The mean of 5 observation is 5 and their variance is 12.4. If three of the observations are 1, 2 and 6; then the mean deviation from the mean of the data is:

Given: 5 observations, mean = 5 ⇒ total sum = \(5\times5=25\). Three values are 1, 2, 6. Let the other two be \(a,b\).

From mean: \(a+b=25-(1+2+6)=16\).

Variance (about mean): \(12.4\) ⇒ \(\sum (x_i-5)^2 = 5\times12.4 = 62\).
Known part: \((1-5)^2+(2-5)^2+(6-5)^2=16+9+1=26\).
Hence \((a-5)^2+(b-5)^2 = 62-26 = 36\).

Let \(u=a-5,\ v=b-5\). Then \(u+v=(a+b)-10=6\) and \(u^2+v^2=36\).
\((u+v)^2 = u^2+v^2+2uv \Rightarrow 36 = 36 + 2uv \Rightarrow uv=0\).
So one of \(u,v\) is 0 ⇒ one of \(a,b\) is 5, the other is \(16-5=11\).

Mean deviation about mean:
\(\displaystyle \text{MD}=\frac{1}{5}\big(|1-5|+|2-5|+|6-5|+|5-5|+|11-5|\big) \) \( =\frac{1}{5}(4+3+1+0+6)=\frac{14}{5}=2.8.\)

Answer: 2.8