How do you Calculate Median of Grouped Frequency Distribution
Median of Grouped Frequency Distribution
Median = ℓ + \(\frac{{\frac{N}{2} – C}}{f}\,\, \times \,\,h\)
where,
ℓ = lower limit of median class interval
C = cumulative frequency preceding to the median class frequency
f = frequency of the class interval to which median belongs
h = width of the class interval
N = f1 + f2 + f3 + … + fn.
Working rule to find median
Step 1: Prepare a table containing less than type cumulative frequency with the help of given frequencies.
Step 2 : Find out the cumulative frequency to which \(\frac{N}{2}\) belongs. Class-interval of this cumulative frequency is the median class-interval.
Step 3 : Find out the frequency f and lower limit l of this median class.
Step 4 : Find the width h of the median class interval
Step 5 : Find the cumulative frequency C of the class preceding the median class.
Step 6 : Apply the formula,
Median = ℓ + \(\frac{{\frac{N}{2} – C}}{f}\,\, \times \,\,h\) to find the median
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Median of Grouped Frequency Distribution Example Problems with Solutions
Example 1: Find the median of the followng distribution :
| Wages (in Rs) | No. of labourers |
| 200 – 300 | 3 |
| 300 – 400 | 5 |
| 400 – 500 | 20 |
| 500 – 600 | 10 |
| 600 – 700 | 6 |
Solution: We have,
| Wages (in Rs) | No. of labours | Less than type cumulative frequency |
| 200 – 300 | 3 | 3 |
| 300 – 400 | 5 | 8 = C |
| 400 – 500 | 20 = f | 28 |
| 500 – 600 | 10 | 38 |
| 600 – 700 | 6 | 44 |
Here, the median class is 400 – 500 as \(\frac{44}{2}\) i.e. 22 belongs to the cumulative frequency of this class interval.
Lower limit of the median class = ℓ = 400
width of the class interval = h = 100
Cumulative frequency preceding median class frequency = C = 8
Frequency of Median class = f =20
Median = ℓ + h \(\left( {\frac{{\frac{N}{2} – C}}{f}} \right)\) = 400 + 100 \(\left( {\frac{{\frac{{44}}{2} – 8}}{{20}}} \right)\,\)
= 400 + 100 \(\left( {\frac{{22 – 8}}{{20}}} \right)\) = 400 + 100 \(\left( {\frac{{14}}{{20}}} \right)\)
= 400 + 70 = 470
Hence, the median of the given frequency distribution is 470.
Example 2: Find the median for the following :
| Class Interval | 0–8 | 8–16 | 16–24 | 24–32 | 32–40 | 40–48 |
| Frequency | 8 | 10 | 16 | 24 | 15 | 7 |
Solution:
| Class interval | Frequency | Less than type cumulative frequency |
| 0 – 8 | 8 | 8 |
| 8 – 16 | 10 | 18 |
| 16 – 24 | 16 | 34 = C |
| 24 – 32 | 24 = f | 58 |
| 32 – 40 | 15 | 73 |
| 40 – 48 | 7 | 80 |
Since \(\frac{80}{2}\) = 40 lies in the cumulative frequency of the class interval 24 – 32, so 24 – 32 belongs to the median class interval.
Lower limit of median class interval = ℓ = 24.
Width of the class interval = h = 8
Total frequency = N = 80
Cumulative frequency preceding median class frequency = C = 34
Frequency of median class = f = 24
Median = ℓ + \(\left( {\frac{{\frac{N}{2} – C}}{f}} \right)\,\,\,h\)
= 24 + \(\left( {\frac{{\frac{{80}}{2} – 34}}{{24}}} \right)\) 8 = 24 + \(\left( {\frac{{40 – 34}}{{24}}} \right)\) 8
= 24 + 2 = 26
Hence, the median of the given frequency distribution = 26.
Example 3: The following table shows the weekly drawn by number of workers in a factory :
| Weekly Wages (in Rs.) | 0–100 | 100–200 | 200–300 | 300–400 |
| No. of workers | 40 | 39 | 34 | 30 |
Find the median income of the workers.
