Question

Calculate the coefficient of variation of the given continuous series.
$\begin{array}{ccccccccc}\text{more than}& 0& 10& 20& 30& 40& 50& 60& 70\\ \text{Cumulative Frequency}& 100& 90& 75& 50& 20& 10& 5& 0\end{array}$

Answer 1

Since, cumulative frequencies are give, we first convert them into simple frequnecies
$\begin{array}{cccccccc}\text{Class Interval}& \text{frequency}& m& \text{d = m-A}& d^{\prime }=\frac{d}{c}& f{d}^{\prime }& (d^{\prime }{)}^2& f(d^{\prime }{)}^2\\ & & & A=35& c=10& & & \\ 0-10& 10& 5& -30& -3& -30& 9& 90\\ 10-20& 15& 15& -20& -2& -30& 4& 60\\ 20-30& 25& 25& -10& -1& -25& 1& 25\\ 30-40& 30& 34& 0& 0& 0& 0& 0\\ 40-50& 10& 45& 10& 1& 10& 1& 10\\ 50-60& 5& 55& 20& 2& 10& 4& 20\\ 60-70& 5& 65& 30& 3& 15& 9& 45\\ 70-80& 0& 75& 40& 4& 0& 16& 0\\ & N=100& & & & \sum f{d}^{\prime }=-50& & \sum f(d^{\prime }{)}^2=250\end{array}$
Thus, mean $\overline{x}=A+\frac{\sum f{d}^{\prime }}{N}\times h$
$=35+\frac{(-50)}{100}\times 10$
$\therefore \overline{x}=35-5=30$
variance ${\sigma }^2=\frac{h^2}{N^2}[N\sum f(d^{\prime }{)}^2-(\sum f{d}^{\prime }{)}^2]$
$=\frac{{10}^2}{(100{)}^2}\left[100\right(250)-(-50{)}^2]$
$=\left(\frac{1}{10}{)}^2\right[25000-250]$
$=\left(0.1{)}^2\right[24750]$
$=0.01\left[247.50\right]$
$\therefore \sigma =\sqrt{247.5}=15.73$
Hence, coefficient of variance$=\frac{\sigma }{\overline{x}}\times 100$

$=\frac{15.73}{30}\times 100$
$=\frac{1573}{30}$
$=52.44$