Question

What is the variance of the first $11$ natural numbers?

• A. $10$
• B. $11$
• C. $12$
• D. $13$

Answer 1

The correct option is A $10$

Sum of first $11$ natural numbers $=\frac{n(n+1)}{2}=\frac{11\times 12}{2}=66$

So, mean $=\frac{66}{11}=6$

Variance $=\frac{(1-6{)}^2+(2-6{)}^2+(3-6{)}^2+........+(11-6{)}^2}{11}$

$=\frac{25+16+9+4+1+0+1+4+9+16+25}{11}=\frac{110}{11}=10$

Hence, A is correct.