Question

The mean and SD of $63$ children on an arithmetic test are respectively $27,6$ and $7.1$. To them are added a new group of $26$ who had less training and whose mean is $19.2$ and SD $6.2$. The values of the combined group differ from the original as to (i) the mean and (ii) the SD is

• A. 25.1,7.8
• B. 2.3,0.8
• C. 1.5,0.9
• D. None of these

Answer 1

Answer: (A)

Mean and $SD \sigma$ of the combined group are
$m=\frac{63 \times 27.6+26 \times 19.2}{63+26}=25.1$
Thus, AM is decreased by $27.6-25.1=2.5$.
$\sigma^{2}=\frac{63 \times(7.1)^{2}+26 \times(6.2)^{2}}{89} $
$+\frac{63(25.1-27.6)^{2}+26(25.1-19.2)^{2}}{89}$
$ \Rightarrow \sigma=7.8 $ (approx.)