Question

the means of two sample of sizes 60 and 120 respectively are 35.4 and 30.9 and the standard deviations 4 and 5. obtain the standard deviation of the sample of size 180 obtained by combining the two sample.​

Answer 1

Answer:

The combined standard deviation is 5.15.

Step-by-step explanation:

Given:

$n_{1}=60\\n_{2}=120\\\bar x_{1}=35.4\\\bar x_{2}=30.9\\s_{1}=4\\s_{2}=5$

The formula to compute combined standard deviation is:

$s_{c} =\sqrt{\frac{n_{1}s^{2}_{1}+n_{2}s^{2}_{2}+n_{1}(\bar x_{1}-\bar x_{c})+n_{2}(\bar x_{2}-\bar x_{c})}{n_{1}+n_{2}} }$

Here $\bar x_{c}$ is the combined mean.

Formula of combined mean is:

$\bar x_{c} = \frac{n_{1}\bar x_{1}+n_{2}\bar x_{2}}{n_{1}+n_{2}}$

First compute the combined mean:

$\bar x_{c} = \frac{n_{1}\bar x_{1}+n_{2}\bar x_{2}}{n_{1}+n_{2}}\\=\frac{(60\times35.4)+(120\times30.9)}{60+120}\\=32.4$

Now compute the combined standard deviation:

$s_{c} =\sqrt{\frac{n_{1}s^{2}_{1}+n_{2}s^{2}_{2}+n_{1}(\bar x_{1}-\bar x_{c})^{2}+n_{2}(\bar x_{2}-\bar x_{c})^{2}}{n_{1}+n_{2}} }\\=\sqrt{\frac{(60\times4^{2})+(120\times5^{2})+[(60\times(35.4-32.4)^{2}]+[(120\times(30.9-32.4)^{2}]}{60+120}}\\= \sqrt{26.5}\\ =5.1478\\\approx5.15$

Thus, the combined standard deviation is 5.15.