Here, you are going to learn formula for mean median and mode and how to find mean, median and mode using those formula with examples.
In statistics, mean is the average of numbers.
The median of a series is the value of middle term of the series when the values are written in ascending order. Therefore median, divided an arranged series into two equal parts.
In a frequency distribution the mode is the value of that values which have the maximum frequency.
Formula for Mean Median and Mode :
The formula for mean median and mode for grouped and ungrouped frequency distribution is given below.
Mean :
(i) For ungrouped distribution : If \(x_1\), \(x_2\), …… \(x_n\) are n values of variate \(x_i\) then their mean \(\bar{x}\) is defined as
\(\bar{x}\) = \(x_1 + x_2, …… + x_n \over n\) = \({\sum_{i=1}^{n}x_i}\over n\)
\(\implies\) \(\sum x_i\) = n\(\bar{x}\)
(ii) For ungrouped and grouped frequency distribution :
If \(x_1\), \(x_2\), …… \(x_n\) are values of variate with corresponding frequencies \(f_1\), \(f_2\), …… \(f_n\), theb their A.M. is given by
\(\bar{x}\) = \(f_1x_1 + f_2x_2 + …… + f_nx_n \over f_1 + f_2 + …… + f_n\) = \({\sum_{i=1}^{n}f_ix_i}\over N\), where N = \({\sum_{i=1}^{n}f_i}\)
Example : Find the mean of the following freq. dist.
| \(x_i\) | 5 | 8 | 11 | 14 | 17 |
| \(f_i\) | 4 | 5 | 6 | 10 | 20 |
Solution : Here N = \(\sum f_i\) = 4 + 5 + 6 + 10 + 20 = 45
\(\sum f_ix_i\) = 606
\(\therefore\) \(\bar{x}\) = \(\sum f_ix_i\over N\) = \(606\over 45\) = 13.47
Median :
(i) For ungrouped distribution : Let n be the number of variate in a series then
Median = \(({n + 1\over 2})^{th}\) term, (when n is odd)
Median = Mean of \(({n\over 2})^{th}\) and \(({n\over 2} + 1)^{th}\) terms, (where n is even)
(ii) For ungrouped frequency distribution : First we prepare the cumulative frequency(c.f.) column and Find value of N then
Median = \(({N + 1\over 2})^{th}\) term, (when N is odd)
Median = Mean of \(({N\over 2})^{th}\) and \(({N\over 2} + 1)^{th}\) terms, (where n is even)
(iii) For grouped frequency distribution : Prepare c.f. column and find value of \(N\over 2\) then find the class which contain value of c.f. is equal or just greater to N/2, this is median class
Median = \(l\) + \(({N\over 2} – F)\over f\)\(\times\)h
where \(l\) – lower limit of median class
f – frequency of median class
F – c.f. of the class preceding median class
h – class interval of median class
Example : Find the median of the following frequency distribution.
| class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| \(f_i\) | 8 | 30 | 40 | 12 | 10 |
Solution :
| class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| \(f_i\) | 8 | 30 | 40 | 12 | 10 |
| c.f. | 8 | 38 | 78 | 90 | 100 |
Here \(N\over 2\) = \(100\over 2\) = 50 which lies in the value of 78 of c.f. hence corresponding class of this c.f. is 20 – 30 is the median class, so
\(l\) = 20, f = 40, f = 38, h = 10
\(\therefore\) Median = \(l\) + \(({N\over 2} – F)\over f\)\(\times\)h =
20 + \((50 – 38)\over 40\)\(\times\)10 = 23
Mode :
(i) For ungrouped distribution : The value of that variate which is repeated maximum number of times.
(ii) For ungrouped distribution : The value of that variate which have maximum frequency.
(iii) For grouped frequency distribution : First we find the class which have maximum frequency, this is model class.
\(\therefore\) Mode = (\(l\) + \(f_0 – f_1\over {2f_0 – f_1 – f_2}\))\(\times\)h
where \(l\) = lower limit of model class
\(f_0\) = freq. of model class
\(f_1\) = freq. of the class preceding model class
\(f_2\) = freq. of the class succeeding model class
h = class interval of model class
Relationship Between Mean Median and Mode
In a moderately asymmetric distribution the following relation between mean, median and mode of a distribution. It is known as impirical formula.
Mode = 3 Median – 2 Mean
Hope you learnt what is the formula for mean median and mode, learn more concepts of statistics and practice more questions to get ahead in the competition. Good luck!