Consider a set of 3n numbers having variance 4.

Question

Consider a set of $3n$ numbers having variance 4. In this set, the mean of first $2n$ numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of first $2n$ numbers and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then $9k$ is equal to

Answer: 68

Accepted Answer

Combined mean: $\bar x = \dfrac{2n(6)+n(3)}{3n}=5$

$$\frac{1}{3n}\sum x_i^2-25=4 \implies \sum x_i^2 = 87n$$

New mean: $\bar x' = \dfrac{2n(7)+n(2)}{3n} = \dfrac{16}{3}$

$$k = \frac{1}{3n}\sum x_i'^2-\left(\frac{16}{3}\right)^2$$

where $\sum x_i'^2 = \sum x_i^2+2\sum_{2n}x_i-2\sum_{n}x_i+3n = 87n+2(12n)-2(3n)+3n$

$$k = \frac{87n+24n-6n+3n}{3n}-\frac{256}{9} = 36-\frac{256}{9}$$

$9k = 324-256=68$

Solutions · 1