Step 1: Understanding the Concept:
For \(n\) observations \(x_1, x_2, \dots, x_n\), the mean is \(\bar{x} = \frac{\sum x_i}{n}\) and the variance is \(\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2\).
We are given partial sums for the first \((n-1)\) terms, which allows us to find expressions for the \(n\)-th term \(x_n\) and its square \(x_n^2\).
Step 2: Key Formula or Approach:
Using the given mean \(\bar{x} = 8\), we get \(\sum_{i=1}^n x_i = 8n\).
Using the given variance \(\sigma^2 = 16\), we get \(\frac{\sum_{i=1}^n x_i^2}{n} - 8^2 = 16\).
This simplifies to \(\sum_{i=1}^n x_i^2 = 80n\).
Step 3: Detailed Explanation:
We are given the sum of the first \((n-1)\) observations.
\[ \sum_{i=1}^{n-1} x_i = 48 \]
This allows us to write the \(n\)-th observation in terms of \(n\).
\[ x_n = \sum_{i=1}^n x_i - \sum_{i=1}^{n-1} x_i = 8n - 48 \]
We are also given the sum of squares of the first \((n-1)\) observations.
\[ \sum_{i=1}^{n-1} x_i^2 = 496 \]
This allows us to write the square of the \(n\)-th observation.
\[ x_n^2 = \sum_{i=1}^n x_i^2 - \sum_{i=1}^{n-1} x_i^2 = 80n - 496 \]
Equating the square of \(x_n\) from the first relation to the second relation gives a quadratic equation.
\[ (8n - 48)^2 = 80n - 496 \]
Factor out constants to heavily simplify the calculation.
\[ [8(n - 6)]^2 = 16(5n - 31) \]
\[ 64(n^2 - 12n + 36) = 16(5n - 31) \]
Divide both sides by 16.
\[ 4(n^2 - 12n + 36) = 5n - 31 \]
\[ 4n^2 - 48n + 144 - 5n + 31 = 0 \]
\[ 4n^2 - 53n + 175 = 0 \]
Now, solve this quadratic equation for \(n\).
\[ n = \frac{53 \pm \sqrt{53^2 - 4(4)(175)}}{2(4)} \]
\[ n = \frac{53 \pm \sqrt{2809 - 2800}}{8} = \frac{53 \pm \sqrt{9}}{8} = \frac{53 \pm 3}{8} \]
This yields two mathematical possibilities.
\(n = \frac{56}{8} = 7\) or \(n = \frac{50}{8} = 6.25\).
Since the number of observations \(n\) must be a positive integer, we conclude that \(n = 7\).
Step 4: Final Answer:
The value of \(n\) is \(7\).