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If the coefficients of variation of two distributions are 40 and 20 and their variances are 144 and 64 respectively, then the mean of their arithmetic means is
(a) 40
(b) 12
(c) 30
(d) 35
Options:
(a) 40
(b) 12
(c) 30
(d) 35
Solution:
2852 Upvotes
Verified Answer
The correct answer is:
35
Let $\bar{x}_1, \bar{x}_2$ be the means and $\sigma_1^2, \sigma_2^2$ be the variances of two distributions.
Then, we have, $\sigma_1^2=144, \sigma_2^2=64$,
$$
\frac{\sigma_1}{\bar{x}_1} \times 100=40 \text { and } \frac{\sigma_2}{\bar{x}_2} \times 100=20
$$
Now,
$$
\begin{gathered}
\sigma_1^2=144 \Rightarrow \sigma=12, \\
\sigma_2^2=64 \Rightarrow \sigma_2=8
\end{gathered}
$$
$$
\Rightarrow \frac{\sigma_1}{\bar{x}_1} \times 100=40 \Rightarrow \frac{12}{\bar{x}_1} \times 100=40
$$
$$
\Rightarrow \quad \bar{x}_1=\frac{1200}{40}=30 \text { and } \frac{\sigma_2}{x_2} \times 100=20
$$
$$
\Rightarrow \frac{8}{x_2} \times 100=20 \Rightarrow x_2=\frac{800}{20}=40
$$
Hence, the mean of $\bar{x}_1$ and $\bar{x}_2=\frac{\bar{x}_1+\bar{x}_2}{2}$
$$
=\frac{30+40}{2}=\frac{70}{2}=35
$$
Then, we have, $\sigma_1^2=144, \sigma_2^2=64$,
$$
\frac{\sigma_1}{\bar{x}_1} \times 100=40 \text { and } \frac{\sigma_2}{\bar{x}_2} \times 100=20
$$
Now,
$$
\begin{gathered}
\sigma_1^2=144 \Rightarrow \sigma=12, \\
\sigma_2^2=64 \Rightarrow \sigma_2=8
\end{gathered}
$$
$$
\Rightarrow \frac{\sigma_1}{\bar{x}_1} \times 100=40 \Rightarrow \frac{12}{\bar{x}_1} \times 100=40
$$
$$
\Rightarrow \quad \bar{x}_1=\frac{1200}{40}=30 \text { and } \frac{\sigma_2}{x_2} \times 100=20
$$
$$
\Rightarrow \frac{8}{x_2} \times 100=20 \Rightarrow x_2=\frac{800}{20}=40
$$
Hence, the mean of $\bar{x}_1$ and $\bar{x}_2=\frac{\bar{x}_1+\bar{x}_2}{2}$
$$
=\frac{30+40}{2}=\frac{70}{2}=35
$$
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