Questions & Answers
Question
Answers
Related Questions
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
![Verified]()
Verified
Verified
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)$/2$
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)$/2$
Like this
Liked
-
LIVECourses
-
V PRO
-
JEE
-
JEE Main & Advanced 2021
-
JEE Main & Advanced 2022
-
-
NEET
-
NEET 2021
-
NEET 2022
-
-
CBSE
-
Class 6
-
Class 7
-
Class 8
-
Class 9
-
Class 10
-
Class 11
-
Class 12
-
-
ICSE
-
Class 9
-
Class 10
-
-
Maharashtra Board
-
Class 9
-
Class 10
-
-
Micro Courses
-
-
PREMIUM1-on-1 Classes
-
Study Material
-
NCERT Solutions
-
Previous Year Papers
-
Mock Tests
-
Sample Papers
-
Reference Book Solutions
-
ICSE Solutions
-
School Syllabus
-
Revision Notes
-
Important Questions
-
Math Formula Sheets
-
-
More
Book your FREE Online counselling session
Share your contact information
Your FREE Online counselling session has been booked!
Vedantu academic counsellor will be calling you shortly for your Online Counselling session.
DONE
Please try again later
OK
