Question

If the variance of $1, 2, 3, 4, 5, ..., 10$ is $ \frac{99}{12}, $ then the standard deviation of $3, 6, 9, 12, ...,30$ is

• A. $ \frac{297}{4} $
• B. $ \frac{3}{2}\sqrt{33} $
• C. $ \frac{3}{2}\sqrt{99} $
• D. $ \sqrt{\frac{99}{12}} $
• E. $ \frac{3\sqrt{3}}{2} $

Answer 1

Answer: (B)

Given, $ \sigma _{10}^{2}=\frac{99}{12}=\frac{33}{4} $
$ \Rightarrow $ $ {{\sigma }_{10}}=\frac{\sqrt{33}}{2} $
SD of required series $=3{{\sigma }_{10}} $ $=\frac{3\sqrt{33}}{2} $