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!