Measures of Central Tendency - Class 9 : Notes

(1) Arithmetic mean (AM), Geometric mean(GM),  Harmonic mean(HM),  Median and Mode are various measures of central tendency.
(i) Arithmetic mean:
For Example: Find the mean of 994, 996, 998, 1002 and 1000.
Given,       No. of values n = 5
We know, mean ($\overline{x}$) = $\frac{x_{1}+x_{2}+...........+x_{n}}{n}$
So, mean$\frac{994+996+998+1002+1000}{5}$
= $\frac{4990}{5}$
= 998
Hence, mean of the given numbers is 998

(ii) Geometric mean:
If we have a series of n positive values such as ${x_1},{x_2},{x_3},.....,{x_k}$ are repeated ${f_1},{f_2},{f_3},.....,{f_k}$ times respectively then geometric mean will become:
G.M of X= $\overline X$= $\root n \of {x_1^{{f_1}}.x_2^{{f_2}}.x_3^{{f_3}}.....x_k^{{f_k}}}$ (For Grouped Data)
Where n= ${f_1} + {f_2} + {f_3} + ...{f_k}$
Example: Find the Geometric mean of the values 10, 5, 15, 8, 12
Solution: Here ${x_1} = 10$, ${x_2} = 5$, ${x_3} = 15$, ${x_4} = 8$, ${x_5} = 12$ and ${n} = 5$
G.M of X= $\overline X$ = $\root 5 \of {10 \times 5 \times 15 \times 8 \times 12}$
$\overline X$ = $\root 5 \of {72000}$ = 9.36

(iii) Harmonic mean: Harmonic mean is quotient of “number of the given values” and “sum of the reciprocals of the given values”.
Harmonic mean in mathematical terms is defined as follows:Example: Calculate the harmonic mean of the numbers: 13.5, 14.5, 14.8, 15.2 and 16.1
Solution: The harmonic mean is calculated as below:H.M of X = $\overline X$ = ${n \over {\sum {\left( {{1 \over x}} \right)} }}$
H.M of X = $\overline X$ = ${5 \over {0.3417}}$ = 14.63

(iv) Median:
For Example: Find the median of this data: 83,37,70,29,45,63,41,70,34,54
Solution: Arrange the data in ascending order, we get-
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Here, the number of observation n = 10 (even)
Now, medians is-
$\Rightarrow&space;\frac{value&space;of&space;the&space;(\frac{n}{2})^{th}observation&space;+&space;value&space;of&space;(\frac{n}{2}+1)^{th}&space;observation)}{2}$
$\Rightarrow \frac{value of the (\frac{10}{2})^{th}observation + value of (\frac{10}{2}+1)^{th} observation)}{2}$
$\Rightarrow \frac{value of the (5)^{th}observation + value of (6)^{th} observation)}{2}$
$\Rightarrow \frac{45 + 54}{2}$
$\Rightarrow \frac{99}{2}$
49.5
Hence the value of median is 49.5.

(v) Mode:
For Example: Find out the mode of the following marks obtained by 15 students in  class:
Marks: 4,6,5,7,9,8,10,4,7,6,5,9,8,7,7.
Solution:Arrange the data in the form of  a frequency table-Since. the value of 7 occurs maximum number of times i.e 4.
Hence, the mode value is 7

(2) (i) If  ${x_1},{x_2},{x_3},.....,{x_n}$ are n values of a variable X,  then the arithmetic mean of these values is given by $\bar X = {{{x_1} + {x_2} + {x_3} + ..... + {x_n}} \over n}$ or, $\bar X = {{\sum\limits_{i = 1}^n {{x_1}} } \over n}$
For Example: Find the mean of 994, 996, 998,1002 and 1000.
Solution: No. of values n = 5
We know, mean ($\overline{x}$) = $\frac{x_{1}+x_{2}+...........+x_{n}}{n}$
So, mean$\frac{994+996+998+1002+1000}{5}$
= $\frac{4990}{5}$

