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Solution for The Following Table Gives the Daily Income of 50 Workers of a Factory Find the Mean, Mode and Median of the Above Data. - CBSE Class 10 - Mathematics

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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 (f1) = 12

Frequency of class succeeding the modal class (f2) = 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.

  Is there an error in this question or solution?
Solution The Following Table Gives the Daily Income of 50 Workers of a Factory Find the Mean, Mode and Median of the Above Data. Concept: Mode of Grouped Data.
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