Dataset

Data: 11,15,11,16,12,17,13,21,14,21,15,22

Find dispersion of given dataset

Correct answer:

s =  13.8889

Step-by-step explanation:

$$\begin{gathered} S=11+15+11+16+12+17+13+21+14+21+15+22=188 \\ m=S\mathrm{/}12=188\mathrm{/}12={\displaystyle \frac{47}{3}}=15\frac{2}{3}\doteq 15.6667 \\ {s}_1=(m-11{)}^2=(15.6667-11{)}^2={\displaystyle \frac{196}{9}}=21\frac{7}{9}\doteq 21.7778 \\ {s}_2=(m-15{)}^2=(15.6667-15{)}^2={\displaystyle \frac{4}{9}}\doteq 0.4444 \\ {s}_3=(m-11{)}^2=(15.6667-11{)}^2={\displaystyle \frac{196}{9}}=21\frac{7}{9}\doteq 21.7778 \\ {s}_4=(m-16{)}^2=(15.6667-16{)}^2\doteq 0.1111 \\ {s}_5=(m-12{)}^2=(15.6667-12{)}^2={\displaystyle \frac{121}{9}}=13\frac{4}{9}\doteq 13.4444 \\ {s}_6=(m-17{)}^2=(15.6667-17{)}^2={\displaystyle \frac{16}{9}}=1\frac{7}{9}\doteq 1.7778 \\ {s}_7=(m-13{)}^2=(15.6667-13{)}^2={\displaystyle \frac{64}{9}}=7\frac{1}{9}\doteq 7.1111 \\ {s}_8=(m-21{)}^2=(15.6667-21{)}^2={\displaystyle \frac{256}{9}}=28\frac{4}{9}\doteq 28.4444 \\ {s}_9=(m-14{)}^2=(15.6667-14{)}^2={\displaystyle \frac{25}{9}}=2\frac{7}{9}\doteq 2.7778 \\ {s}_{10}=(m-21{)}^2=(15.6667-21{)}^2={\displaystyle \frac{256}{9}}=28\frac{4}{9}\doteq 28.4444 \\ {s}_{11}=(m-15{)}^2=(15.6667-15{)}^2={\displaystyle \frac{4}{9}}\doteq 0.4444 \\ {s}_{12}=(m-22{)}^2=(15.6667-22{)}^2={\displaystyle \frac{361}{9}}=40\frac{1}{9}\doteq 40.1111 \\ s=(s_1+s_2+s_3+s_4+s_5+s_6+s_7+s_8+s_9+s_{10}+s_{11}+s_{12})\mathrm{/}12=(21.7778+0.4444+21.7778+0.1111+13.4444+1.7778+7.1111+28.4444+2.7778+28.4444+0.4444+40.1111)\mathrm{/}12=13.8889 \end{gathered}$$

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