Mean Absolute Deviation

10 =   $\frac{7+10+11+15+a+b}{6}\Rightarrow a+b=17$  . (i)

$\Rightarrow \frac{20}{3}=\frac{7^2+10^2+11^2+15^2+a^2+b^2}{6}-10^2$

a² + b² = 145       . (ii)

solving (i) and (ii) we get a = 8, b = 9 or a = 9, b = 8

|a – b|= 1

a, b, c, d, e be 5 unknown

n = 7, mean = 8, variance = 16

sum of observations = 7 * 8 = 56

mean of 5 remaining observation = $\frac{56-8-6}{25}=\frac{42}{5}$

$16=\frac{\sum x_i^2}{7}-64$

$\sum x_i^2=560$

$\begin{array}{l}{a}^2+b^2+c^2+d^2+e^2=460\\ =460-{\left(\frac{42}{5}\right)}^2\end{array}$

$=\frac{536}{25}$