If 2 sets have 10 and 20 observations have coefficients of variation 50 and 60 respectively and arithmetic means 30 and 25 respectively, then the combined variance of those 30 observations is
Step-by-Step Solution
Key Concept: Combined variance involves pooling individual variances and squared deviations from the combined mean weighted by sample sizes.
Given $n_1 = 10, \bar{x}_1 = 30, CV_1 = 50$; $n_2 = 20, \bar{x}_2 = 25, CV_2 = 60$. First, find the standard deviations: $\sigma_1 = \frac{CV_1}{100} \times \bar{x}_1 = \frac{50}{100} \times 30 = 15$ and $\sigma_2 = \frac{CV_2}{100} \times \bar{x}_2 = \frac{60}{100} \times 25 = 15$. The combined mean is $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{10(30) + 20(25)}{30} = \frac{800}{30} = \frac{80}{3}$. Calculate deviations: $d_1 = 30 - \frac{80}{3} = \frac{10}{3}$ and $d_2 = 25 - \frac{80}{3} = -\frac{5}{3}$. The combined variance is $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2} = \frac{10(225 + \frac{100}{9}) + 20(225 + \frac{25}{9})}{30} = \frac{2475 + \frac{1000}{9}}{30} = \frac{225}{9} = 25$. Therefore, $CV = \frac{5}{\frac{80}{3}} \times 100 = \frac{9}{25}$.
Correct Answer: 2