Solution:
| Weekly Wages (in Rs.) | No. of workers | Less than type cumulative frequency |
| 0–100 | 40 | 40 |
| 100–200 | 39 | 79 = C |
| 200–300 | 34 = f | 113 |
| 300–400 | 30 | 143 |
| 400 – 500 | 45 | 188 |
Since \(\frac{188}{2}\) = 94 belongs to the cumulative frequency of the median class interval (200 – 300), so 200 – 300 is the median class.
Lower limit of the median class interval = ℓ = 200.
Width of the class interval = h = 100
Total frequency = N = 188
Frequency of the median class = f = 34
Cumulative frequency preceding median class
= C = 79
Median = ℓ + \(\left( {\frac{{\frac{N}{2} – C}}{f}} \right)\,\,\,h\) = 200 + \(\left( {\frac{{\frac{{188}}{2} – 79}}{{34}}} \right)\) 100
= 200 + \(\left( {\frac{{94 – 79}}{{34}}} \right)\) 100 = 200 + 44.117
= 244.117
Hence, the median of the given frequency distribution = 244.12.
Example 4: The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median and mode of the data and compare them.
| Monthly consumption | Number of consumers |
| 65 – 85 | 4 |
| 85 – 105 | 5 |
| 105 – 125 | 13 |
| 125 – 145 | 20 |
| 145 – 165 | 14 |
| 165 – 185 | 8 |
| 185 – 205 | 4 |
Solution:
| Monthly consumption | Number of consumers | Less than type cumulative frequency |
| 65 – 85 | 4 | 4 |
| 85 – 105 | 5 | 9 |
| 105 – 125 | 13 | 22 =C |
| 125 – 145 | 20 = f | 42 |
| 145 – 165 | 14 | 56 |
| 165 – 185 | 8 | 64 |
| 185 – 205 | 4 | 68 |
Since \(\frac{68}{2}\) belongs to the cumulative frequency (42) of the class interval 125 – 145, therefore 125 – 145 is the median class interval
Lower limit of the median class interval = ℓ = 125.
Width of the class interval = h = 20
Total frequency = N = 68
Cumulative frequency preceding median class frequency = C = 22
Frequency of the median class = f = 20
Median = ℓ + \(\left( {\frac{{\frac{N}{2} – C}}{f}} \right)\,\,\,h\) = 125 + \(\left( {\frac{{\frac{{68}}{2} – 22}}{{20}}} \right)\) 20
= 125 + \(\frac{{12 \times 20}}{{20}}\) = 125 + 12 = 137
The frequency of class 125 – 145 is maximum i.e., 20, this is the modal class,
xk = 125, fk = 20, fk-1 = 13, fk+1 = 14, h = 20
Mode = xk + \(\frac{{f – {f_{k – 1}}}}{{2f – {f_{k – 1}} – {f_{k + 1}}}}\)
= 125 + \(\frac{{20 – 13}}{{40 – 13 – 14}}\) × 20
= 125 + \(\frac{7}{{40 – 27}}\) × 20 = 125 + \(\frac{7}{{13}}\) × 20
= 125 + 10.77 = 135.77
Example 5: Compute the median from the marks obtained by the students of class X.
| Marks | Number of Students |
| 40 – 49 | 5 |
| 50 – 59 | 10 |
| 60 – 69 | 20 |
| 70 – 79 | 30 |
| 80 – 89 | 20 |
| 90 – 99 | 15 |
Solution: First we will form the less than type cumulative frequency distribution and we make the distribution continuous by subtracting 0.5 from the lower limits and adding 0.5 to the upper limits.
| Marks | Number of students | Less than type cumulative frequency |
| 39.5 – 49.5 | 5 | 5 |
| 49.5 – 59.5 | 10 | 15 |
| 59.5 – 69.5 | 20 | 35 = C |
| 69.5 – 79.5 | 30 = f | 65 |
| 79.5 – 89.5 | 20 | 85 |
| 89.5 – 99.5 | 15 | 100 |
Since \(\frac{100}{2}\) belongs to the cumulative frequency (65) of the class interval 69.5 – 79.5, therefore 69.5 – 79.5 is the median class.
Lower limit of the median class = ℓ = 69.5.
