Chapter Notes – Statistics – Measures of Central Tendency
(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 x1,x2,x3,.....,xk are repeated f1,f2,f3,.....,fk times respectively then geometric mean will become:
G.M of X= X¯¯¯¯= xf11.xf22.xf33.....xfkk−−−−−−−−−−−−−−−√n (For Grouped Data)
Where n= f1+f2+f3+...fk
Example: Find the Geometric mean of the values 10, 5, 15, 8, 12
Solution: Here x1=10, x2=5, x3=15, x4=8, x5=12 and n=5
G.M of X= X¯¯¯¯ = 10×5×15×8×12−−−−−−−−−−−−−−−−−√5
X¯¯¯¯ = 72000−−−−−√5 = 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 = X¯¯¯¯ = n∑(1x)
H.M of X = X¯¯¯¯ = 50.3417 = 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-
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 x1,x2,x3,.....,xn are n values of a variable X, then the arithmetic mean of these values is given by X¯=x1+x2+x3+.....+xnn or, X¯=∑i=1nx1n
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 x1,x2,x3,.....,xn with corresponding frequencies f1,f2,f3,.....,fn respectively, then the arithmetic mean of these values is given by X¯=f1x1+f2x2+.....+fnxnf1+f2+.....+fn or, X¯=∑i=1nfixiN, where N=∑i=1nfi
For Example: Calculate the mean for the following distribution
Solution: Calculation of the Arithmetic mean-
Now, mean = =
= 7.025
Hence, value of mean is 7.025
(3) If X¯ is the mean of n observations x1,x2,....xn, then
(i) The algebraic sum of the deviations about X¯ is 0, i.e. ∑i=1n(xi−X¯)=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 ∑i=110(xi−X¯)=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 observationsx1±a,x2±a,.....,xn±a is X¯±a
Given, x¯=x1+x2+......+xnn …..(i)
The observation are (x1+a),(x2+a),......,(xn+a).
Mean =(x1+a)+(x2+a)+......+(xn+a)n
=x1+x2+......+xn+n×an
=x1+x2+......+xnn+n×an
From Equation (1); we get
Mean =(x¯+a)
So that given statement is true.
(iii) Prove that the mean of the observations ax1,ax2,....,axn is aX¯
For Example: Mean of x1,x2,.....,xn=x1+x2+......+xnn
But mean =x¯ (given)
x¯=x1+x2+......+xnn ……..(1)
The observation are ax1,ax2,.....,axn
Mean =ax1+ax2+......+axnn
=a(x1+x2+......+xn)n
=ax¯
Thus, the given statement is true.
(iv) Prove that the mean of the observations x1a,x2a,....,xna is x¯n
Proof: We have, X¯=1n(∑i=1nxi)
Let X¯′ be the mean of x1a,x2a,.....,xna. Then,
X¯′=1n(x1a+x2a+.....+xna)
X¯′=1n(x1+x2+.....+xna)
X¯′=1a(x1+x2+.....+xnn)
X¯′=1a[1n(∑i=1nxi)]=1a(X¯)
X¯′=X¯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 =(1×2+3+(−2)1+3,1×8+3×21+3)
=(−1,72)
Coordinates of Q =(2+(−2)2,2+82)
=(0,5)
Coordinates of R =(3×2+1×(−2)3+1,3×8+1×23+1)
=(1,132)
(5) lf x1,x2,....,xn are n values of a variable arranged in ascending or descending order, then
Median = value of (n+12)th observation, if n is odd
Median = (value of (n2)th observation + value of (n2+1)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.