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Question: If the first five elements of the set \({x_i} + 5,i = 1,2,3...5\) and the next five elements are rep...

If the first five elements of the set xi+5,i=1,2,3...5{x_i} + 5,i = 1,2,3...5 and the next five elements are replaced by xj5,j=6...10{x_j} - 5,j = 6...10 then the mean will change by
(A) 0{\text{(A) 0}}
(B) 5{\text{(B) 5}}
(C) 10{\text{(C) 10}}
(D) 25{\text{(D) 25}}

Explanation

Solution

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=sum of termsnumber of termsMean = \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 55 terms of which the first five terms are xi+5,i=1,2,3...5{x_i} + 5,i = 1,2,3...5 and the next five terms are replaced by xj5,j=6...10{x_j} - 5,j = 6...10.
On elaborating the first set we get the first five terms as:
x1+5,x2+5,x3+5,x4+5,x5+5{x_1} + 5,{x_2} + 5,{x_3} + 5,{x_4} + 5,{x_5} + 5, which is the initial distribution.
And since there are 55 terms in the distribution the mean will be:
\Rightarrow Mean=x1+5+x2+5+x3+5+x4+5+x5+55Mean = \dfrac{{{x_1} + 5 + {x_2} + 5 + {x_3} + 5 + {x_4} + 5 + {x_5} + 5}}{5},
On simplifying we get:
\Rightarrow Mean=x1+x2+x3+x4+x5+2510Mean = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + 25}}{{10}}, which is the mean of distribution 11.
Also, we elaborating the next set we get the next five terms as:
x65,x75,x85,x95,x105{x_6} - 5,{x_7} - 5,{x_8} - 5,{x_9} - 5,{x_{10}} - 5, which is the other distribution
And since there are 55 terms in the distribution the mean will be:
\Rightarrow Mean=x65+x75+x85+x95+x1055Mean = \dfrac{{{x_6} - 5 + {x_7} - 5 + {x_8} - 5 + {x_9} - 5 + {x_{10}} - 5}}{5}
On simplifying we get:
\Rightarrow Mean=x6+x7+x8+x9+x102510Mean = \dfrac{{{x_6} + {x_7} + {x_8} + {x_9} + {x_{10}} - 25}}{{10}}, which is the mean of distribution 22.
Now, the distribution of the 1010 terms can be written as:
\Rightarrow x1+5,x2+5,x3+5,x4+5,x5+5,x65,x75,x85,x95,x105{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=x1+5+x2+5+x3+5+x4+5+x5+5+x65+x75+x85+x95+x10510Mean = \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=x1+x2+x3+x4+x5+25+x6+x7+x8+x9+x102510Mean = \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=x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+252510Mean = \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=x1+x2+x3+x4+x5+x6+x7+x8+x9+x1010=xˉMean = \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 00.

Therefore, the correct option is option (A)(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 aa is added to all the values in the distribution, the change in mean will be the value aa and
Whenever a constant bb is multiplied to all the terms in a distribution, the change in the mean will be bb times the original mean.