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.

-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

= 4^{th} 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 4^{th} item and 5^{th} item

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

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

= 22

\(\therefore\) Median = 22

####
Find the median of the given data:

x | 10 | 15 | 20 | 25 | 30 |

f | 2 | 4 | 6 | 5 | 4 |

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 11^{th} item

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

\(\therefore\) Median is 20.

####
Find the median from the given data:

x |
10 |
16 |
20 |
25 |
30 |
35 |
50 |

f |
4 |
6 |
10 |
15 |
20 |
12 |
8 |

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 38^{th} item

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

\(\therefore\) Median = 30

####
Find the median from the given data:

Marks |
80 |
75 |
60 |
55 |
50 |
40 |
35 |

no. of students |
3 |
7 |
10 |
6 |
4 |
2 |
1 |

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 17^{th} item

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

\(\therefore\) Median = 60

####
Find the median from the given data:

height (cm) |
110 |
115 |
120 |
124 |
128 |
130 |

no. of girls |
3 |
5 |
10 |
7 |
3 |
2 |

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

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

= value of 20.5^{th} item

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

\(\therefore\) Median = 120

####
Find the median from the given data:

x |
10 |
20 |
30 |
40 |
50 |

f |
4 |
6 |
10 |
8 |
5 |

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 17^{th} item

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

\(\therefore\) Median = 30

####
Find the median from the given data:

Age (in yrs) | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 |

No. of students | 2 | 4 | 8 | 6 | 2 |

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

####
Find the median from the given data:

Marks | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 |

Frequency | 20 | 30 | 50 | 40 | 10 |

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

####
Find the median from the given data:

x | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |

f | 3 | 6 | 8 | 10 | 15 | 12 | 6 |

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

####
Find the median from the given data:

x |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |

f |
5 |
8 |
11 |
15 |
20 |
12 |
9 |

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 (\(\frac{N}{2}\))^{th}item

= (40)^{th}item

= (40-50)

Here,

*l*= 40, (\(\frac{N}{2}\))^{th}item = 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

####
Find the median from the given data:

class interval |
0-5 |
5-10 |
10-15 |
15-20 |
20-25 |
25-30 |

frequency |
5 |
9 |
15 |
22 |
18 |
11 |

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 (\(\frac{N}{2}\))^{th}item

= (40)^{th}item

= (15-20)

Here,

*l*= 15, (\(\frac{N}{2}\))^{th}item = 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