The mean of 5 observations is 5 and their variance is 9.2. If three of the observations are 1, 2, and 6, then the mean deviation from the mean of the data, is

Let x1=1, x2=2, x3=6,x4,x5 be five observations. It is given that     Mean =5 and Variance =9.2⇒ x1+x2+x3+x4+x55=5and, x12+x22+x32+x42+x525−52=9.2⇒ 1+2+6+x4+x55=5and, 12+22+62+x42+x525−25=9.2⇒ x4+x5=16  and        x42+x52=130⇒ x4+x52=256  and    x42+x52=130⇒ 130+2x4x5=256 and  x42+x52=130⇒ x4x5=63   and   x42+x52=130∴ x4−x52=x4+x52−4x4x5=256−252=4            x4−x5=2Thus, we have      x4+x5=16 and, x4−x5=2⇒ x4=9 and,   x5=7The mean deviation M from the mean is given by              M=|1−5|+|2−5|+|6−5|+|9−5|+|7−5|5⇒ M=4+3+1+4+25=2.8