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If the first five elements of the set ${x_i} + 5,i = 1,2,3...5$ and the next five elements are replaced by ${x_j} - 5,j = 6...10$ then the mean will change by
${\text{(A) 0}}$
${\text{(B) 5}}$
${\text{(C) 10}}$
${\text{(D) 25}}$

seo-qna
Last updated date: 09th Apr 2024
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MVSAT 2024
Answer
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Hint: Here we will first calculate the original mean from the data given and then find the changed mean from the mean formula, and then find the change in mean. Finally we get the required answer.

Formula used: $Mean = \dfrac{{{\text{sum of terms}}}}{{{\text{number of terms}}}}$

Complete step by step solution:
From the question it is given that the distribution has total $5$ terms of which the first five terms are ${x_i} + 5,i = 1,2,3...5$ and the next five terms are replaced by ${x_j} - 5,j = 6...10$.
On elaborating the first set we get the first five terms as:
${x_1} + 5,{x_2} + 5,{x_3} + 5,{x_4} + 5,{x_5} + 5$, which is the initial distribution.
And since there are $5$ terms in the distribution the mean will be:
$\Rightarrow$$Mean = \dfrac{{{x_1} + 5 + {x_2} + 5 + {x_3} + 5 + {x_4} + 5 + {x_5} + 5}}{5}$,
On simplifying we get:
$\Rightarrow$$Mean = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + 25}}{{10}}$, which is the mean of distribution $1$.
Also, we elaborating the next set we get the next five terms as:
${x_6} - 5,{x_7} - 5,{x_8} - 5,{x_9} - 5,{x_{10}} - 5$, which is the other distribution
And since there are $5$ terms in the distribution the mean will be:
$\Rightarrow$$Mean = \dfrac{{{x_6} - 5 + {x_7} - 5 + {x_8} - 5 + {x_9} - 5 + {x_{10}} - 5}}{5}$
On simplifying we get:
$\Rightarrow$$Mean = \dfrac{{{x_6} + {x_7} + {x_8} + {x_9} + {x_{10}} - 25}}{{10}}$, which is the mean of distribution $2$.
Now, the distribution of the $10$ terms can be written as:
$\Rightarrow$${x_1} + 5,{x_2} + 5,{x_3} + 5,{x_4} + 5,{x_5} + 5,{x_6} - 5,{x_7} - 5,{x_8} - 5,{x_9} - 5,{x_{10}} - 5$
Now, the net mean of the new distribution will be:
$\Rightarrow$$Mean = \dfrac{{{x_1} + 5 + {x_2} + 5 + {x_3} + 5 + {x_4} + 5 + {x_5} + 5 + {x_6} - 5 + {x_7} - 5 + {x_8} - 5 + {x_9} - 5 + {x_{10}} - 5}}{{10}}$
On simplifying the equation, we get:
$\Rightarrow$$Mean = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + 25 + {x_6} + {x_7} + {x_8} + {x_9} + {x_{10}} - 25}}{{10}}$
This could be further simplified as:
$\Rightarrow$$Mean = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} + {x_7} + {x_8} + {x_9} + {x_{10}} + 25 - 25}}{{10}}$
This can be further simplified as:
$\Rightarrow$$Mean = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} + {x_7} + {x_8} + {x_9} + {x_{10}}}}{{10}} = \bar x$, which the changed mean.
Since there is no change in the mean and the net mean, the total value of which the mean has changed is $0$.

Therefore, the correct option is option $(A)$.

Note: Now the change in mean refers to the change in mean between one distribution and the changed version of the same distribution. The change can be linear or scalar.
Property to be remembered about change in mean are:
Whenever a constant $a$ is added to all the values in the distribution, the change in mean will be the value $a$ and
Whenever a constant $b$ is multiplied to all the terms in a distribution, the change in the mean will be $b$ times the original mean.