Let $$\alpha, \beta \in {R}$$. Let the mean and the variance of 6 observations −3, 4, 7, −6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is:

Mean = 2: (-3+4+7-6+α+β)/6 = 2 → α+β = 10

Variance = 23: (1/6)∑(xi-2)² = 23

∑(xi-2)² = 25+4+25+64+(α-2)²+(β-2)² = 138

118 + (α-2)² + (β-2)² = 138

(α-2)² + (β-2)² = 20

With α+β=10: let α=2+a, β=2+b, then a+b=6 and a²+b²=20

2ab = 36-20 = 16, ab = 8. So a,b are roots of t²-6t+8=0 → t=2,4

α=4,β=6 or α=6,β=4

Mean deviation = (1/6)(|{-3}-2|+|4-2|+|7-2|+|{-6}-2|+|4-2|+|6-2|)

= (1/6)(5+2+5+8+2+4) = 26/6 = 13/3

The correct answer is Option 1: 13/3.