The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are 2, 3, 5, 10, 11 , 13, 15, 21, then the mean deviation about the median of all the 10 observations is

We need to find the mean deviation about the median of 10 observations.

Mean = 9, Variance = 34.2. Known 8 observations: 2, 3, 5, 10, 11, 13, 15, 21.

Sum of known 8 = 2+3+5+10+11+13+15+21 = 80

Let the two unknown values be a and b.

$$80 + a + b = 90 \implies a + b = 10 \quad \cdots (1)$$

For variance: $$\sum x_i^2/n - \bar{x}^2 = 34.2$$

$$\sum x_i^2/10 - 81 = 34.2 \implies \sum x_i^2 = 1152$$

Sum of squares of known 8: $$4+9+25+100+121+169+225+441 = 1094$$

$$a^2 + b^2 = 1152 - 1094 = 58 \quad \cdots (2)$$

From (1) and (2): $$a + b = 10$$ and $$a^2 + b^2 = 58$$

$$(a+b)^2 = a^2 + 2ab + b^2 = 100 \implies 2ab = 42 \implies ab = 21$$

$$a, b$$ are roots of $$t^2 - 10t + 21 = 0 \implies t = 3, 7$$

So the two missing values are 3 and 7.

2, 3, 3, 5, 7, 10, 11, 13, 15, 21

Median = $$(7 + 10)/2 = 8.5$$

$$MD = \frac{1}{10}\sum|x_i - 8.5|$$

$$= \frac{|2-8.5|+|3-8.5|+|3-8.5|+|5-8.5|+|7-8.5|+|10-8.5|+|11-8.5|+|13-8.5|+|15-8.5|+|21-8.5|}{10}$$

$$= \frac{6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5}{10} = \frac{50}{10} = 5$$

Therefore, the mean deviation is Option 1: 5.