Consider a set of 3n numbers having variance | vedclass

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(B) Let the numbers be $x_1, x_2, \ldots, x_{2n}$ and $y_1, y_2, \ldots, y_n$.The mean of the first $2n$ numbers is $\bar{x} = 6$,so $\sum x_i = 12n$.

Vedclass · Accepted Answer

$68$

Vedclass · Answer

(B) Let the numbers be $x_1, x_2, \ldots, x_{2n}$ and $y_1, y_2, \ldots, y_n$.The mean of the first $2n$ numbers is $\bar{x} = 6$,so $\sum x_i = 12n$.The mean of the remaining $n$ numbers is $\bar{y} = 3$,so $\sum y_i = 3n$.The overall mean $\bar{X} = \frac{12n + 3n}{3n} = 5$.The variance is given by $\sigma^2 = \frac{\sum x_i^2 + \sum y_i^2}{3n} - (\bar{X})^2 = 4$.$4 = \frac{\sum x_i^2 + \sum y_i^2}{3n} - 25 \implies \sum x_i^2 + \sum y_i^2 = 87n$.In the new set,the numbers are $(x_i + 1)$ and $(y_i - 1)$.The new mean $\bar{X}' = \frac{\sum (x_i + 1) + \sum (y_i - 1)}{3n} = \frac{12n + 2n + 3n - n}{3n} = \frac{16n}{3n} = \frac{16}{3}$.The new variance $k = \frac{\sum (x_i + 1)^2 + \sum (y_i - 1)^2}{3n} - (\bar{X}')^2$.$k = \frac{\sum x_i^2 + 2\sum x_i + 2n + \sum y_i^2 - 2\sum y_i + n}{3n} - (\frac{16}{3})^2$.$k = \frac{87n + 2(12n) + 2n - 2(3n) + n}{3n} - \frac{256}{9} = \frac{108n}{3n} - \frac{256}{9} = 36 - \frac{256}{9} = \frac{324 - 256}{9} = \frac{68}{9}$.Thus,$9k = 68$.

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