The mean and variance of $$n$$ observations are 8 and 16, respectively. If the sum of the first $$(n-1)$$ observations is 48 and the sum of squares of the first $$(n-1)$$ observations is 496, then the value of $$n$$ is :

Let the $$n$$ observations be $$x_1,\,x_2,\,\ldots,\,x_{n-1},\,x_n$$.

Step 1 : Convert mean and variance into sums
Mean $$\bar{x}=8 \Longrightarrow \displaystyle\sum_{i=1}^{n}x_i = n\bar{x}=8n$$ $$-(1)$$

Variance $$\sigma^2 = 16$$ is defined as
$$\sigma^2=\frac1n\displaystyle\sum_{i=1}^{n}(x_i-\bar{x})^{2}$$

Using the identity $$\displaystyle\sum(x_i-\bar{x})^{2}= \sum x_i^{2}-n\bar{x}^{2}$$, we get
$$16=\frac1n\left(\sum x_i^{2}-n\cdot8^{2}\right)$$
$$\Longrightarrow \sum_{i=1}^{n}x_i^{2}=16n+64n=80n$$ $$-(2)$$

Step 2 : Separate the last observation
Given $$\displaystyle\sum_{i=1}^{n-1}x_i = 48$$, so the last observation is
$$x_n = \left(\sum_{i=1}^{n}x_i\right)-\left(\sum_{i=1}^{n-1}x_i\right)=8n-48$$ $$-(3)$$

Similarly, the sum of squares of the first $$n-1$$ observations is 496, hence
$$x_n^{2}= \left(\sum_{i=1}^{n}x_i^{2}\right)-496 = 80n-496$$ $$-(4)$$

Step 3 : Form and solve the equation for $$n$$
Square of $$x_n$$ from (3) must equal the expression in (4):
$$(8n-48)^{2}=80n-496$$

Expand and simplify:
$$(8n-48)^{2}=64n^{2}-768n+2304$$
$$64n^{2}-768n+2304=80n-496$$
$$64n^{2}-848n+2800=0$$
Divide by 8:
$$8n^{2}-106n+350=0$$

Solve the quadratic:
Discriminant $$D=106^{2}-4\cdot8\cdot350=11236-11200=36$$
$$n=\dfrac{106\pm\sqrt{36}}{16}= \dfrac{106\pm6}{16}$$
$$n_1=\dfrac{112}{16}=7,\qquad n_2=\dfrac{100}{16}=6.25$$

Since $$n$$ must be a positive integer, $$n=7$$.

Hence the required value of $$n$$ is $$7$$.

Option D which is: $$7$$