Statistics And Probability Question 569
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\le i\le n $ is
[Pb. CET 1988]
Options:
A) $ \bar{x}+2n $
B) $ \bar{x}+n+1 $
C) $ \bar{x}+2 $
D) $ \bar{x}+n $
Show Answer
Answer:
Correct Answer: B
Solution:
We know that $ \bar{x}=\frac{\sum\limits_{i=1}^{n}{x_{i}}}{n} $ i.e., $ \sum\limits_{i=1}^{n}{x_{i}}=n\bar{x} $ \ $ \frac{\sum\limits_{i=1}^{n}{(x_{i}+2i)}}{n}=\frac{\sum\limits_{i=1}^{n}{x_{i}}+2\sum\limits_{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 $ .
BETA
