Question

If 1 is added to first 10 natural numbers, then variance of the numbers so obtained is

• A. 8.25
• B. 3.87
• C. 6.5
• D. 2.87

Hint:

Never waste time adding constants and recalculating means during exams! Adding a constant shifts the whole data block but keeps the internal gaps identical, so simply calculate $\frac{n^2-1}{12}$ for the original count.

Answer 1

The Correct Option is A

Step 1: Use the shift rule.
Adding a constant to every value does not change the variance. So adding 1 to each number leaves the variance the same as for the first 10 natural numbers.

Step 2: Apply the formula.
Variance of first $n$ natural numbers is $\dfrac{n^2 - 1}{12}$.

Step 3: Compute.
\[ \frac{10^2 - 1}{12} = \frac{99}{12} = 8.25 \]
\[ \boxed{8.25,\ \text{option 1}} \]