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(D) Let $\bar{x}_1$ and $\bar{x}_2$ be the means and $\sigma_1^2$ and $\sigma_2^2$ be the variances of the two distributions.Given $\sigma_1^2 = 144$

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$35$

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(D) Let $\bar{x}_1$ and $\bar{x}_2$ be the means and $\sigma_1^2$ and $\sigma_2^2$ be the variances of the two distributions.Given $\sigma_1^2 = 144$ and $\sigma_2^2 = 64$,we have $\sigma_1 = 12$ and $\sigma_2 = 8$.The coefficient of variation $(CV)$ is given by $CV = \frac{\sigma}{\bar{x}} 	imes 100$.For the first distribution: $\frac{12}{\bar{x}_1} 	imes 100 = 40 \Rightarrow \bar{x}_1 = \frac{1200}{40} = 30$.For the second distribution: $\frac{8}{\bar{x}_2} 	imes 100 = 20 \Rightarrow \bar{x}_2 = \frac{800}{20} = 40$.The mean of the arithmetic means is $\frac{\bar{x}_1 + \bar{x}_2}{2} = \frac{30 + 40}{2} = \frac{70}{2} = 35$.

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
$(A)$ $40$
$(B)$ $12$
$(C)$ $30$
$(D)$ $35$

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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)$ $35$