Measures of Central Tendency - Class 9 : Notes



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(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 () = 
                  So, mean
                                    = 
            = 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

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(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-





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 () = 
                  So, mean
                                    = 

                                    = 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 = = 7.025
Hence, value of mean is 7.025

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(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
Taking LHS,
                   
               
 (9.6+ 5.2+3.5+1.5+1.6+2.4+2.6+8.4+10.3+10.9) - 10 x 
56 - 10 x 5.6
56 - 56
 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)

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(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-





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 =
=
=
=
                      = 16
Hence, value of median is 16.

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