Question

If the mean of the set of numbers $x_{1},x_{2},x_{3},.....,x_{n}$ is $\bar{x}$, then the mean of the numbers $x_{i} + 2i$, $1 \leq i \leq n$ is

• A. $\bar{x} + 2n$
• B. $\bar{x} + n + 1$
• C. $\bar{x} + 2$
• D. $\bar{x} + n$

Answer 1

Answer: (B)

$\bar{x} + n + 1$

Answer 2

We know that $\bar{x} = \frac{\sum_{i = 1}^{n}x_{i}}{n}$ i.e., $\sum_{i = 1}^{n}x_{i} = n\bar{x}$

∴ $\frac{\sum_{i = 1}^{n}{(x_{i} + 2i)}}{n} = \frac{\sum_{i = 1}^{n}x_{i} + 2\sum_{i = 1}^{n}i}{n} = \frac{n\bar{x} + 2(1 + 2 + ...n)}{n} = \frac{n\bar{x} + 2\frac{n(n + 1)}{2}}{n} = \bar{x} + (n + 1)$

$= \frac{n\bar{x} + 2\frac{n(n + 1)}{2}}{n} = \bar{x} + n + 1$.