Part 1
To find the range, we subtract the lowest value from the greatest value, ignoring the others.
So, in this case, we have:
• Lowest value: 20.3
,
• Greatest value: 110.4
[tex]\text{Range = }110.4-20.3=90.1[/tex]
Therefore, the range is 90.1
Part 2The formula to find the variance is
[tex]\begin{gathered} \sigma^2=\frac{\sum ^{}_{}(x-\mu)^2}{n} \\ \text{ Where} \\ x=\text{ data values} \\ \mu=\text{ mean} \\ n=\text{ number of data points} \end{gathered}[/tex]
The formula to find the mean is
[tex]\mu=\frac{\sum ^{}_{}x}{n}[/tex]
So, as you can see, we first find the mean, and with this value, we find the variance of the data set.
• Mean
[tex]\begin{gathered} \mu=\frac{20.3+33.5+21.8+58.2+23.2+110.4+30.9+24.4+74.6+60.4+40.8}{11} \\ \mu=\frac{498.5}{11} \\ \mu=45.32 \end{gathered}[/tex]
• Variance
[tex]\begin{gathered} \sigma=\frac{(20.3-45.32)^2+(33.5-45.32)^2+(21.8-45.32)^2+(58.2-45.32)^2+(23.2-45.32)^2+(110.4-45.32)^2+(30.9-45.32)^2+(24.4-45.32)^2+(74.6-45.32)^2+(60.4-45.32)^2+(40.8-45.32)^2}{11} \\ \sigma=\frac{(-25.02)^2+(-11.82)^2+(-23.52)^2+(12.88)^2+(-22.12)^2+(65.08)^2+(-12.42)^2+(-20.92)^2+(29.28)^2+(15.08)^2+(-4.52)^2}{11} \\ \sigma=\frac{625.91+139.67+553.10+165.94+489.21+4325.64+207.88+437.57+857.42+227.46+20.41}{11} \\ \sigma=\frac{7960.24}{11} \\ $\boldsymbol{\sigma=723.66}$ \end{gathered}[/tex]
Therefore, the variance is 723.66.
