# When calculating to find the mean and standard deviation of these two sets of numbers: (a)4 0 1 4 3 6 (b)5 3 1 3 4 2 Which data set is more spread out?

Sep 16, 2016

Data-set (a) is more spread out.

#### Explanation:

Both the given data sets have exactly $6$ data points and their sum is too $18$, hence mean of both is $\frac{18}{6} = 3$.

As regards which data is more spread out, there are mainly three measures for the same.

First is Range, which is the difference of larges and smallest data point. As the range of data-set (a) is $6 - 0 = 6$ and that of data-set (b) is $5 - 1 = 4$, hence data-set (a) is more spread out.

Another measure is mean deviation, which is the average of difference (i.e. absolute value) between data point and mean.

In data-set (a) absolute value of deviations (note it makes deviations positive and hence does not cancel out, which will happen if we keep positive or negative sign of differences) is $\left\{| 4 - 3 | , | 0 - 3 | . | 1 - 3 | , | 4 - 3 | , | 3 - 3 | , | 6 - 3 |\right\}$ i.e. $\left\{1 , 3 , 2 , 1 , 0 , 3\right\}$ and their mean is $\frac{1 + 3 + 2 + 1 + 0 + 3}{6} = \frac{10}{6} = \frac{5}{3}$. In data-set (b) absolute value of deviations is $\left\{| 5 - 3 | , | 3 - 3 | . | 1 - 3 | , | 3 - 3 | , | 4 - 3 | , | 2 - 3 |\right\}$ i.e. $\left\{2 , 0 , 2 , 0 , 1 , 1\right\}$ and their mean is $\frac{2 + 0 + 2 + 0 + 1 + 1}{6} = \frac{6}{6} = 1$. As mean deviation in data-set (a) is higher, it is more spread out.

The third measure is standard deviation, for which we square the deviations (again to make them positive), take their average and then take square root.

We have already worked out deviations above. In data-set (a) squares of deviations are$\left\{1 , 9 , 4 , 1 , 0 , 9\right\}$ and their mean is $\frac{1 + 9 + 4 + 1 + 0 + 9}{6} = \frac{24}{6} = 4$ and standard deviation is $\sqrt{4} = 2$. In data-set (b) squares of deviations are $\left\{4 , 0 , 4 , 0 , 1 , 1\right\}$ and their mean is $\frac{4 + 0 + 4 + 0 + 1 + 1}{6} = \frac{10}{6} = \frac{5}{3}$ and standard deviation is $\sqrt{\frac{5}{3}} = 1.291$. Again as standard deviation in data-set (a) is higher, it is more spread out.