The mean of n values of a distribution is bar | vedclass

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(C) Let the original mean be $\bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n}$.The new values are $(x_1 + 1), (x_2 + 2), \dots, (x_n + n)$.The new mean is

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$\bar{x} + \left( \frac{n + 1}{2} \right)$

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(C) Let the original mean be $\bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n}$.The new values are $(x_1 + 1), (x_2 + 2), \dots, (x_n + n)$.The new mean is $\bar{x}_{new} = \frac{(x_1 + 1) + (x_2 + 2) + \dots + (x_n + n)}{n}$.$\bar{x}_{new} = \frac{(x_1 + x_2 + \dots + x_n) + (1 + 2 + \dots + n)}{n}$.$\bar{x}_{new} = \frac{\sum x_i}{n} + \frac{\sum_{i=1}^{n} i}{n}$.Since $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$,we have $\bar{x}_{new} = \bar{x} + \frac{n(n+1)}{2n}$.$\bar{x}_{new} = \bar{x} + \frac{n+1}{2}$.

The mean of $n$ values of a distribution is $\bar{x}$. If the first value is increased by $1$,the second value by $2$,and so on,what will be the mean of the new values?

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