Given:
[tex]\sigma_A=3[/tex]
[tex]n_A=6[/tex]
[tex]\sigma_B=5[/tex]
[tex]n_B=4[/tex]
[tex]\overline{x}_A=\overline{x}_B[/tex]
To find:
The variance. of combined set.
Solution:
Formula for variance is
[tex]\sigma^2=\dfrac{\sum (x_i-\overline{x})^2}{n}[/tex] ...(i)
Using (i), we get
[tex]\sigma_A^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{n_A}[/tex]
[tex](3)^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}[/tex]
[tex]9=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}[/tex]
[tex]54=\sum (x_i-\overline{x}_A)^2[/tex]
Similarly,
[tex]\sigma_B^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{n_B}[/tex]
[tex](5)^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}[/tex]
[tex]25=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}[/tex]
[tex]100=\sum (x_i-\overline{x}_B)^2[/tex]
Now, after combining both sets, we get
[tex]\sigma^2=\dfrac{\sum (x_i-\overline{x}_A)^2+\sum (x_i-\overline{x}_B)^2}{n_A+n_B}[/tex]
[tex]\sigma^2=\dfrac{54+100}{6+4}[/tex]
[tex]\sigma^2=\dfrac{154}{10}[/tex]
[tex]\sigma^2=15.4[/tex]
Therefore, the variance of combined set is 15.4.
