Topic:
Mean, Mode and Median
By:
Tania Ilham
Arithmetic Mean
Definition:
 Arithmetic Mean is the average and is computed as sum of all the
observed outcomes from the data divided by the total number of events.
 It is usually denoted by the X
 Formula:
Sum of all the observations
X = Number of the observations
Example 01
 The marks obtained by 6 students in a class test are 20, 22, 24, 26, 28, 30. Find
the mean ?
Solution?
Arithmetic Mean = 20+22+24+26+28+30
6
Athematic Mean = 25
Conclusion:
Hence, Arithmetic Mean of the given data is 25.
Example 02
 Information regarding the sale of a shop for seven days of a particular week are given
below, find Arithmetic Mean?
= 67+69+66+68+72+76+54/ 7
= 67.42 Answer
Days Sales
Monday 67
Tuesday 69
Wednesday 66
Thursday 68
Friday 72
Saturday 76
Sunday 54
Mode
Definition:
 The mode of a set of data is the number with the highest frequency.
 Formula: For Grouped Data Continuous Series
 l = lower class boundary of the modal class
 fm= frequency of the modal class.
 f1 = frequency of the class preceding the modal class.
 f2 = frequency of the class following modal class.
 h= length of class interval of the modal class.
   
hx
ffff
ff
1Xˆ
2m1m
1m



Example 01
 From the given below data find Mode.
 Solution:
Because, 3 repeats 14 times and it is most frequent hence 3 is
the mode OR model Value.
Value No. of items (frequency)
1 3
2 8
3 14
4 3
5 4
6 2
7 1
Example 02
 Frequency Distribution of Child-Care Managers Age given below, find
mode.
Class Interval Frequency
20-29 6
30-39 18
40-49 11
50-59 11
60-69 3
70-79 1
 Putting values:
   
hx
ffff
ff
1Xˆ
2m1m
1m



   
18 6ˆ 29.5 10
18 6 18 11
12
29.5 10
12 7
120
29.5
19
29.5 6.3 35.8
X

  
  
  

 
  
MEDIAN
Definition:
 The median is the middle score.
 If we have an even number of events we take the average of
the two middles.
 The median is defined as a value which divides a set of data
into two halves, one half comprising of observations greater
than and the other half smaller than it. More precisely, the
median is a value at or below which 50% of the data lie.
 Formula:
Median= n+1/2
Example 01
Calculate the median of the following data.
Solution: let us arrange the data in ascending order then form cumulative frequency.
Here Σf = n = 43
Median = n+1/2 = 43+1/2
Median = 22nd Value
No. of
students
6 16 7 4 2 8
Marks 20 25 50 9 80 40
Marks 9 20 25 40 50 80
Frequency 4 6 16 8 7 2
CF 4 10 26 34 41 43
Example 02
 Find the median and median class of the data given below.
Class
Boundary
15-25 25-35 35-45 45-55 55-65 65-75
Frequency
4 11 19 14 0 2
Solution:
Hence,
Putting Values: n/2 = 50/2 = 25
L = 35, f = 19, cf = 15, h = 10
Median = 35 + 25-15 10
19
Median = 35 + 5.263 = 40.263 Answer
Class
Boundary 15-25 25-35 35-45 45-55 55-65 65-75
Midvalue (v) 20 30 40 50 60 70
Frequency 4 11 19 14 0 2
Cumulative
Frequency 4 15 34 48 48 50

Mean Mode & Median

  • 1.
    Topic: Mean, Mode andMedian By: Tania Ilham
  • 2.
  • 3.
    Definition:  Arithmetic Meanis the average and is computed as sum of all the observed outcomes from the data divided by the total number of events.  It is usually denoted by the X  Formula: Sum of all the observations X = Number of the observations
  • 4.
    Example 01  Themarks obtained by 6 students in a class test are 20, 22, 24, 26, 28, 30. Find the mean ? Solution? Arithmetic Mean = 20+22+24+26+28+30 6 Athematic Mean = 25 Conclusion: Hence, Arithmetic Mean of the given data is 25.
  • 5.
    Example 02  Informationregarding the sale of a shop for seven days of a particular week are given below, find Arithmetic Mean? = 67+69+66+68+72+76+54/ 7 = 67.42 Answer Days Sales Monday 67 Tuesday 69 Wednesday 66 Thursday 68 Friday 72 Saturday 76 Sunday 54
  • 6.
  • 7.
    Definition:  The modeof a set of data is the number with the highest frequency.  Formula: For Grouped Data Continuous Series  l = lower class boundary of the modal class  fm= frequency of the modal class.  f1 = frequency of the class preceding the modal class.  f2 = frequency of the class following modal class.  h= length of class interval of the modal class.     hx ffff ff 1Xˆ 2m1m 1m   
  • 8.
    Example 01  Fromthe given below data find Mode.  Solution: Because, 3 repeats 14 times and it is most frequent hence 3 is the mode OR model Value. Value No. of items (frequency) 1 3 2 8 3 14 4 3 5 4 6 2 7 1
  • 9.
    Example 02  FrequencyDistribution of Child-Care Managers Age given below, find mode. Class Interval Frequency 20-29 6 30-39 18 40-49 11 50-59 11 60-69 3 70-79 1
  • 10.
     Putting values:    hx ffff ff 1Xˆ 2m1m 1m        18 6ˆ 29.5 10 18 6 18 11 12 29.5 10 12 7 120 29.5 19 29.5 6.3 35.8 X                
  • 11.
  • 12.
    Definition:  The medianis the middle score.  If we have an even number of events we take the average of the two middles.  The median is defined as a value which divides a set of data into two halves, one half comprising of observations greater than and the other half smaller than it. More precisely, the median is a value at or below which 50% of the data lie.  Formula: Median= n+1/2
  • 13.
    Example 01 Calculate themedian of the following data. Solution: let us arrange the data in ascending order then form cumulative frequency. Here Σf = n = 43 Median = n+1/2 = 43+1/2 Median = 22nd Value No. of students 6 16 7 4 2 8 Marks 20 25 50 9 80 40 Marks 9 20 25 40 50 80 Frequency 4 6 16 8 7 2 CF 4 10 26 34 41 43
  • 14.
    Example 02  Findthe median and median class of the data given below. Class Boundary 15-25 25-35 35-45 45-55 55-65 65-75 Frequency 4 11 19 14 0 2
  • 15.
    Solution: Hence, Putting Values: n/2= 50/2 = 25 L = 35, f = 19, cf = 15, h = 10 Median = 35 + 25-15 10 19 Median = 35 + 5.263 = 40.263 Answer Class Boundary 15-25 25-35 35-45 45-55 55-65 65-75 Midvalue (v) 20 30 40 50 60 70 Frequency 4 11 19 14 0 2 Cumulative Frequency 4 15 34 48 48 50