Variance

We have ten observations: $$2,4,6,8,10,12,14,16,18,x$$ whose mean is given to be $$11$$.

1. Determine the unknown observation $$x$$.
If the mean of the ten numbers is $$11$$, then their total sum must be
$$\text{Sum}= (\text{mean}) \times (\text{number of observations}) = 11 \times 10 = 110.$$

The sum of the first nine known observations is an arithmetic-progression sum:
$$2+4+6+\dots+18 = \frac{9}{2}\,(2+18)=\frac{9}{2}\times20 = 90.$$

Hence
$$x = 110 - 90 = 20.$$

The complete data set is now $$\{2,4,6,8,10,12,14,16,18,20\}.$$

2. Compute the variance.
For a set of $$n$$ observations $$(x_1,x_2,\dots,x_n)$$ with mean $$\bar{x}$$, the variance is
$$\sigma^{2}= \frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^{2}.$$

Here, $$n=10$$ and $$\bar{x}=11$$. Calculate each deviation $$x_i-\bar{x}$$ and then square it:

$$\begin{aligned} 2-11 &= -9 &\Rightarrow&\; (-9)^2 = 81\\ 4-11 &= -7 &\Rightarrow&\; (-7)^2 = 49\\ 6-11 &= -5 &\Rightarrow&\; (-5)^2 = 25\\ 8-11 &= -3 &\Rightarrow&\; (-3)^2 = 9\\ 10-11 &= -1 &\Rightarrow&\; (-1)^2 = 1\\ 12-11 &= 1 &\Rightarrow&\; 1^2 = 1\\ 14-11 &= 3 &\Rightarrow&\; 3^2 = 9\\ 16-11 &= 5 &\Rightarrow&\; 5^2 = 25\\ 18-11 &= 7 &\Rightarrow&\; 7^2 = 49\\ 20-11 &= 9 &\Rightarrow&\; 9^2 = 81\\ \end{aligned}$$

Add these squared deviations:
$$81+49+25+9+1+1+9+25+49+81 = 330.$$

Finally, divide by $$n=10$$ to get the variance:
$$\sigma^{2}= \frac{330}{10}=33.$$

Therefore, the variance of the given data is $$33$$.

Option A which is: $$33$$