Question

f the mean of a set of observation ${x_1},{x_2},...{x_{10}}$ is $20$ then the mean of ${x_1} + 4,{x_2} + 8,{x_3} + 12,...,{x_{10}} + 40$ is?
(A) $34$ (B) $42$ (C) $38$ (D) $40$

Answer 1

Hint- In this question we will use two main concepts to find the mean of given terms.
1) First one is definition of mean i.e. mean is equal to sum of observations divided by number of observations.
2) And the second one is Arithmetic progression. It is a sequence of numbers such that the difference between the consecutive terms is constant.

Complete Step by step solution:
The mean of the given set ${x_1},{x_2},...{x_{10}}$ is$20$.
We know that the formula for mean $m = \dfrac{{{x_1} + {x_2} + {x_3}......... + {x_n}}}{n}$ where, ${x_1},{x_2},...{x_n}$ are the observations and $n$ is number of observations. By applying the formula,
 \[\dfrac{{({x_1} + {x_2} + ... + {x_{10}})}}{{10}} = 20\] [As given that mean is $20$]
\[ \Rightarrow {x_1} + {x_2} + ... + {x_{10}} = 200\] equation (1)
We have to find the mean of ${x_1} + 4,{x_2} + 8,{x_3} + 12,...{x_{10}} + 40$, therefore,
$\dfrac{{({x_1} + 4 + {x_2} + 8 + {x_3} + 12 + ... + {x_{10}} + 40)}}{{10}}$
$ = \dfrac{{({x_1} + {x_2} + {x_3} + ... + {x_{10}}) + (4 + 8 + 12 + ... + 40)}}{{10}}$
\[ = \dfrac{{200 + (4 + 8 + 12 + ... + 40)}}{{10}}\] [from equation (1)] equation (2)
We can observe that $4 + 8 + 12 + ... + 40$ is an Arithmetic progression,
 $8 - 4 = 4$
$12 - 8 = 4$
It means that the difference between two consecutive terms of the sequence is common i.e. $4$.
We know that, formula for sum of finite number of terms of an A.P. is
${S_n} = \dfrac{n}{2}\left\{ {2a + (n - 1)d} \right\}$
Where $a$ is first term, $d$ is common difference and $n$ is total number of term in the A.P.
For $4 + 8 + 12 + ... + 40$,
$a = 4$, $d = 4$ and $n = 10$
On substituting the values,
${S_{10}} = \dfrac{{10}}{2}\left\{ {2 \times 4 + \left( {10 - 1} \right)4} \right\}$
$ \Rightarrow {S_{10}} = 5\left\{ {8 + 9 \times 4} \right\}$
$ \Rightarrow {S_{10}} = 5\left\{ {8 + 36} \right\}$
$ \Rightarrow {S_{10}} = 5\left\{ {44} \right\}$
$ \Rightarrow {S_{10}} = 220$
$\therefore 4 + 8 + 12 + ... + 40 = 220$
On substituting the value in equation (2),
\[\dfrac{{({x_1} + 4 + {x_2} + 8 + {x_3} + 12 + ... + {x_{10}} + 40)}}{{10}} = \dfrac{{200 + 220}}{{10}}\]
$ \Rightarrow \dfrac{{({x_1} + 4 + {x_2} + 8 + {x_3} + 12 + ... + {x_{10}} + 40)}}{{10}} = \dfrac{{420}}{{10}}$
$ \Rightarrow \dfrac{{({x_1} + 4 + {x_2} + 8 + {x_3} + 12 + ... + {x_{10}} + 40)}}{{10}} = 42$
Hence the mean of ${x_1} + 4,{x_2} + 8,{x_3} + 12,...,{x_{10}} + 40 = 42$
Answer- option (B)


Note: Formula for the mean for grouped data when class intervals are not given
Mean = $\overline x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}{f_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}$ where ${f_i}$ is the frequency of ${i_{th}}$ observation ${x_i}$.
Formula for the mean for grouped data when class intervals are given
Mean = $\overline x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}{f_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}$ where ${f_i}$ is the frequency of ${i_{th}}$ class whose class mark is ${x_i}$.
Class mark $ = $ (Upper class limit $ + $ Lower class limit)$/2$