Find Value of N Given Mean and Variance

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

We are given:

1. Mean of $n$ observations, $\bar{x} = 8$
2. Variance of $n$ observations, $\sigma^2 = 16$
3. Sum of the first $(n-1)$ observations, $\sum_{i=1}^{n-1} x_i = 48$
4. Sum of squares of the first $(n-1)$ observations, $\sum_{i=1}^{n-1} x_i^2 = 496$

Using the formula for the mean of $n$ observations: $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = 8$ $\implies \sum_{i=1}^{n} x_i = 8n$

Since $\sum_{i=1}^{n} x_i = \sum_{i=1}^{n-1} x_i + x_n$, we substitute the sum of the first $(n-1)$ observations: $48 + x_n = 8n$ $\implies x_n = 8n - 48 = 8(n - 6) \quad \text{--- (Equation 1)}$

Next, using the formula for variance: $\sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - (\bar{x})^2$

Substitute the given values of $\sigma^2 = 16$ and $\bar{x} = 8$: $16 = \frac{\sum_{i=1}^{n} x_i^2}{n} - 8^2$ $16 = \frac{\sum_{i=1}^{n} x_i^2}{n} - 64$ $\implies \frac{\sum_{i=1}^{n} x_i^2}{n} = 80$ $\implies \sum_{i=1}^{n} x_i^2 = 80n$

We know that: $\sum_{i=1}^{n} x_i^2 = \sum_{i=1}^{n-1} x_i^2 + x_n^2$

Substitute the given sum of squares of the first $(n-1)$ observations: $80n = 496 + x_n^2$

Now, substitute $x_n = 8(n - 6)$ from Equation 1: $80n = 496 + [8(n - 6)]^2$ $80n = 496 + 64(n - 6)^2$

Dividing the entire equation by $16$: $5n = 31 + 4(n - 6)^2$ $5n = 31 + 4(n^2 - 12n + 36)$ $5n = 31 + 4n^2 - 48n + 144$ $4n^2 - 53n + 175 = 0$

Solving this quadratic equation for $n$: $n = \frac{-(-53) \pm \sqrt{(-53)^2 - 4(4)(175)}}{2(4)}$ $n = \frac{53 \pm \sqrt{2809 - 2800}}{8}$ $n = \frac{53 \pm \sqrt{9}}{8}$ $n = \frac{53 \pm 3}{8}$

This gives two values for $n$: $n = \frac{56}{8} = 7 \quad \text{or} \quad n = \frac{50}{8} = 6.25$

Since $n$ represents the number of observations, it must be a positive integer. Thus, $n = 7$.

Therefore, the correct option is D.