Question

For two data sets, each size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is

• A. $\frac{5}{2}$
• B. $\frac{11}{2}$
• C. $\sqrt{\frac{11}{2}}$
• D. $\frac{13}{2}$

Answer 1

Answer: (B)

$\frac{11}{2}$

We have the formulas

${\sigma }^2=\frac{1}{n}\sum _{i=1}^n{x}_i^2-(\sum _{i=1}^n\frac{x_i}{n}{)}^2$ and $\overline{x}=\sum _{i=1}^n\frac{x_i}{n}$.

For the first data we have

$2=\sum _{i=1}^5\frac{x_i}{5}\Rightarrow \sum _{i=1}^5{x}_i=10$

$4={\sigma }_1^2=\frac{1}{5}\sum _{i=1}^5{x}_i^2-2^2\Rightarrow \sum _{i=1}^5{x}_i^2=40$

Similarly for the second data

$4=\sum _{i=1}^5\frac{y_i}{5}\Rightarrow \sum _{i=1}^5{y}_i=20$

$5={\sigma }_2^2=\frac{1}{5}\sum _{i=1}^5{y}_i^2-4^2\Rightarrow \sum _{i=1}^5{y}_i^2=105$

Hence for the combined data

${\sigma }^2=\frac{1}{10}\sum _{i=1}^5(x_i^2+y_i^2)-(\sum _{i=1}^5\frac{x_i+y_i}{10}{)}^2=\frac{145}{10}-9=\frac{11}{2}.$