Two sets each of 20 observations have the same

Given two sets each of 20 observations, have the same standard derivation 5 . The first set has a mean 17 and the second a mean 22 .

Now we have to show that the standard deviation of the set obtained by

combining the given two sets

As per given criteria, for first set

Number of observations, $n_{1}=20$

Standard deviation, $\mathrm{s}_{1}=5$

And mean, $\bar{x}_{1}=17$

For second set, number of observations, $\mathrm{n}_{2}=20$

Standard deviation, $\mathrm{S}_{2}=5$

And mean, $\bar{x}_{2}=22$

We know the standard deviation for combined two series is

S. D $(\sigma)=\sqrt{\frac{\mathrm{n}_{1} \mathrm{~s}_{1}^{2}+\mathrm{n}_{2} \mathrm{~s}_{2}^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}\left(\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right)^{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)^{2}}}$

Substituting the corresponding values, we get

S. D $(\sigma)=\sqrt{\frac{(20)(5)^{2}+(20)(5)^{2}}{20+20}+\frac{(20 \times 20)(17-22)^{2}}{(20+20)^{2}}}$

On simplifying we get

S. D $(\sigma)=\sqrt{\frac{(20) 25+(20) 25}{40}+\frac{(400)(-5)^{2}}{(40)^{2}}}$

S. $D(\sigma)=\sqrt{\frac{2(500)}{40}+\frac{(400)(25)}{1600}}$

S. D $(\sigma)=\sqrt{\frac{1000}{40}+\frac{10000}{1600}}$

S. D $(\sigma)=\sqrt{25+\frac{25}{4}}$

Taking LCM and simplifying,

S. D $(\sigma)=\sqrt{\frac{100+25}{4}}$

S. D $(\sigma)=\sqrt{\frac{125}{4}}$

$\mathrm{Or}, \sigma=5.59$

Hence the standard deviation of the set obtained by combining the given two sets is $5.59$