Width of the class interval = h = 10
Total frequency = N = 100
Cumulative frequency preceding median class frequency = C = 35
Frequency of median class = f = 30
Median = ℓ + \(\left( {\frac{{\frac{N}{2} – C}}{f}} \right)\,\,\,h\) = 69.5 + \(\left( {\frac{{\frac{{100}}{2} – 35}}{{30}}} \right)\) 10
= 69.5 + \(\left( {\frac{{50 – 35}}{{30}}} \right)\) 10 = 69.5 + \(\frac{{10 \times 15}}{{30}}\)
= 69.5 + 5 = 74.5
Hence, the median of given frequency distribution is 74.50.
Example 6: An incomplete frequency distribution is given as follows :
| Variable | Frequency |
| 10 – 20 | 12 |
| 20 – 30 | 30 |
| 30 –40 | ? |
| 40 – 50 | 65 |
| 50 – 60 | ? |
| 60 – 70 | 25 |
| 70 – 80 | 18 |
| Total | 229 |
Given that the median value is 46, determine the missing frequencies using the median formula.
Solution: Let the frequency of the class 30 – 40 be f1 and that of 50 – 60 be f2.
From the last item of the third column, we have
150 + f1 + f2 = 229
⇒ f1 + f2 = 229 – 150
⇒ f1 + f2 = 79
Since, the median is given to be 46, the class 40 – 50 is median class
Therefore, ℓ = 40, C = 42 + f1, N = 299, h = 10
Median = 46, f = 65
Median = ℓ + \(\left( {\frac{{\frac{N}{2} – C}}{f}} \right)\,\,\,h\) = 46
46 = 40 + 10 \(\frac{{\left( {\frac{{229}}{2} – 42 – {f_1}} \right)}}{{65}}\)
⇒ 6 = \(\frac{{10}}{{65}}\left( {\frac{{229}}{2} – 42 – {f_1}} \right)\)
⇒ 6 = \(\frac{2}{{13}}\left( {\frac{{229 – 84 – 2{f_1}}}{2}} \right)\)
⇒ 78 = 229 – 84 – 2f1 ⇒ 2f1 = 229 – 84 – 78
⇒ 2f1 = 67 ⇒ f1 = \(\frac{{67}}{2}\) = 33.5 = 34
Putting the value of f1 in (1), we have
34 + f2 = 79
⇒ f2 = 45
Hence, f1 = 34 and f2 = 45.
Example 7: Recast the following cumulative table in the form of an ordinary frequency distribution and determine the median.
| No. of days absent | No. of students |
| less than 5 | 29 |
| less than 10 | 224 |
| less than 15 | 465 |
| less than 20 | 582 |
| less than 25 | 634 |
| less than 30 | 644 |
| less than 35 | 650 |
| less than 40 | 653 |
| less than 45 | 655 |
Solution:
| No. of days | No. of students | No. of days absent | No. of students | Less than type cumulative frequency |
| less than 5 | 29 | 0 – 5 | 29 | 29 |
| less than 10 | 224 | 5 – 10 | 195 | 224 = C |
| less than 15 | 465 | 10 – 15 | 241 = f | 465 |
| less than 20 | 582 | 15 – 20 | 117 | 582 |
| less than 25 | 634 | 20 – 25 | 52 | 634 |
| less than 30 | 644 | 25 – 30 | 10 | 644 |
| less than 35 | 650 | 30 – 35 | 6 | 650 |
| less than 40 | 653 | 35 – 40 | 3 | 653 |
| less than 45 | 655 | 40 – 45 | 2 | 655 |
Since \(\frac{655}{2}\) belongs to the cumulative frequency (465) of the class interval 10 – 15, therefore 10 – 15 is the median class.
Lower limit of the median class = ℓ = 10.
Width of the class interval = h = 5
Total frequency = N = 655
Cumulative frequency preceding median class frequency = C = 224
Frequency of median class = f = 241
Median = ℓ + \(\left( {\frac{{\frac{N}{2} – C}}{f}} \right)\,\,\,h\) = 10 + 5 \(\left( {\frac{{\frac{{655}}{2} – 224}}{{241}}} \right)\)
= 10 + 5 \(\left( {\frac{{327.5 \times 224}}{{241}}} \right)\) = 10 + \(\frac{{5 \times 103.5}}{{241}}\)
= 10 + 2.147 = 12.147
Hence, the median of given frequency distribution is 12.147.