The data sets of sizes 6 and 9 have standard deviation 3 and 4, respectively, and arithmetic means 4 and 4, respectively. The standard deviation of combined data set of size 15 is:

The correct answer is \(\mathbf{\sqrt{\frac{66}{5}}}\)

Key Points

• The formula used to derive this formula is \(\mathbf{s}=\mathbf{\sqrt{\frac{n_{1}*s^2_{1}+n_{2}*s^2_{2}+n_{1}*d^2_{1}+n_{2}*d^2_{2}}{n_{1}+n_{2}}}}\), where \(d_{1}=\overline{x}_{1}-\overline{x}\) and \(d_{2}=\overline{x}_{2}-\overline{x}\) also, \(\overline{x}=\frac{n_{1}*\overline{x}_{1}+n_{2}*\overline{x}_{2}}{n_{1}+n_{2}}\) where \(n_{1}\)is the total observation of the first data set, \(n_{2}\) is the total observation of the second data set. Also,\(s_{1},x_{1} \text{ and } s_{2},x_{2}\) are the standard deviation and mean for data set 1 and data set 2 respectively.
• Here \(n_{1}=6,n_{2}=9 \text{ and } s_{1}=3 , s_{2}=4 \text{ and } \overline{x}_{1}=4 , \overline{x}_{2}=4\). The following calculations are done:

​\(\overline{x}=\frac{6*4+9*4}{6+9} \\ \overline{x}=4 \\ {d}_{1}=4-4=0 \\d_{2}=4-4=0\)

• Substituting the values in the expression s, we get:

​ \(s=\sqrt{\frac{6*9+9*16+6*0+9*0}{15}} \\s=\sqrt{\frac{198}{15}}\\s=\sqrt{\frac{66}{5}}\)

Additional Information​

• Combined Standard Deviation is the result of calculating the standard deviation of two or more series.
• The number of series that can be included in the Combined Standard Deviation calculation is N.
• If we know the mean and quantity of items in each group of the data, we can calculate the combined arithmetic mean.

Hence, the standard deviation of data sets of sizes 6 and 9 have a standard deviation 3 and 4, respectively, and arithmetic means 4 and 4, respectively. The standard deviation of the combined data set of size 15 is \(\mathit{\mathbf{ \sqrt{\frac{66}{5}}}} \)