Median Archives - CBSE Library

How do you Calculate the Median

Median of a distribution is the value of the variable which divides the distribution into two equal parts i.e. it is the value of the variable such that the number of observations above it is equal to the number of observations below it.
If the values xi in the raw data. are arranged in order of increasing or decreasing magnitude, then the middle, most value in the arrangement is called the median.
Algorithm :
Step I : Arrange the observations (values of the variate) in ascending or descending order of magnitude.
Step II : Determine the total number of observations, say, n.
Step III : If n is odd, then
Median = value of \({\left( {\frac{{n + 1}}{2}} \right)^{th}}\) observation
If n is even, then
Median = \(\frac{{Value\;of\;{{\left( {\frac{n}{2}} \right)}^{th}}observation + \;Value\;of\,{{\left( {\frac{n}{2} + 1} \right)}^{th}}observation}}{2}\)

The median can be calculated graphically while mean cannot be.
The sum of the absolute deviations taken from the median is less than the sum of the absolute deviations taken from any other observation in the data.
Median is not affected by extreme values.

Median Example Problems with Solutions

Example 1: Find the median of the following data :
25, 34, 31, 23, 22, 26, 35, 28, 20, 32
Solution: Arranging the data in ascending order, we get20, 22, 23, 25, 26, 28, 31, 32, 34, 35
Here, the number of observations n = 10 (even).
∴ Median = \(\frac{{Value\;of\;{{\left( {\frac{10}{2}} \right)}^{th}}observation + \;Value\;of\,{{\left( {\frac{10}{2} + 1} \right)}^{th}}observation}}{2}\)
⇒ Median = \(\frac{{Value\;of\;{5^{th}}observation\; + Value\;of\;{6^{th}}observation}}{2}\)
∴ Median = \(\frac{{26 + 28}}{2}\) = 27
Hence, median of the given data is 27.

Example 2: Find the median of the following values :
37, 31, 42, 43, 46, 25, 39, 45, 32
Solution: Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46
Here, the number of observations n = 9 (odd)
∴ Median = Value of \({\left( {\frac{{9 + 1}}{2}} \right)^{th}}\) observation
= Value of 5th observation = 39.

Example 3: The median of the observations 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the value of x.
Solution: Here, the number of observations n = 10. Since n is even, therefore
Median = \(\frac{{{{\left( {\frac{n}{2}} \right)}^{th}}observation\; + {{\left( {\frac{n}{2} + 1} \right)}^{th}}observation}}{2}\)
⇒   24 = \(\frac{{{5^{th}}observation + {6^{th}}observation}}{2}\)
⇒   24 = \(\frac{{(x + 2) + (x + 4)}}{2}\)
⇒   24 = \(\frac{{2x + 6}}{2}\)
⇒   24 = x + 3 ⇒ x = 21.
Hence, x = 21.

Example 4: Find the median of the following data : 19, 25, 59, 48, 35, 31, 30, 32, 51. If 25 is replaced by 52, what will be the new median.
Solution: Arranging the given data in ascending order, we have 19, 25, 30, 31, 32, 35, 48, 51, 59
Here, the number of observations n = 9 (odd)
Since the number of observations is odd. Therefore.
Median = Value of \({\left( {\frac{{9 + 1}}{2}} \right)^{th}}\) observations
⇒ Median = value of 5^th observation = 32.
Hence, Median = 32
If 25 is replaced by 52, then the new observations arranged in ascending order are :
19, 30, 31, 32, 35, 48, 51, 52, 59
∴ New median = Value of 5^th observation = 35.

Example 5: Calculate the median for the following distribution.

