Question

Let n be an odd natural number such that the variance of 1, 2, 3, 4, ..., n is 14. Then n is equal to _________.

Hint:

For any arithmetic progression with $n$ terms and common difference $d$, the variance is $d^2 \left( \frac{n^2 - 1}{12} \right)$.

Answer 1

Correct Answer: 13

Step 1: Understanding the Concept:
The variance of the first $n$ natural numbers is a standard statistical formula given by $\sigma^2 = \frac{n^2 - 1}{12}$.
Step 2: Detailed Explanation:
Given variance $\sigma^2 = 14$.
The set of observations is $\{1, 2, 3, ..., n\}$.
Applying the formula for variance of first $n$ natural numbers:
\[ \sigma^2 = \frac{n^2 - 1}{12} \]
\[ 14 = \frac{n^2 - 1}{12} \]
\[ n^2 - 1 = 14 \times 12 = 168 \]
\[ n^2 = 169 \]
\[ n = \sqrt{169} = 13 \quad (\text{since } n \text{ is a natural number}) \]
Since 13 is an odd natural number, it satisfies the given condition.
Step 3: Final Answer:
The value of $n$ is 13.