Median

Median

The median is a variate value which divides the given data into two equal parts. If the data are arranged either in ascending or descending order, the middlemost number is median.
For example,
1, 3, 8, \(\underline{10}\), 13, 16, 20
The middle number 10 is median.

.

Calculation of median for individual series:
The formula for calculating median for individual series is given by;
Median (Md) = Position of \((\frac{N + 1}{2})\)th item
where, N = No. of observation or items.

The steps for calculation of median for individual series:

  1. Arrange the data in ascending or descending order.
  2. Count the number of terms (n).

Median for a discrete data:

Let, x1, x2, x3..................xn be the variable values and f1, f2, f3,............ fn be their corresponding frequency respectively.
Then, Median (Md) = Position of \((\frac{N + 1}{2})\)th item
where, N = total sum of a frequency

Median for a continuous data:

To calculate the median for a continuous data, the following steps are applied:

  1. Prepare cumulative frequency table
  2. Find the value of \((\frac{N}{2})\)th items which gives the median class.
  3. Use the formula, Median = l + \(\frac{\frac{N}{2}− c.f}{f}\) × i
    where,
    l = lower limit of median class
    c.f = cumulative frequency
    f = frequency
    i = class interval

  • Calculation of median for individual series = Position of \((\frac{N + 1}{2})\)th item
  • Median for a discrete data = Position of \((\frac{N + 1}{2})\)th item
  • Median for a continuous data = L + \(\frac{\frac{N}{2}− c.f}{F}\) × i

First, arrange the numbers from least to greatest. Remember that with negative numbers, larger numbers like -9 (if you ignore the minus sign) are less than smaller numbers.

-9 -8 -5 -5 -5 -4 -3 0

There is an even number of numbers, so there are two numbers in the middle.

-9 -8 -5 -5 -5 -4 -3 0

The median is the mean of the two middle numbers. Find the mean of -5 and -5.

-5 + -5 = -10
-10 ÷ 2 = -5

Hence, The median is -5.

First, arrange the numbers from least to greatest:

9.1 9.1 9.3 9.3 9.3 9.5 9.8
9.8 9.9

Now find the number in the middle.

9.1 9.1 9.3 9.3 9.3 9.5 9.8
9.8 9.9

The number in the middle is 9.3.

Hence, The median voltage was 9.3 volts.

Solution:

Here, Given data is 10 ,14, 16,. 20, 22, 25, 28

Numbers of terms(n) = 7

We have, Median =( \(\frac{n+1}{2}\))th item

( \(\frac{7+1}{2}\))th item

= (\(\frac{8}{2}\))th item

= 4th item

\(\therefore\) The median of the given data is 20.

Solution:

Arranging the data in ascending order

10, 12, 16, 20, 24, 28 , 30, 32

Number of terms(n) = 8

We have,

Median = (\(\frac{n+1}{2}\))th item

= (\(\frac{8+1}{2}\))th item

= (\(\frac{9}{2}\))th item

= (4.5)th item

Now,

Median = mean of 4th item and 5th item

= \(\frac{20+24}{2}\)

= \(\frac{44}{2}\)

= 22

\(\therefore\) Median = 22

Solution:

x f c.f
10 2 2
15 4 6
20 6 12
25 5 17
30 4 21
N = 21

We have,

Median = value of (\(\frac{n+1}{2}\))th item

= value of (\(\frac{21+1}{2}\))th item

= value of 11th item

In c.f. just greater than 11 is 12 and its corresponding value is 20.

\(\therefore\) Median is 20.

Solution:

x

f

c.f.

10

4

4

16

6

10

20

10

20

25

15

35

30

20

55

35

12

67

50

8

75

N=75

We have, = value of (\(\frac{N+1}{2}\))th item

= value of (\(\frac{75+1}{2}\))th item

= value of 38th item

In c.f. column, c.f. just greater then 38 is 55 and its corresponding value is 30.

\(\therefore\) Median = 30

Solution:

Marks

f

c.f.

80

3

3

75

7

10

60

10

20

55

6

26

50

4

30

40

2

32

35

1

33

N=33

We have, = value of (\(\frac{N+1}{2}\))th item

= value of (\(\frac{33+1}{2}\))th item

= value of 17th item

In c.f. column, c.f. just greater then 17 is 20 and its corresponding value is 60.

