Question

The mean of a set of observations is $\overline {\text{x}} $. If each observation is divided by $\alpha $, $\alpha \ne 0$ then it is increased by $10$. Then find the new mean of the new set?
A. $\dfrac{{\overline x }}{\alpha }$
B. $\dfrac{{\overline x + 10}}{\alpha }$
C. $\dfrac{{\overline x + 10\alpha }}{\alpha }$
D. $\alpha \overline x + 10$

Answer 1

Hint: From the question, we have to find the mean of the given data. First, we have to frame a mathematical expression for the given data. Then, we have to find the new mean by data handling.

Formula used: The mean is the meaning of an average. The process of sharing out equally is the basis of average. We call the averaged or quantity as the arithmetic average (or arithmetic mean or simply average or mean).
The mean of several items is the value equally shared out among the items.
\[{\text{Mean = }}\dfrac{{{\text{Total of all items}}}}{{{\text{Number of items}}}}\].
\[ \Rightarrow {\text{Total of all items = Mean}} \times {\text{Number of items}}\]
Let denote the mean as $\overline {\text{x}} $.
Let us consider the set of observations are${{\text{x}}_{\text{1}}}{\text{,}}{{\text{x}}_{\text{2}}}{\text{,}}{{\text{x}}_{\text{3}}}{\text{,}}........{\text{,}}{{\text{x}}_{\text{n}}}$.
Now, we have to find the mean for the above set of observations by using the mean formula.
Let the total of a set of observations are ${{\text{x}}_1} + {{\text{x}}_2} + {{\text{x}}_3} + .......... + {{\text{x}}_{\text{n}}}$.
Given, each observation is divided by $\alpha $ ,$\alpha \ne 0$ then is increased by $10$ . So, now we are going to frame new means with these conditions.
Let the first observation can be divided by $\alpha $,$\alpha \ne 0$ then is increased by $10$as \[\dfrac{{{{\text{x}}_{\text{1}}}}}{\alpha } + 10\].
Let the second observation can be divided by $\alpha $,$\alpha \ne 0$ then is increased by $10$as \[\dfrac{{{{\text{x}}_2}}}{\alpha } + 10\].
Let the last observation can be divided by $\alpha $,$\alpha \ne 0$ then is increased by $10$as \[\dfrac{{{{\text{x}}_{\text{n}}}}}{\alpha } + 10\] .
 Let the new set of observations are \[\dfrac{{{{\text{x}}_{\text{1}}}}}{\alpha } + 10,\dfrac{{{{\text{x}}_2}}}{\alpha } + 10,............,\dfrac{{{{\text{x}}_{\text{n}}}}}{\alpha } + 10\] .
Thus, the total of a new set of observations are\[\left( {\dfrac{{{{\text{x}}_{\text{1}}}}}{\alpha } + 10} \right) + \left( {\dfrac{{{{\text{x}}_2}}}{\alpha } + 10} \right) + \left( {\dfrac{{{{\text{x}}_3}}}{\alpha } + 10} \right) + ........ + \left( {\dfrac{{{{\text{x}}_{\text{n}}}}}{\alpha } + 10} \right)\] .

Complete step-by-step solution:
Let us consider the set of observations are ${{\text{x}}_{\text{1}}}{\text{,}}{{\text{x}}_{\text{2}}}{\text{,}}{{\text{x}}_{\text{3}}}{\text{,}}........{\text{,}}{{\text{x}}_{\text{n}}}$.
Now, we have to find the mean for the above set of observations by using the mean formula.
Here, the number of observations is ${\text{n}}$.
\[{\text{Mean = }}\dfrac{{{\text{Total of set of observations}}}}{{{\text{Number of observations}}}}\]
\[\overline {\text{x}} = \dfrac{{{{\text{x}}_1} + {{\text{x}}_2} + {{\text{x}}_3} + .......... + {{\text{x}}_{\text{n}}}}}{{\text{n}}}\]
Now, again use the mean formula to find the mean for the new set of observations.
 \[{\text{Mean = }}\dfrac{{{\text{Total of a new set of observations}}}}{{{\text{Number of observations}}}}\]
${\text{Mean}} = \dfrac{{\left( {\dfrac{{{{\text{x}}_{\text{1}}}}}{\alpha } + 10} \right) + \left( {\dfrac{{{{\text{x}}_2}}}{\alpha } + 10} \right) + \left( {\dfrac{{{{\text{x}}_3}}}{\alpha } + 10} \right) + ........ + \left( {\dfrac{{{{\text{x}}_{\text{n}}}}}{\alpha } + 10} \right)}}{{\text{n}}}$
Now, we write ${\text{n}}$ times of $10$ as$10{\text{n}}$ .
$ \Rightarrow $\[{\text{Mean}} = \dfrac{{\left( {\dfrac{{{{\text{x}}_{\text{1}}}}}{\alpha } + \dfrac{{{{\text{x}}_2}}}{\alpha } + ...... + \dfrac{{{{\text{x}}_{\text{n}}}}}{\alpha }} \right) + 10{\text{n}}}}{{\text{n}}}\]
$ \Rightarrow $\[{\text{Mean}} = \dfrac{1}{{\text{n}}}\left( {\dfrac{{{{\text{x}}_{\text{1}}}}}{\alpha } + \dfrac{{{{\text{x}}_2}}}{\alpha } + ...... + \dfrac{{{{\text{x}}_{\text{n}}}}}{\alpha }} \right) + \dfrac{{10{\text{n}}}}{{\text{n}}}\]
$ \Rightarrow $\[{\text{Mean}} = \dfrac{1}{{{\alpha }}}\left( {\dfrac{{{{\text{x}}_{\text{1}}} + {{\text{x}}_2} + ........ + {{\text{x}}_{\text{n}}}}}{{\text{n}}}} \right) + 10\]
Now, substitute the original mean of a set of observation, \[\overline x = \dfrac{{{{\text{x}}_1} + {{\text{x}}_2} + {{\text{x}}_3} + .......... + {{\text{x}}_{\text{n}}}}}{{\text{n}}}\] in the above term, we get
$ \Rightarrow $\[{\text{Mean}} = \dfrac{1}{\alpha }\left( {\overline {\text{x}} } \right) + 10\]
$ \Rightarrow $ \[{\text{Mean}} = \dfrac{{\overline {\text{x}} }}{\alpha } + 10\]
Now, taking Least Common Multiple (LCM) of \[\alpha \] and $1$ is \[\alpha \] .
\[{\text{Mean}} = \dfrac{{\overline {\text{x}} }}{\alpha } + 10 \times \left( {\dfrac{\alpha }{\alpha }} \right)\]
$ \Rightarrow $\[{\text{Mean}} = \dfrac{{\overline {\text{x}} + 10\alpha }}{\alpha }\]
Hence, the mean of a new set of observations are\[\dfrac{{\overline {\text{x}} + 10\alpha }}{\alpha }\] .

$\therefore $ The correct answer is option${\text{C}}$ .

Note: Data handling is an art. Sometimes the raw data (data as they are) will not be useful to get the required information. In order to get proper useful information, we have to process very important useful information from the given data.
The given problem is easy to solve. The students should concentrate on expressing the mathematical terms and on simple calculations. Particularly, adding the terms in the mean formula and on further calculations. The students should do calculations carefully on each and every step.