Question

If $A$ and $B$ are the variances of the $1^{st}$ $n$ even numbers and $1^{st}$ $'n'$ odd numbers respectively then

• A. A = B
• B. A > B
• C. A < B
• D. A = B + 1

Answer 1

Answer: (A)

We know that variance of $n$ number which are in A.P. whose first terms is 'a' and common difference $d$ is
$\sigma^{2}=\frac{d^{2}\left(n^{2}-1\right)}{12}$
$\therefore \operatorname{Var}\left(\sigma_{1}\right)=\frac{2^{2}\left(n^{2}-1\right)}{12}=\frac{\left(n^{2}-1\right)}{3}=A$
$\operatorname{Var}\left(\sigma_{2}\right)=\frac{2^{2}\left(n^{2}-1\right)}{12}-\frac{n^{2}-1}{3}=B$
$\therefore A=B$