The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is:

We have a total of five observations whose arithmetic mean is given to be $$\mu = 4$$. By definition of the mean,

$$\mu \;=\;\dfrac{\text{Sum of all observations}}{n},\qquad n = 5.$$

Substituting the known values,

$$4 \;=\;\dfrac{\text{Sum of all observations}}{5}\;\;\Longrightarrow\;\;\text{Sum of all observations}=4\times 5=20.$$

Out of the five observations, three are already known: $$3,\;4,\;4.$$ Let the remaining two unknown observations be denoted by $$x$$ and $$y.$$ Then, from the sum obtained above,

$$x + y + 3 + 4 + 4 = 20$$

$$\Longrightarrow\;x + y = 20 - (3 + 4 + 4) = 20 - 11 = 9.$$

Next, we use the information about the variance. For a set of $$n$$ observations, the (population) variance is defined as

$$\sigma^{2} = \dfrac{\displaystyle\sum_{i=1}^{n}(x_{i} - \mu)^{2}}{n}.$$

Here $$\sigma^{2}=5.20$$ and $$n = 5,$$ so

$$\sum_{i=1}^{5}(x_{i}-4)^{2}=n\sigma^{2}=5\times 5.20=26.$$

We now separate this total into the contributions from the known observations and the unknown ones.

The squared deviations of the known observations are

$$\begin{aligned} (3-4)^{2} &= (-1)^{2}=1,\\ (4-4)^{2} &= 0,\\ (4-4)^{2} &= 0. \end{aligned}$$

Hence the combined contribution from the known observations is $$1+0+0=1.$$ Therefore, the contribution from the unknown observations must be

$$26 - 1 = 25.$$

That is,

$$(x-4)^{2} + (y-4)^{2}=25.$$

We now expand the left-hand side algebraically. Using the identity $$(a-b)^{2}=a^{2}-2ab+b^{2},$$ we get

$$(x-4)^{2}=x^{2}-8x+16,\qquad (y-4)^{2}=y^{2}-8y+16.$$

Adding these two,

$$(x-4)^{2} + (y-4)^{2}=x^{2}+y^{2} - 8(x+y) + 32.$$

But we have already shown that $$(x-4)^{2} + (y-4)^{2}=25,$$ and we also know $$x+y=9.$$ Substituting $$x+y=9$$ into the expanded expression gives

$$x^{2}+y^{2} - 8(9) + 32 = 25.$$

Simplifying step by step,

$$x^{2}+y^{2} - 72 + 32 = 25,$$

$$x^{2}+y^{2} - 40 = 25,$$

$$x^{2}+y^{2} = 25 + 40 = 65.$$

So far we have

$$x + y = 9, \qquad x^{2} + y^{2} = 65.$$

To find the absolute difference $$|x - y|,$$ we employ the identity

$$(x - y)^{2} = x^{2} + y^{2} - 2xy.$$

Therefore we must first obtain $$xy.$$ Squaring the sum $$x + y = 9$$ gives

$$(x + y)^{2} = 9^{2} \;\;\Longrightarrow\;\;x^{2} + 2xy + y^{2} = 81.$$

We already have $$x^{2} + y^{2} = 65,$$ so substituting this value into the equation above yields

$$65 + 2xy = 81 \;\;\Longrightarrow\;\;2xy = 81 - 65 = 16 \;\;\Longrightarrow\;\;xy = 8.$$

Now we are ready to compute $$(x - y)^{2}:$$

$$(x - y)^{2} = x^{2} + y^{2} - 2xy = 65 - 16 = 49.$$

Taking the positive square root (because we are asked for the absolute value),

$$|x - y| = \sqrt{49} = 7.$$

Hence, the correct answer is Option C.