If the mean and variance of six observations 7, 10, 11, 15, a b, are 10 and $\frac{20}{3}$, respectively, then the value of |a − b| is equal to

Given,

Mean = 10

i.e. $\frac{7+10+11+15+\mathrm{a}+\mathrm{b}}{6}=10$

$\Rightarrow \mathrm{a}+\mathrm{b}=17$                ...(i)

Also, variance $=\frac{20}{3}$

i.e. $\frac{(7{)}^2+(10{)}^2+(11{)}^2+(15{)}^2+{\mathrm{a}}^2+{\mathrm{b}}^2}{6}-(\text{ Mean }{)}^2=\frac{20}{3}$

$\Rightarrow \frac{495+{\mathrm{a}}^2+{\mathrm{b}}^2}{6}-100=\frac{20}{3}$

$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=145$           ...(ii)

Using Eqs. (i) and (ii), put b = 17 - a

$\begin{array}{l}{\mathrm{a}}^2+(17-\mathrm{a}{)}^2=145\\ \Rightarrow {\mathrm{a}}^2+289+{\mathrm{a}}^2-34\mathrm{a}=145\\ \Rightarrow 2{\mathrm{a}}^2-34\mathrm{a}+144=0\\ \Rightarrow {\mathrm{a}}^2-17\mathrm{a}+72=0\\ \Rightarrow (\mathrm{a}-9)(\mathrm{a}-8)=0\\ \therefore \mathrm{a}=9\text{ or }\mathrm{a}=8\end{array}$

Using b = 17 - a

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}=8\text{ or }\mathrm{b}=9\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}|\mathrm{a}-\mathrm{b}|=1\end{array}$