Weight (in kg)	Number of students
46	3
47	2
48	4
49	6
50	5
51	2
52	1

Solution: The cumulative frequency table is constructed as shown below :

Weight (x_i)	Number of students (f_i)	Cumulative frequency
46	3	3
47	2	5
48	4	9
49	6	15
50	5	20
51	2	22
52	1	23

Here, n = 23, which is odd
Median = \({t_{\frac{{23 + 1}}{2}}}\) = t₁₂ = 49
(i.e. weight of the 12th student when the weights have been arranged in order)

Example 6: The following data have been arranged in desending orders of magnitude 75, 70, 68, x + 2, x – 2, 50, 45, 40
If the median of the data is 60, find the value of x.
Solution: The number of observations are 8, the median will be the average of 4th and 5th number
⇒ Median = \(\frac{{(x + 2) + (x-2)}}{2}\)
⇒ 60 = \(\frac{{2x}}{2}\) ⇒ x = 60

Example 7: Find the median of the following data
(i)   17, 27, 37, 13, 18, 25, 32, 34, 23
(ii) 24, 37, 19, 41, 28, 32, 29, 31, 33, 21
Solution: (i)   The scores when arranged in ascending order are
13, 17, 18, 23, 25, 27, 32, 34, 37
Here, the number of scores n = 9 (odd)
∴    Median = \({t_{\frac{{9 + 1}}{2}}}\) = t₅ = 25
(ii) The scores when arraged in ascending order are
19, 21, 24, 28, 29, 31, 33, 34, 37, 41.
Total number of scores = 10, which is even. So there will be two middle-terms which are t₅ = 29 and t₆ = 31.
∴ Median = \(\frac{{{t_5} + {t_6}}}{2} = \frac{{29 + 31}}{2}\) = 30

Example 8: Find the median of the following data :
(i)   8, 10, 5, 7, 12, 15, 11
(ii) 12, 14, 10, 7, 15, 16
Solution: (i) 8, 10, 5, 7, 12, 15, 11
These numbers are arranged in an order
5, 7, 8, 10, 11, 12, 15
The number of observations = 7 (odd)
⇒   Median = \(\frac{7+1}{2}\) = 4th term
⇒   Median = 10
(ii) 12, 14, 10, 7, 15, 16
These numbers are arranged in an order
7, 10, 12, 14, 15, 16
The number of observations = 6 (even)
The medians will be mean of = 3rd and 4th terms i.e., 12 and 14
⇒ The median = \(\frac{12+14}{2}\) = 13

Example 9: The following data have been arranged in desending orders of magnitude 75, 70, 68, x + 2, x – 2, 50, 45, 40
If the median of the data is 60, find the value of x.
Solution: The number of observations are 8, the median will be the average of 4th and 5th number
⇒ Median = \(\frac{{(x + 2) + (x–2)}}{2}\)
⇒ 60 = \(\frac{2x}{2}\)
⇒ x = 60

Example 10: Find the median of 6, 8, 9, 10, 11, 12 and 13.
Solution: Total number of terms = 7
The middle terms = \(\frac{1}{2}\) (7 + 1) = 4th
Median = Value of the 4th term = 10.
Hence, the median of the given series is 10.

Example 11: Find the median of 21, 22, 23, 24, 25, 26, 27 and 28.
Solution: Total number of terms = 8
Median
= Value of \(\frac{1}{2}\left[ {\frac{8}{2}th\,\,term + \,\left( {\frac{8}{2} + 1} \right)th\,term} \right]\)
= Value of \(\frac{1}{2}\) [4th term + 5th term]
= \(\frac{1}{2}\) [24 + 25] = \(\frac{49}{2}\) = 24.5

Example 12: The number of runs scored by 11 players of a cricket team of school are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27. Find the median.
Solution: Let us arrange the value in ascending order
0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52
∴ Median M = \({\left( {\frac{{n + 1}}{2}} \right)^{th}}\) value
= \({\left( {\frac{{11 + 1}}{2}} \right)^{th}}\) value = 6th value
Now 6th value in data is 27
∴ Median = 27 runs.

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 + … + f_n.
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

Read More:

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,
x_k = 125, f_k = 20, f_k-1 = 13, f_k+1 = 14, h = 20
Mode = x_k + \(\frac{{f – {f_{k – 1}}}}{{2f – {f_{k – 1}} – {f_{k + 1}}}}\)
= 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.