Question

A number equal to $2$ times the mean and with a frequency equal to $k$ is inserted in a data having $n$ observations. If the new mean is $\frac{4}{3}$ times the old mean, then the value of $\frac{k}{n}$ is

Answer 1

Answer: 0.5

${\bar{x}}_{new}=\frac{4}{3}\bar{x}$
From given information
${\bar{x}}_{new}=\frac{n\bar{x}+2k\bar{x}}{n+k}=\frac{4}{3}\bar{x}$
$\Rightarrow 3n+6k=4n+4k$
$\Rightarrow 2k=n$
$\Rightarrow \frac{k}{n}=0.5$