\(\therefore\) Median = 60


We have, = value of (\(\frac{N+1}{2}\))th item

= value of (\(\frac{40+1}{2}\))th item

= value of 20.5th item

In c.f. column, c.f. just greater then 20.5 is 28 and its corresponding value is 120.

\(\therefore\) Median = 120

Solution:

x

f

c.f.

10

4

4

20

6

10

30

10

20

40

8

28

50

5

33

N=33

We have, = value of (\(\frac{N+1}{2}\))th item

= value of (\(\frac{33+1}{2}\))th item

= value of 17th item

In c.f. column, c.f. just greater then 17 is 20 and its corresponding value is 30.

\(\therefore\) Median = 30

Solution:

Class Frequency c.f.
0-4 2 2
4-8 4 6
8-12 8 14
12-16 6 20
16-20 2 22
N=22

Here, \(\frac{N}{2}\) = \(\frac{22}{2}\) =11

Median Class = value of (\(\frac{N}{2}\))th item

= (11)th item

= (8-12)

Here,

l = 8, (\(\frac{N}{2}\))th item = 11, c.f.= 6, f = 8, i= 4

Now,

median = l + \(\frac{\frac{N}{2}− c.f}{f}\) × i

= 8 + \(\frac{11−6}{8}\) ×4

= 8+\(\frac{5}{8}\) ×4

= 8+ 2.5

= 10.5

\(\therefore\) median =10.5

Solution:

Marks Frequency c.f.
5-10 20 20
10-15 30 50
15-20 50 100
20-25 40 140
25-30 10 150
N=150

Here, \(\frac{N}{2}\) = \(\frac{150}{2}\) =75

Median Class = value of (\(\frac{N}{2}\))th item

= (75)th item

= (15-20)

l = 15, (\(\frac{N}{2}\))th item = 75, c.f.= 50, f = 50, i= 5

Now,

median = l + \(\frac{\frac{N}{2}− c.f}{f}\) × i

=15 + \(\frac{75−50}{50}\) ×5

= 15+\(\frac{25}{50}\) ×5

= 15+2.5

= 17.5

\(\therefore\) median =17.5

Solution:

x f c.f.
0-10 3 3
10-20 6 9
20-30 8 17
30-40 10 27
40-50 15 42
50-60 12 54
60-70 6 60
N=60

Here, \(\frac{N}{2}\) = \(\frac{60}{2}\) =30

Median Class = value of (\(\frac{N}{2}\))th item

= (30)th item

= (40-50)

Here,

l = 40, (\(\frac{N}{2}\))th item = 30, c.f.= 27, f =15, i=10

Now,

median = l + \(\frac{\frac{N}{2}− c.f}{f}\) × i

=40 + \(\frac{30−27}{15}\) ×10

= 40+\(\frac{3}{15}\) ×10

= 40+2

= 42

\(\therefore\) median = 42

Solution:

x

f

c.f.

0-10

5

5

10-20

8

13

20-30

11

24

30-40

15

39

40-50

20

59

50-60

12

71

60-70

9

80

N=80

Here, \(\frac{N}{2}\) = \(\frac{80}{2}\) =40

Median Class = value of (\(\frac{N}{2}\))thitem

= (40)thitem

= (40-50)

Here,

l= 40, (\(\frac{N}{2}\))thitem = 40, c.f.= 39, f =20, i=10

Now,

median = l + \(\frac{\frac{N}{2}− c.f}{f}\) × i

= 40 + \(\frac{40−39}{20}\) ×10

= 40+\(\frac{1}{20}\) ×10

= 40+0.5

= 40.5

\(\therefore\) median =40.5

Solution:

x

f

c.f.

0-5

5

5

5-10

9

14

10-15

15

29

15-20

22

51

20-25

18

69

25-30

11

80

N=80

Here, \(\frac{N}{2}\) = \(\frac{80}{2}\) =40

Median Class = value of (\(\frac{N}{2}\))thitem

= (40)thitem

= (15-20)

Here,

l= 15, (\(\frac{N}{2}\))thitem = 40, c.f.= 29, f =22, i=5

Now,

median = l + \(\frac{\frac{N}{2}− c.f}{f}\) × i

= 40 + \(\frac{40−29}{22}\) ×5

= 40+\(\frac{11}{22}\) ×5

=40+2.5

=42.5

\(\therefore\) median = 42.5

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    4
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    5
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    3
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    8
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    10
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    4
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    20
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