= 998
Hence, mean of the given numbers is 998

(ii)  If a variate X take values ${x_1},{x_2},{x_3},.....,{x_n}$ with corresponding frequencies ${f_1},{f_2},{f_3},.....,{f_n}$  respectively,  then the arithmetic mean of these values is given by $\bar X = {{{f_1}{x_1} + {f_2}{x_2} + ..... + {f_n}{x_n}} \over {{f_1} + {f_2} + ..... + {f_n}}}$ or,  $\bar X = {{\sum\limits_{i = 1}^n {{f_i}{x_i}} } \over N}$, where  $N = \sum\limits_{i = 1}^n {{f_i}}$
For Example: Calculate the mean for the following distributionSolution: Calculation of the Arithmetic mean-Now, mean = $\Sigma&space;\frac{f_{i}x_{i}}{f_{1}}$$\frac{281}{40}$= 7.025
Hence, value of mean is 7.025

(3) If ${\bar X}$ is the mean of n observations ${x_{1,}}{x_2},....{x_n}$, then
(i) The algebraic sum of the deviations about ${\bar X}$ is 0, i.e. $\sum\limits_{i = 1}^n {\left( {{x_i} - \bar X} \right)} = 0$
For Example: Duration of sunshine (in hours) in Amritsar for first 10 days of August 1997 as reported by the Meteorological Department are given below:
9.6,5.2,3.5,1.5,1.6,2.4,2.6,8.4,10.3,10.9
Verify that  $\sum\limits_{i = 1}^{10} {\left( {{x_i} - \bar X} \right)} = 0$
Solution:We have to verify that $\sum_{i=1}^{10}(x_{i}-\overline{x})=0$
Taking LHS,
$\Rightarrow$                   $\sum_{i=1}^{10}(x_{i}-\overline{x})$
$\Rightarrow$               $\sum_{i=1}^{10}x_{i}-\sum_{i=1}^{10}\overline{x}$
$\Rightarrow$ (9.6+ 5.2+3.5+1.5+1.6+2.4+2.6+8.4+10.3+10.9) - 10 x $\overline{x}$
$\Rightarrow$56 - 10 x 5.6
$\Rightarrow$56 - 56
$\Rightarrow$ 0 = RHS   Hence Proved

(ii) Prove that the mean of the observations${x_1} \pm a,{x_2} \pm a,.....,{x_n} \pm a$ is $\bar X \pm a$
Given, $\bar x = {{{x_1} + {x_2} + ...... + {x_n}} \over n}$   …..(i)
The observation are $\left( {{x_1} + a} \right),\left( {{x_2} + a} \right),......,\left( {{x_n} + a} \right)$.
Mean $= {{\left( {{x_1} + a} \right) + \left( {{x_2} + a} \right) + ...... + \left( {{x_n} + a} \right)} \over n}$
$= {{{x_1} + {x_2} + ...... + {x_n} + n \times a} \over n}$
$= {{{x_1} + {x_2} + ...... + {x_n}} \over n} + {{n \times a} \over n}$
From Equation (1); we get
Mean $= \left( {\bar x + a} \right)$
So that given statement is true.

(iii) Prove that the mean of the observations $a{x_{1,}}a{x_2},....,a{x_n}$ is $a\bar X$
For Example: Mean of  ${x_1},{x_2},.....,{x_n} = {{{x_1} + {x_2} + ...... + {x_n}} \over n}$
But mean $= \bar x$ (given)
$\bar x = {{{x_1} + {x_2} + ...... + {x_n}} \over n}$ ……..(1)
The observation are $a{x_1},a{x_2},.....,a{x_n}$
Mean $= {{a{x_1} + a{x_2} + ...... + a{x_n}} \over n}$
$= {{a\left( {{x_1} + {x_2} + ...... + {x_n}} \right)} \over n}$
$= a\bar x$
Thus, the given statement is true.

