The mean of n observations is bar{x}. If thre | vedclass

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(C) Let the sum of $n$ observations be $S_n = n \bar{x}$.When three observations $n+1, n-1, 2n-1$ are added,the new sum becomes $S_{new} = n \bar{x} +

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$\frac{3 \bar{x}+1}{4}$

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(C) Let the sum of $n$ observations be $S_n = n \bar{x}$.When three observations $n+1, n-1, 2n-1$ are added,the new sum becomes $S_{new} = n \bar{x} + (n+1) + (n-1) + (2n-1) = n \bar{x} + 4n - 1$.The new number of observations is $n+3$.Given that the mean remains the same,we have:$\bar{x} = \frac{n \bar{x} + 4n - 1}{n+3}$Multiplying both sides by $(n+3)$:$\bar{x}(n+3) = n \bar{x} + 4n - 1$$n \bar{x} + 3 \bar{x} = n \bar{x} + 4n - 1$Subtracting $n \bar{x}$ from both sides:$3 \bar{x} = 4n - 1$$4n = 3 \bar{x} + 1$$n = \frac{3 \bar{x} + 1}{4}$

Value	$1$	$2$	$3$	$4$
Frequency	$5$	$4$	$6$	$f$

The mean of $n$ observations is $\bar{x}$. If three observations $n+1, n-1, 2n-1$ are added such that the mean remains the same,then the value of $n$ is

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