## Mean, median and mode for grouped data

Let `$(x_i,f_i), i=1,2, \cdots , n$`

be given frequency distribution.

## Formula

## Sample mean

The mean of $X$ is denoted by $\overline{x}$ and is given by

`$\overline{x} =\dfrac{1}{N}\sum_{i=1}^{n}f_ix_i$`

In case of continuous frequency distribution, $x_i$'s are the mid-values of the respective classes.

## Sample median

The median is given by

`$\text{Median } = l + \bigg(\dfrac{\frac{N}{2} - F_<}{f}\bigg)\times h$`

where,

- $N$, total number of observations
- $l$, the lower limit of the median class
- $f$, frequency of the median class
- $F_<$, cumulative frequency of the pre median class
- $h$, the class width

## Sample mode

The mode of the distribution is given by

`$\text{Mode } = l + \bigg(\dfrac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h$`

where,

- $l$, the lower limit of the modal class
- $f_m$, frequency of the modal class
- $f_1$, frequency of the class pre-modal class
- $f_2$, frequency of the class post-modal class
- $h$, the class width

## Example 1

Following is the data about the daily number of car accidents during a month

No. of car accidents | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|

No. of days | 2 | 4 | 4 | 10 | 7 | 2 | 1 |

Caclulate mean, median and mode.

### Solution

x | f | f*x | cf | |
---|---|---|---|---|

1 | 2 | 2 | 2 | |

2 | 4 | 8 | 6 | |

3 | 4 | 12 | 10 | |

4 | 10 | 40 | 20 | |

5 | 7 | 35 | 27 | |

6 | 2 | 12 | 29 | |

7 | 1 | 7 | 30 | |

Total | 30 | 116 |

**Mean**

The mean number of accidents is

`$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{116}{30}\\ &=3.8667 \end{aligned} $$`

**Median**

Median number of accidents is
`$$ \begin{aligned} \text{Median} &=\bigg(\dfrac{N}{2}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{30}{2}\bigg)^{th}\text{ value}\\ &=\big(15\big)^{th}\text{ value} \end{aligned} $$`

The cumulative frequency just greater than or equal to $15$ is $20$. The corresponding value of $x$ is median. That is, $M =4$.

Thus, median number of accidents $M$ = $4$.

**Mode**

Mode is the value of $x$ with maximum frequency.

The maximum frequency is $10$. The value of $x$ corresponding to the maximum frequency $10$ is $\text{Mode }=4$.

## Example 2

The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students. Compute five number summary for the following frequency distribution.

Time spent on Internet ($x$) | 10-12 | 13-15 | 16-18 | 19-21 | 22-24 |
---|---|---|---|---|---|

No. of students ($f$) | 3 | 12 | 15 | 24 | 2 |

### Solution

The classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

Class Interval | Class Boundries | mid-value (x) | Freq (f) | f*x | cf | |
---|---|---|---|---|---|---|

10-12 | 9.5-12.5 | 11 | 3 | 33 | 3 | |

13-15 | 12.5-15.5 | 14 | 12 | 168 | 15 | |

16-18 | 15.5-18.5 | 17 | 15 | 255 | 30 | |

19-21 | 18.5-21.5 | 20 | 24 | 480 | 54 | |

22-24 | 21.5-24.5 | 23 | 2 | 46 | 56 | |

Total | 56 | 982 |

**Mean**

The mean time spent on internet is

`$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{982}{56}\\ &=17.5357 \text{ minutes} \end{aligned} $$`

**Median**

Median time spent on internet by the students is
`$$ \begin{aligned} \text{Median} &=\bigg(\dfrac{N}{2}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{56}{2}\bigg)^{th}\text{ value}\\ &=\big(28\big)^{th}\text{ value} \end{aligned} $$`

The cumulative frequency just greater than or equal to $28$ is $30$. The corresponding class $15.5-18.5$ is the median class.

Thus

- $N=56$, total number of observations
- $l = 15.5$, the lower limit of the median class
- $f =15$, frequency of the median class
- $F_< = 15$, cumulative frequency of the pre median class
- $h =3$, the class width

The median can be computed as follows:

`$$ \begin{aligned} \text{Median } &= l + \bigg(\frac{\frac{N}{2} - F_<}{f}\bigg)\times h\\ &= 15.5 + \bigg(\frac{\frac{56}{2} - 15}{15}\bigg)\times 3\\ &= 15.5 + \bigg(\frac{28 - 15}{15}\bigg)\times 3\\ &= 15.5 + \big(0.8667\big)\times 3\\ &= 15.5 + 2.6\\ &= 18.1 \text{ minutes} \end{aligned} $$`

**Mode**

The maximum frequency is $24$, the corresponding class $18.5-21.5$ is the modal class.

Mode of the given frequency distribution is:
`$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ \end{aligned} $$`

where,

- $l = 18.5$, the lower limit of the modal class
- $f_m =24$, frequency of the modal class
- $f_1 = 15$, frequency of the pre-modal class
- $f_2 = 2$, frequency of the post-modal class
- $h =3$, the class width

Thus mode of a frequency distribution is

`$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ &= 18.5 + \bigg(\frac{24 - 15}{2\times24 - 15 - 2}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{9}{31}\bigg)\times 3\\ &= 18.5 + \big(0.2903\big)\times 3\\ &= 18.5 + \big(0.871\big)\\ &= 19.371 \text{ minutes} \end{aligned} $$`