#### Question

The following table gives the daily income of 50 workers of a factory:

Daily income (in Rs) | 100 - 120 | 120 - 140 | 140 - 160 | 160 - 180 | 180 - 200 |

Number of workers: | 12 | 14 | 8 | 6 | 10 |

Find the mean, mode and median of the above data.

#### Solution

Class interval |
Mid value(x) | Frequency(f) | fx | Cumulative frequency |

100 - 120 | 110 | 12 | 1320 | 12 |

120 - 140 | 130 | 14 |
1820 |
26 |

140 - 160 | 150 | 8 | 1200 | 34 |

160 - 180 | 170 | 6 | 1020 | 40 |

180 - 200 | 190 | 10 | 1900 | 50 |

N = 50 | `sumfx=7260` |

**Mean** `=(sumfx)/N=7260/50=145.2`

Thus, the mean of the given data is 145.2.

It can be seen in the data table that the maximum frequency is 14. The class corresponding to this frequency is 120−140.

∴ Modal class = 120 − 140

Lower limit of modal class (*l*) = 120

Class size (*h)* = 140 − 120 = 20

Frequency of modal class (*f*) = 14

Frequency of class preceding the modal class (*f*_{1}) = 12

Frequency of class succeeding the modal class (*f*_{2}) = 8

**Mode** `=l+(f-f1)/(2f-f1-f2)xxh`

`=120+(14-12)/(2xx14-12-8)xx20`

`=120+2/(28-20)xx20`

`=120+2/8xx20`

`=120+40/8`

= 120 + 5

= 125

Thus, the mode of the given data is 125.

Here, number of observations N = 50

So, N/2 = 50/2 = 25

This observation lies in class interval 120−140.

Therefore, the median class is 120−140.

Lower limit of median class (*l*) = 120

Cumulative frequency of class preceding the median class(*c*.*f*.) or (F) = 12

Frequency of median class(*f*) = 14

**Median** `=l+(N/2-F)/fxxh`

`=120+(25-12)/14xx20`

`=120+13/14xx20`

`=120+13/7xx10`

`=120+130/7`

= 120 + 18.57

= 138.57

Thus, the median of the given data is 138.57.