Table shows the marks of 130 students of class 10th.

Median of given data is:

Candidate marks 20-30 30-40 40-50 50-60 60-70 70-80

Number of candidates

0 4 18 60 33 15
  

  1. 61.11
  2. 57.16
  3. 47.47
  4. 54.17

Answer (Detailed Solution Below)

Option 2 : 57.16
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Detailed Solution

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Formula used: 

Where the median class is the class interval whose cumulative frequency ≥ N/2.

N = sum of frequencies, L = lower limit of the median class,

f = frequency of median class, h = width of the class 

C = cumulative frequency of the class just preceding the median class.
Calculation:

Class Frequency (f)

Cumulative

  frequency (C)

20 - 30 0 0
30 - 40 4 4
40 - 50 18 22
50 - 60 60 82
60 - 70 33 115
70 - 80 15 130

We can see that, the sum of frequency

N = 130

N/2 = 65

Hence, 50 - 60 will be the median class, where

L = 50, f = 60, h = 10 and c = 22

\(Median = L + \frac{\frac{N}{2}\ -\ C}{f}\times h\)

\(⇒ Median = 50 + \frac{\frac{130}{2}\ -\ 22}{60}\times 10\)

⇒ Median = 57.16

∴ The median of the data is 57.16

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