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: Use the formula of mean.
Since the mean of \(n\) observations is 8, we have:
\[ \frac{x_1+x_2+x_3+\cdots+x_n}{n}=8 \] It is given that:
\[ \sum_{i=1}^{n-1}x_i=48 \] So, if the last observation is \(x_n\), then:
\[ \frac{48+x_n}{n}=8 \] \[ 48+x_n=8n \] \[ x_n=8n-48 \]
Step 2: Use the formula of variance.
Variance is given as:
\[ \sigma^2=\frac{\sum x_i^2}{n}-\bar{x}^{\,2} \] Here, variance \(=16\) and mean \(=8\). Hence:
\[ 16=\frac{\sum x_i^2}{n}-64 \] \[ \frac{\sum x_i^2}{n}=80 \] \[ \sum x_i^2=80n \] Now, it is 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 the equation.
\[ 496+64n^2-768n+2304=80n \] \[ 64n^2-848n+2800=0 \] Divide by 8:
\[ 8n^2-106n+350=0 \] Now factorizing:
\[ 8n^2-56n-50n+350=0 \] \[ 8n(n-7)-50(n-7)=0 \] \[ (n-7)(8n-50)=0 \] \[ (n-7)(4n-25)=0 \] So,
\[ n=7 \quad \text{or} \quad n=\frac{25}{4} \] Since \(n\) must be a natural number, we get:
\[ n=7 \]
Step 4: Conclusion.
Hence, the required number of observations is \(7\).
Final Answer: \(7\)