The mean of 5 observations 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:

The correct option is C
2.8
Let the other two numbers be x and y.
Given, Variance, σ²=12.4
Mean x̄=5
n=5
∴1+2+6+x+y/5=5
⇒x+y=16
Variance, σ²=∑(xi-x̄)²/n=12.4
∑(xi-5)²=12.4*5=62
(1-5)²+(2-5)²+(6-5)²+(x-5)²+(y-5)²=62
16+9+1+(x-5)²+(y-5)²=62
(x-5)²+(y-5)²=36
Let x=11 and y=5
Then (11-5)²+(5-5)²=36
Observations are 1,2,5,6,11
Mean=25/5=5
Mean deviation from mean = |1-5|+|2-5|+|5-5|+|6-5|+|11-5|/5
=4+3+0+1+6/5=14/5=2.8