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Q. 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

AP EAMCETAP EAMCET 2019

Solution:

According to the given information,
Coefficient of variance $(C V)=\frac{\sigma}{\mu} \times 100$
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 $\mu^{'}=35+15=50$ but
$\Sigma\left|\mu^{'}-x_{i}^{'}\right|=\Sigma\left|\mu-x_{i}\right|,\left[\right.$ where $\left.x_{i}^{'}=x_{i}+15\right]$
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
$\sigma^{'}=\sqrt{\frac{\Sigma\left(\mu^{'}-x_{i}^{'}\right)^{2}}{n}}=\sqrt{\frac{(2 l )^{2} \times n}{n}}=21$
$\therefore $ New coefficient of variance
$=\frac{\sigma^{\prime}}{\mu^{\prime}} \times 100=\frac{21}{50} \times 100=21 \times 2=42$