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