(iv) Prove that the mean of the observations ${{{x_1}} \over a},{{{x_2}} \over a},....,{{{x_n}} \over a}$ is ${{\bar x} \over n}$
Proof: We have,    $\bar X = {1 \over n}\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)$
Let ${\bar X'}$ be the mean of ${{{x_1}} \over a},{{{x_2}} \over a},.....,{{{x_n}} \over a}$. Then,
${\bar X'} = {1 \over n}\left( {{{{x_1}} \over a} + {{{x_2}} \over a} + ..... + {{{x_n}} \over a}} \right)$
${\bar X'} = {1 \over n}\left( {{{{x_1} + {x_2} + ..... + {x_n}} \over a}} \right)$
${\bar X'} = {1 \over a}\left( {{{{x_1} + {x_2} + ..... + {x_n}} \over n}} \right)$
${\bar X'} = {1 \over a}\left[ {{1 \over n}\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)} \right] = {1 \over a}\left( {\bar X} \right)$
${\bar X'} = {{\bar X} \over a}$

(4) Media of a distribution is the value of the variable which divides the distribution into two equal parts.
For Example: Find the coordinates of the points which divide the line segment A(-2, 2) & B(2, 8) into four equal parts.
Solution: From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.
Coordinates of P $= \left( {\frac{{1 \times 2 + 3 + ( - 2)}}{{1 + 3}},\frac{{1 \times 8 + 3 \times 2}}{{1 + 3}}} \right)$
$= \left( { - 1,\frac{7}{2}} \right)$
Coordinates of Q $= \left( {\frac{{2 + ( - 2)}}{2},\frac{{2 + 8}}{2}} \right)$
$= \left( {0,5} \right)$
Coordinates of R $= \left( {\frac{{3 \times 2 + 1 \times ( - 2)}}{{3 + 1}},\frac{{3 \times 8 + 1 \times 2}}{{3 + 1}}} \right)$
$= \left( {1,\frac{{13}}{2}} \right)$

(5) lf ${x_{1,}}{x_2},....,{x_n}$ are n values of a variable arranged in ascending or descending order, then
Median = value of ${\left( {{{n + 1} \over 2}} \right)^{th}}$ observation, if n is odd
Median = (value of ${{{\left( {{n \over 2}} \right)}^{th}}}$ observation + value of ${{{\left( {{n \over 2} + 1} \right)}^{th}}}$ observation)/2, if n is even

For Example:
(i) Find the median of this data : 83,37,70,29,45,63,41,70,34,54
Solution:Arrange the data in ascending order, we get-
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Here, the number of observation n = 10 (even)
Now, medians is-
$\Rightarrow&space;\frac{value&space;of&space;the&space;(\frac{n}{2})^{th}observation&space;+&space;value&space;of&space;(\frac{n}{2}+1)^{th}&space;observation)}{2}$
$\Rightarrow \frac{value of the (\frac{10}{2})^{th}observation + value of (\frac{10}{2}+1)^{th} observation)}{2}$
$\Rightarrow \frac{value of the (5)^{th}observation + value of (6)^{th} observation)}{2}$
$\Rightarrow \frac{45 + 54}{2}$
$\Rightarrow \frac{99}{2}$
49.5
Hence the value of median is 49.5.

(ii) Find the median of this data : 15,6,16,8,22,21,9,18,,25
Solution: Arrange the data in the ascending order, we get-
6, 8, 9, 15, 16, 18, 21, 22, 25

Here, the number of observation n=9 (odd)
Now, median = $Value&space;of&space;(\frac{n+1}{2})^{th}&space;observation$
=$Value&space;of&space;(\frac{9+1}{2})^{th}&space;observation$
=$Value of (\frac{10}{2})^{th} observation$
=$Value of (5)^{th} observation$
= 16
Hence, value of median is 16.