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The table below gives an incomplete frequency distribution with two missing frequencies $f_{1}$ and $f_{2}$ \begin{array}{|c|c|}
\hline Value of x & Frequency \\
\hline 0 & f_{1} \\
1 & f_{2} \\
2 & 4 \\
3 & 4 \\
4 & 3 \\
\hline
\end{array} The total frequency is 18 and the arithmetic mean of $x$ is 2 .
What is the coefficient of variance? $\quad$
Options:
\hline Value of x & Frequency \\
\hline 0 & f_{1} \\
1 & f_{2} \\
2 & 4 \\
3 & 4 \\
4 & 3 \\
\hline
\end{array} The total frequency is 18 and the arithmetic mean of $x$ is 2 .
What is the coefficient of variance? $\quad$
Solution:
1036 Upvotes
Verified Answer
The correct answer is:
$\frac{200}{3}$
\begin{array}{|c|c|c|}
\hlinex & f & x f \\
\hline 0 & f_{1} & 0 \\
1 & f_{2} & f_{2} \\
2 & 4 & 8 \\
3 & 4 & 12 \\
4 & 3 & 12 \\
\hline Total & f_{1}+f_{2}+11 & 32+f_{2} \\
\hline
\end{array} Since, total frequency is 18
$f_{1}+f_{2}+11=18$
$\Rightarrow f_{1}+f_{2}=7$
As we have, Mean $=\frac{\Sigma x f}{\Sigma f}=2$
(i)
$\frac{32+f_{2}}{18}=2 \Rightarrow f_{2}=36-32=4$
On putting the value of $f_{2}$ in Eq. (i), we get $f_{1}=7-4=3$
Coefficient of variance $=\frac{\sigma}{\bar{x}} \times 100$ where $\sigma=S . D$
$=\frac{4}{3} \times \frac{1}{2} \times 100=\frac{200}{3}$
\hlinex & f & x f \\
\hline 0 & f_{1} & 0 \\
1 & f_{2} & f_{2} \\
2 & 4 & 8 \\
3 & 4 & 12 \\
4 & 3 & 12 \\
\hline Total & f_{1}+f_{2}+11 & 32+f_{2} \\
\hline
\end{array} Since, total frequency is 18
$f_{1}+f_{2}+11=18$
$\Rightarrow f_{1}+f_{2}=7$
As we have, Mean $=\frac{\Sigma x f}{\Sigma f}=2$
(i)
$\frac{32+f_{2}}{18}=2 \Rightarrow f_{2}=36-32=4$
On putting the value of $f_{2}$ in Eq. (i), we get $f_{1}=7-4=3$
Coefficient of variance $=\frac{\sigma}{\bar{x}} \times 100$ where $\sigma=S . D$
$=\frac{4}{3} \times \frac{1}{2} \times 100=\frac{200}{3}$
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