The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14 then the absolute difference of the remaining two observations is:

We have a total of 7 observations whose arithmetic mean is given to be $$8$$. By definition, the mean $$\bar x$$ of $$n$$ observations satisfies the relation

$$\bar x=\dfrac{\sum_{i=1}^{n}x_i}{n}\;.$$

Putting $$\bar x=8$$ and $$n=7$$ we obtain

$$\sum_{i=1}^{7}x_i \;=\; 7 \times 8 \;=\; 56.$$

Five of the observations are already known, namely $$2,4,10,12,14$$. Let the remaining two observations be denoted by $$a$$ and $$b$$. Summing the known five values,

$$2+4+10+12+14 \;=\; 42.$$

Substituting this in the previous result gives

$$42 + a + b \;=\; 56 \;\;\Longrightarrow\;\; a + b = 14.$$

Next, we use the information about the variance. For a set of $$n$$ observations, the (population) variance $$\sigma^{2}$$ is defined as

$$\sigma^{2}= \dfrac{\sum_{i=1}^{n}(x_i-\bar x)^{2}}{n}.$$

The question states that the variance equals $$16$$, so

$$16 \;=\; \dfrac{\sum_{i=1}^{7}(x_i-\bar x)^{2}}{7}\;.$$

Multiplying by $$7$$ we get the total of squared deviations:

$$\sum_{i=1}^{7}(x_i-\bar x)^{2}=16 \times 7 = 112.$$

A very convenient identity connects this quantity with the sum of squares of the observations:

$$\sum_{i=1}^{n}(x_i-\bar x)^{2}=\sum_{i=1}^{n}x_i^{2}-\dfrac{\left(\sum_{i=1}^{n}x_i\right)^{2}}{n}.$$

We already know $$\sum_{i=1}^{7}x_i=56$$, so

$$112 = \sum_{i=1}^{7}x_i^{2}-\dfrac{56^{2}}{7}.$$

Calculating the second term on the right,

$$\dfrac{56^{2}}{7}=\dfrac{3136}{7}=448.$$

Substituting this back, we have

$$112 = \sum_{i=1}^{7}x_i^{2}-448 \;\;\Longrightarrow\;\; \sum_{i=1}^{7}x_i^{2}=112+448=560.$$

Now we list the squares of the five known observations:

$$2^{2}=4,\;4^{2}=16,\;10^{2}=100,\;12^{2}=144,\;14^{2}=196.$$

Adding them gives

$$4+16+100+144+196 = 460.$$

Hence, the sum of the squares of the two unknown observations is

$$a^{2}+b^{2}=560-460=100.$$

At this point we have the simultaneous relations

$$a+b = 14,\qquad a^{2}+b^{2}=100.$$

To find the absolute difference $$|a-b|$$, we use the identity

$$a^{2}+b^{2}=(a+b)^{2}-2ab.$$

Substituting the known values,

$$100 = 14^{2} - 2ab \;\;\Longrightarrow\;\; 100 = 196 - 2ab$$

and therefore

$$2ab = 196 - 100 = 96 \;\;\Longrightarrow\;\; ab = 48.$$

Another standard identity is

$$(a-b)^{2} = (a+b)^{2} - 4ab.$$

Substituting $$a+b = 14$$ and $$ab = 48$$ we obtain

$$(a-b)^{2} = 14^{2} - 4 \times 48 = 196 - 192 = 4.$$

Taking the positive square root gives the absolute difference:

$$|a-b| = \sqrt{4} = 2.$$

Hence, the correct answer is Option C.