Answer
Verified
401.7k+ views
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$$Mean = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} + {x_7} + {x_8} + {x_9} + {x_{10}}}}{{10}} = \bar x$, which the changed mean.
Since there is no change in the mean and the net mean, the total value of which the mean has changed is $0$.
Therefore, the correct option is option $(A)$.
Note: Now the change in mean refers to the change in mean between one distribution and the changed version of the same distribution. The change can be linear or scalar.
Property to be remembered about change in mean are:
Whenever a constant $a$ is added to all the values in the distribution, the change in mean will be the value $a$ and
Whenever a constant $b$ is multiplied to all the terms in a distribution, the change in the mean will be $b$ times the original mean.
Formula used: $Mean = \dfrac{{{\text{sum of terms}}}}{{{\text{number of terms}}}}$
Complete step by step solution:
From the question it is given that the distribution has total $5$ terms of which the first five terms are ${x_i} + 5,i = 1,2,3...5$ and the next five terms are replaced by ${x_j} - 5,j = 6...10$.
On elaborating the first set we get the first five terms as:
${x_1} + 5,{x_2} + 5,{x_3} + 5,{x_4} + 5,{x_5} + 5$, which is the initial distribution.
And since there are $5$ terms in the distribution the mean will be:
$\Rightarrow$$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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE