Question

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

• A. 30
• B. 42
• C. 40
• D. 20

Answer 1

Answer: (B)

According to the given information,
Coefficient of variance $(CV)=\frac{\sigma }{\mu }\times 100$
Where $\sigma$ is standard deviation and $\mu$ is mean of an ungrouped data.
$\because \frac{\sigma }{\mu }\times 100=60$ and $\sigma =21$
So, $\mu =\frac{21\times 10}{6}=35$
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, $\sigma =\sqrt{\frac{\Sigma {\left(\mu -x_i\right)}^2}{n}}=21$
$\Rightarrow \Sigma {\left(\mu -x_i\right)}^2=(21{)}^2\times n$
So, new standard deviation
${\sigma }^{{}^{\prime }}=\sqrt{\frac{\Sigma {\left({\mu }^{{}^{\prime }}-x_i^{{}^{\prime }}\right)}^2}{n}}=\sqrt{\frac{(2l{)}^2\times n}{n}}=21$
$\therefore$ New coefficient of variance
$=\frac{{\sigma }^{\prime }}{{\mu }^{\prime }}\times 100=\frac{21}{50}\times 100=21\times 2=42$