Question

Find the combined standard deviation of two groups $X$ and $Y$ from the data gives below:

• A. 5
• B. 5.8
• C. 5.2
• D. 6

Answer 1

Answer: (C)

Combined standard deviation $X$ and $Y$ is: $\sigma_{x y}=\sqrt{\frac{N_x \sigma_x^2+N_y \sigma_y^2+N_x d_x^2+N_y d_y^2}{N_x+N_y}} $ $N _{ x }=10, N _{ y }=10, \overline{X_x}=15, \overline{X_y}=20, \sigma_x=4,$ and $\sigma_y=5$ $\overline{X_{x y}}=\frac{\overline{X_x} N_x+\overline{X_y} N_y}{N_x+N_y} $ $\text { Or } \overline{X_{x y}}=\frac{(15 \times 10)+(20 \times 15)}{10+15}=\frac{150+300}{25} $ $=\frac{450}{25}=18 $ $d _x=(\overline{X_{x y}}-\overline{X_x})=18-15=3 $ $d _y=(\overline{X_{x y}}-\overline{X_y})=18-20=-2$ On substituting all the values in the formula for combined standard deviation, we get. $\sigma_{x y}= \sqrt{\frac{10(4)^2+15(5)^2+10(18-15)^2+15(18-20)^2}{10+15}} $ $ =\sqrt{\frac{(10 \times 16)+(15 \times 25)+(10 \times 9)+(15 \times 4)}{25}} $ $ =\sqrt{\frac{160+375+90+60}{25}} $ $ =\sqrt{\frac{68.5}{25}}=5.2 \text { (approximately) }$