation:
the standrard deviation is the square root of the variance.
Hence, we have
[tex]\begin{gathered} standard\text{ deviation = }\sqrt{variance\text{ }}=\sqrt{18.76}=\text{ }4.3312816 \\ \implies standard\text{ deviation}\approx4.3\text{ \lparen1 decimal place\rparen} \end{gathered}[/tex]Hence, the standard deviation is 4.3 (1 decimal place)
