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Question: The mean of the series \({x_1},{x_2},{x_3},....,{x_n}\) is \(\bar X\). If \({x_2}\) is replaced by \...

The mean of the series x1,x2,x3,....,xn{x_1},{x_2},{x_3},....,{x_n} is Xˉ\bar X. If x2{x_2} is replaced by λ\lambda , then what is the new mean?
A) Xˉ−x2+λ\bar X - {x_2} + \lambda
B) Xˉ−x2−λn\dfrac{{\bar X - {x_2} - \lambda }}{n}
C) (n−1)Xˉ+λn\dfrac{{\left( {n - 1} \right)\bar X + \lambda }}{n}
D) nXˉ−x2+λn\dfrac{{n\bar X - {x_2} + \lambda }}{n}

Explanation

Solution

In this question we are given certain numbers along with their mean, and we have been asked the new mean if one of the given numbers is replaced by another number. First, find the sum of the given observations in the terms of nn and Xˉ\bar X. Then subtract the term which was to be replaced from the old mean and add the new term. Put this new sum in the formula and you will have your answer.

Formula used: Mean = Sum of observationsTotal observations\dfrac{{{\text{Sum of observations}}}}{{{\text{Total observations}}}}

Complete step-by-step solution:
We are given nn numbers and we are also given the mean of these numbers. We have been asked the new mean when one number- x2{x_2} is replaced by another number -λ\lambda . Let us put the information that we are given in the question in the formula,
Mean = Sum of observationsTotal observations\dfrac{{{\text{Sum of observations}}}}{{{\text{Total observations}}}}
We are given that, mean = Xˉ\bar X
Total observation = nn
⇒\Rightarrow Xˉ=x1+x2+x3+...........+xnn\bar X = \dfrac{{{x_1} + {x_2} + {x_3} + ........... + {x_n}}}{n}
Shifting n to the other side to find the sum of the observations,
⇒nXˉ=x1+x2+x3+......+xn\Rightarrow n\bar X = {x_1} + {x_2} + {x_3} + ...... + {x_n}
Since an observation x2{x_2} has to be replaced, we will subtract x2{x_2} from both the sides.
⇒nXˉ−x2=x1+x2+x3+......+xn−x2\Rightarrow n\bar X - {x_2} = {x_1} + {x_2} + {x_3} + ...... + {x_n} - {x_2}
Simplifying RHS,
⇒nXˉ−x2=x1+x3+......+xn\Rightarrow n\bar X - {x_2} = {x_1} + {x_3} + ...... + {x_n}
x2{x_2} Had to be replaced by λ\lambda , so now we will add λ\lambda to both the sides.
⇒nXˉ−x2+λ=λ+x1+x3+......+xn\Rightarrow n\bar X - {x_2} + \lambda = \lambda + {x_1} + {x_3} + ...... + {x_n} ….. (1)
Now, we have our new sum of observations. Let us put the sum in the formula.
⇒\RightarrowNew Mean = λ+x1+x3+....+xnn\dfrac{{\lambda + {x_1} + {x_3} + .... + {x_n}}}{n} …. (2)
From (1), we know that λ+x1+x3+.....+xn\lambda + {x_1} + {x_3} + ..... + {x_n} = nM−x1+anM - {x_1} + a. We will put this in equation (2).
⇒\RightarrowNew Mean = nXˉ−x2+λn\dfrac{{n\bar X - {x_2} + \lambda }}{n}

Therefore, our new mean is option (D) nXˉ−x2+λn\dfrac{{n\bar X - {x_2} + \lambda }}{n}.

Note: We find the sum of the given observations in terms of nn and Xˉ\bar X because the answer given in the options is in that terms. We could have skipped the step where we substitute λ+x1+x3+.....+xn\lambda + {x_1} + {x_3} + ..... + {x_n} = nM−x1+anM - {x_1} + a but since the options are in those terms.