The mean of a set of 10 observations is 5 and | vedclass

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(B) Let the first set have $n_1 = 10$ observations with mean $\bar{x}_1 = 5$ and standard deviation $\sigma_1 = 2\sqrt{6}$.Variance $\sigma_1^2 = (2\s

Vedclass · Accepted Answer

$2\sqrt{5}$

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(B) Let the first set have $n_1 = 10$ observations with mean $\bar{x}_1 = 5$ and standard deviation $\sigma_1 = 2\sqrt{6}$.Variance $\sigma_1^2 = (2\sqrt{6})^2 = 4 	imes 6 = 24$.Since $\sigma_1^2 = \frac{\sum x_i^2}{n_1} - \bar{x}_1^2$,we have $24 = \frac{\sum x_i^2}{10} - 5^2$,so $\sum x_i^2 = 10(24 + 25) = 490$.Let the second set have $n_2 = 20$ observations with mean $\bar{x}_2 = 5$ and standard deviation $\sigma_2 = 3\sqrt{2}$.Variance $\sigma_2^2 = (3\sqrt{2})^2 = 9 	imes 2 = 18$.Since $\sigma_2^2 = \frac{\sum y_i^2}{n_2} - \bar{x}_2^2$,we have $18 = \frac{\sum y_i^2}{20} - 5^2$,so $\sum y_i^2 = 20(18 + 25) = 860$.The combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{10(5) + 20(5)}{30} = 5$.The combined variance $\sigma^2 = \frac{\sum x_i^2 + \sum y_i^2}{n_1 + n_2} - \bar{x}^2 = \frac{490 + 860}{30} - 5^2 = \frac{1350}{30} - 25 = 45 - 25 = 20$.The combined standard deviation $\sigma = \sqrt{20} = 2\sqrt{5}$.

Classes	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
Frequency $({f_i})$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

Classes	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Frequency	$11$	$29$	$18$	$4$	$5$	$3$

The mean of a set of $10$ observations is $5$ and the standard deviation is $2\sqrt{6}$. If the mean of another set of $20$ observations is $5$ and the standard deviation is $3\sqrt{2}$,what is the standard deviation of the combined set of $30$ observations?

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