Question

Let \(x_1, x_2, x_3, \ldots, x_n\) be ‘\(n\)’ observations such that \( \sum_{i=1}^{n-1} x_i = 48 \) and \( \sum_{i=1}^{n-1} x_i^2 = 496 \). If mean and variance of the distribution are 8 and 16 respectively then value of \(n\) is:

• A. 7
• B. 9
• C. 8
• D. 12

Hint:

When mean and variance are given, first express the missing observation using the mean formula, then substitute it into the variance formula. This is the fastest way to find the unknown number of observations.

Answer 1

The Correct Option is A

Step 1: Apply the formula for the mean.
The mean of \( n \) observations is given as 8. Therefore, \[ \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = 8 \] It is given that the sum of the first \( n-1 \) observations is: \[ \sum_{i=1}^{n-1} x_i = 48 \] Let the last observation be \( x_n \). Then, \[ \frac{48 + x_n}{n} = 8 \] Multiplying both sides by \( n \): \[ 48 + x_n = 8n \] \[ x_n = 8n - 48 \]
Step 2: Apply the formula for variance.
The variance of the observations is given by: \[ \sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^{\,2} \] Here, variance \( = 16 \) and mean \( = 8 \). Substituting these values: \[ 16 = \frac{\sum x_i^2}{n} - 64 \] \[ \frac{\sum x_i^2}{n} = 80 \] \[ \sum x_i^2 = 80n \] It is also given that: \[ \sum_{i=1}^{n-1} x_i^2 = 496 \] Therefore, \[ 496 + x_n^2 = 80n \] Substituting \( x_n = 8n - 48 \): \[ 496 + (8n - 48)^2 = 80n \]
Step 3: Simplify and solve the quadratic equation.
Expanding the square: \[ 496 + 64n^2 - 768n + 2304 = 80n \] Combining like terms: \[ 64n^2 - 848n + 2800 = 0 \] Dividing the entire equation by 8: \[ 8n^2 - 106n + 350 = 0 \] Factorizing: \[ 8n^2 - 56n - 50n + 350 = 0 \] \[ 8n(n - 7) - 50(n - 7) = 0 \] \[ (n - 7)(8n - 50) = 0 \] \[ (n - 7)(4n - 25) = 0 \] Thus, \[ n = 7 \quad \text{or} \quad n = \frac{25}{4} \] Since the number of observations must be a natural number, we select: \[ n = 7 \]
Step 4: Conclusion.
Therefore, the total number of observations in the data set is \( 7 \).
Final Answer: \( 7 \)