Question

The variance of first $100$ odd natural numbers is

• A. $2222$
• B. $3333$
• C. $4444$
• D. $5555$

Answer 1

Answer: (B)

Variance $=\frac{1^2+3^2+5^2+\dots ..+{\left(199\right)}^2}{100}-{\left(\frac{1+3+5+\dots ..+199}{100}\right)}^2$
Sum $=\sum _{n=1}^n{\left(2n-1\right)}^2=4\Sigma n^2-4\Sigma n+n$
Sum $=\frac{4n\left(n+1\right)\left(2n+1\right)}{6}-\frac{4n\left(n+1\right)}{2}+n$
Put $n=100$
$=\frac{4\times 100\times 101\times 201}{6}-2\times 100\times 101+100$
$=100\left(\frac{2\times 101\times 201}{3}-2\times 101+1\right)=100\left(2\times 6767-202+1\right)$
$=100\left(13534-201\right)=100\times 13333=1333300$
Then, Variance $=\frac{1333300}{100}-{\left(100\right)}^2=13333-10000=3333$