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The mean and standard deviation of a set $n_1$ observations are $\bar{x}_1$ and $s_1$, respectively while the mean and standard deviation of another set of $n_2$ observations are $\overline{\mathrm{x}}_2$ and $s_2$, respectively. Show that the standard deviation of the
combined set of $\left(n_1+n_2\right)$ observations is given by
$$
\text { S.D. }=\sqrt{\frac{n_1\left(s_1\right)^2+n_2\left(s_2\right)^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}-\bar{x}_2\right)^2}{\left(n_1+n_2\right)}}
$$
combined set of $\left(n_1+n_2\right)$ observations is given by
$$
\text { S.D. }=\sqrt{\frac{n_1\left(s_1\right)^2+n_2\left(s_2\right)^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}-\bar{x}_2\right)^2}{\left(n_1+n_2\right)}}
$$
Solution:
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Verified Answer
Let $d_1=\left(\bar{x}_1-\bar{x}_{12}\right), d_2=\left(\bar{x}_2-\bar{x}_{12}\right)$
$\therefore$ Standard deviation (combined set) is given as
$\mathrm{SD}=\sqrt{\frac{n_1\left(s_1^2+d_1^2\right)+n_2\left(s_2^2+d_2^2\right)}{n_1+n_2}}$
$\Rightarrow \quad \mathrm{SD}=\sqrt{\frac{\mathrm{n}_1 \mathrm{~s}_1^2+\mathrm{n}_2 \mathrm{~s}_2^2}{\mathrm{n}_1+\mathrm{n}_2}+\frac{\mathrm{n}_1 \mathrm{~d}_1^2+\mathrm{n}_2 \mathrm{~d}_2^2}{\mathrm{n}_1+\mathrm{n}_2}}$
Now $\frac{n_1 d_1^2+n_2 d_2^2}{n_1+n_2}=\frac{n_1\left(\bar{x}_1-\bar{x}_{12}\right)^2+n_2\left(\bar{x}_2-\bar{x}_{12}\right)^2}{n_1+n_2}$
$=\left(\frac{1}{n_1+n_2}\right)\left[n_1\left\{\bar{x}_1-\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\right\}^2+n_2\left\{\bar{x}_2-\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\right\}^2\right]$
$=\left(\frac{1}{n_1+n_2}\right)\left[n_1\left\{\frac{n_2\left(\bar{x}_1-\bar{x}_2\right)}{n_1+n_2}\right\}^2+n_2\left\{\frac{n_1\left(\bar{x}_2-\bar{x}_1\right)}{n_1+n_2}\right\}^2\right]$
$=\left(\frac{1}{n_1+n_2}\right)\left(\frac{\bar{x}_1-\bar{x}_2}{n_1+n_2}\right)^2\left(n_1 n_2^2+n_2 n_1^2\right)$
$=\left(\frac{1}{n_1+n_2}\right)\left(\frac{\bar{x}_1-\bar{x}_2}{n_1+n_2}\right)^2\left(n_1 n_2\right)\left(n_2+n_1\right)$
$=\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}$, put in (i) we have :
$$
\mathrm{SD}=\sqrt{\frac{n_1 s_1^2+n_2 s_2^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)}}
$$
$\therefore$ Standard deviation (combined set) is given as
$\mathrm{SD}=\sqrt{\frac{n_1\left(s_1^2+d_1^2\right)+n_2\left(s_2^2+d_2^2\right)}{n_1+n_2}}$
$\Rightarrow \quad \mathrm{SD}=\sqrt{\frac{\mathrm{n}_1 \mathrm{~s}_1^2+\mathrm{n}_2 \mathrm{~s}_2^2}{\mathrm{n}_1+\mathrm{n}_2}+\frac{\mathrm{n}_1 \mathrm{~d}_1^2+\mathrm{n}_2 \mathrm{~d}_2^2}{\mathrm{n}_1+\mathrm{n}_2}}$
Now $\frac{n_1 d_1^2+n_2 d_2^2}{n_1+n_2}=\frac{n_1\left(\bar{x}_1-\bar{x}_{12}\right)^2+n_2\left(\bar{x}_2-\bar{x}_{12}\right)^2}{n_1+n_2}$
$=\left(\frac{1}{n_1+n_2}\right)\left[n_1\left\{\bar{x}_1-\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\right\}^2+n_2\left\{\bar{x}_2-\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\right\}^2\right]$
$=\left(\frac{1}{n_1+n_2}\right)\left[n_1\left\{\frac{n_2\left(\bar{x}_1-\bar{x}_2\right)}{n_1+n_2}\right\}^2+n_2\left\{\frac{n_1\left(\bar{x}_2-\bar{x}_1\right)}{n_1+n_2}\right\}^2\right]$
$=\left(\frac{1}{n_1+n_2}\right)\left(\frac{\bar{x}_1-\bar{x}_2}{n_1+n_2}\right)^2\left(n_1 n_2^2+n_2 n_1^2\right)$
$=\left(\frac{1}{n_1+n_2}\right)\left(\frac{\bar{x}_1-\bar{x}_2}{n_1+n_2}\right)^2\left(n_1 n_2\right)\left(n_2+n_1\right)$
$=\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}$, put in (i) we have :
$$
\mathrm{SD}=\sqrt{\frac{n_1 s_1^2+n_2 s_2^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)}}
$$
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