The coefficient of variation and standard deviation of an ungrouped data are 60 and 21 respectively. If 15…
The coefficient of variation and standard deviation of an ungrouped data are 60 and 21 respectively. If 15 is added to every observation of the data, then the coefficient of variation of the new data is
30
42
40
20
Solution
According to the given information,
Coefficient of variance $(\mathrm{CV})=\frac{\sigma}{\mu} \times 100$
Where $\sigma$ is standard deviation and $\mu$ is mean of an ungrouped data.
$\because \quad \frac{\sigma}{\mu} \times 100=60$ and $\sigma=21$
So, $\quad \mu=\frac{21 \times 10}{6}=35$
After adding 15 to each observation of the data, the new mean $\mu^{\prime}=35+15=50$ but
$\Sigma\left|\mu^{\prime}-x_i^{\prime}\right|=\Sigma\left|\mu-x_i\right|$, [where $x_i^{\prime}=x_i+15$ ]
Now, $\sigma=\sqrt{\frac{\sum\left(\mu-x_i\right)^2}{n}}=21$
$\Rightarrow \Sigma\left(\mu-x_i\right)^2=(21)^2 \times n$
So, new standard deviation
$
\sigma^{\prime}=\sqrt{\frac{\sum\left(\mu^{\prime}-x_i^{\prime}\right)^2}{n}}=\sqrt{\frac{(21)^2 \times n}{n}}=21
$
$\therefore$ New coefficient of variance
$
=\frac{\sigma^{\prime}}{\mu^{\prime}} \times 100=\frac{21}{50} \times 100=21 \times 2=42
$
Hence, option (b) is correct