Question

Let $\bar x$ be the mean of ${x_1},{x_2},{x_3}.....{x_n}$. If ${x_i} = a + c{y_i}$ for some constants a and c, then what will be the mean of ${y_1},{y_2},{y_3},......{y_n}$?
$\left( a \right)a + c\bar x$
$\left( b \right)a - \dfrac{1}{c}\bar x$
$\left( c \right)\dfrac{1}{c}\bar x - a$
$\left( d \right)\dfrac{{\bar x - a}}{c}$

Answer 1

Hint: In this particular question use the concept that the mean of n number of terms is the sum of all the terms divided by the number of terms i.e. $\bar x = \dfrac{{{x_1} + {x_2} + {x_3} + ..... + {x_n}}}{n}$, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
$\bar x$ be the mean of ${x_1},{x_2},{x_3}.....{x_n}$.
As we see that there are n number of terms, so the mean of n number of terms is the sum of all the terms divided by the number of terms.
So, $\bar x = \dfrac{{{x_1} + {x_2} + {x_3} + ..... + {x_n}}}{n}$.................... (1)
Now it is also given that ${x_i} = a + c{y_i}$ for some constants a and c.
So first find out the value of ${y_i}$ in terms of ${x_i}$, a and c we have,
$ \Rightarrow {y_i} = \dfrac{{{x_i} - a}}{c}$, where i = 1, 2, 3,.......n
Now for, i = 1, 2, 3, .......n the value of ${y_i}$are
$ \Rightarrow {y_1} = \dfrac{{{x_1} - a}}{c},{y_2} = \dfrac{{{x_2} - a}}{c},{y_3} = \dfrac{{{x_3} - a}}{c},.........{y_n} = \dfrac{{{x_n} - a}}{c}$
So the mean of ${y_1},{y_2},{y_3},......{y_n}$ is the sum of all the terms divided by the number of terms i.e. n, so we have,
$ \Rightarrow {\text{mean}}\left( {{y_1},{y_2},{y_3},......{y_n}} \right) = \dfrac{{{y_1} + {y_2} + {y_3} + ..... + {y_n}}}{n}$
Now substitute the values we have,
$ \Rightarrow {\text{mean}}\left( {{y_1},{y_2},{y_3},......{y_n}} \right) = \dfrac{{\dfrac{{{x_1} - a}}{c} + \dfrac{{{x_2} - a}}{c} + \dfrac{{{x_3} - a}}{c} + ..... + \dfrac{{{x_n} - a}}{c}}}{n}$
Now simplify this we have,
$ \Rightarrow {\text{mean}}\left( {{y_1},{y_2},{y_3},......{y_n}} \right) = \dfrac{{\left( {{x_1} - a} \right) + \left( {{x_2} - a} \right) + \left( {{x_3} - a} \right) + ..... + \left( {{x_n} - a} \right)}}{{nc}}$
$ \Rightarrow {\text{mean}}\left( {{y_1},{y_2},{y_3},......{y_n}} \right) = \dfrac{{\left( {{x_1} + {x_2} + {x_3} + ..... + {x_n}} \right) - \left( {a + a + a + ..... + a} \right)}}{{nc}}$
Now the addition of a is n times, so a + a + a +......+ a = na, so we have,
$ \Rightarrow {\text{mean}}\left( {{y_1},{y_2},{y_3},......{y_n}} \right) = \dfrac{{\left( {{x_1} + {x_2} + {x_3} + ..... + {x_n}} \right) - \left( {na} \right)}}{{nc}}$
$ \Rightarrow {\text{mean}}\left( {{y_1},{y_2},{y_3},......{y_n}} \right) = \dfrac{{\dfrac{{{x_1} + {x_2} + {x_3} + ..... + {x_n}}}{n}}}{c} - \dfrac{{na}}{{nc}}$
$ \Rightarrow {\text{mean}}\left( {{y_1},{y_2},{y_3},......{y_n}} \right) = \dfrac{{\dfrac{{{x_1} + {x_2} + {x_3} + ..... + {x_n}}}{n}}}{c} - \dfrac{a}{c}$
Now from equation (1) we have,
$ \Rightarrow {\text{mean}}\left( {{y_1},{y_2},{y_3},......{y_n}} \right) = \dfrac{{\bar x}}{c} - \dfrac{a}{c} = \dfrac{{\bar x - a}}{c}$
So this is the required mean of ${y_1},{y_2},{y_3},......{y_n}$.

So, the correct answer is “Option d”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the definition of the mean which is stated above, so first write of mean of ${x_1},{x_2},{x_3}.....{x_n}$, then write the value of ${y_i}$ in terms of ${x_i}$, a and c as above, then write the mean of ${y_1},{y_2},{y_3},......{y_n}$ as above and simplify then substitute the value of mean of ${x_1},{x_2},{x_3}.....{x_n}$ in the mean of ${y_1},{y_2},{y_3},......{y_n}$ we will get the required answer.