Accessing S Chand Class 9 Maths Solutions ICSE Chapter 15 Mean, Median and Frequency Polygon Ex 15(B) can be a valuable tool for students seeking extra practice.

## S Chand Class 9 ICSE Maths Solutions Chapter 15 Mean, Median and Frequency Polygon Ex 15(B)

Find the median of the following :

Question 1.

2, 3, 5, 7, 9

Solution:

2, 3, 5, 7, 9

Here n = 5 which is odd

∵ The median is the middle most term

∴ Median = \(\frac { n+1 }{ 2 }\) term = \(\frac { 5+1 }{ 2 }\) th

= 3rd term which is 5

∴ Median = 5

Question 2.

4, 8, 12, 16, 20, 24, 28, 32

Solution:

Here n = 8 which is even

∵ The median is the middle most term

Median = mean of \(\frac { n }{ 2 }\) th and (\(\frac { n+1 }{ 2 }\)) th term

= \(\frac { 1 }{ 2 }\) (4th + 5th term)

= \(\frac { 1 }{ 2 }\) (16 + 20) = \(\frac { 36 }{ 2 }\) = 18

Hence median = 18

Question 3.

60, 33, 63, 61, 44, 48, 51

Solution:

Here number of terms (n) = 7 which is odd Arranging in ascending order,

33, 44, 48, 51, 60, 61, 63

∵ Median is the middle most term

∴ Median = \(\frac { n+1 }{ 2 }\) th term = \(\frac { 7+1 }{ 2 }\) th term

= 4th term which is 51

Hence median = 51

Question 4.

13, 22, 25, 8, 11, 19, 17, 31, 16, 10

Solution:

Arranging in ascending order

8, 10, 11, 13, 16, 17, 19, 22, 25, 31

Here number of terms (n) = 10 which is even

∵ The median is the middle most term

∴ Median = \(\frac{1}{2}\left[\frac{n}{2} \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }\right]\)

= \(\frac{1}{2}\left[\frac{10}{2} \text { th term }+\left(\frac{10}{2}+1\right) \text { th term }\right]\)

= \(\frac { 1 }{ 2 }\) [5th term + 6th term]

= \(\frac { 1 }{ 2 }\) [16+ 17] = \(\frac { 33 }{ 2 }\) = 16.5

∴ Median = 16.5

Question 5.

First 10 prime numbers.

Solution:

First 10 prime number are

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

∴ Here n = 10 which is even

∴ Median = \(\frac{1}{2}\left[\frac{n}{2} \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }\right]\)

= \(\frac{1}{2}\left[\frac{10}{2} \text { th term }+\left(\frac{10}{2}+1\right) \text { th term }\right]\)

= \(\frac { 1 }{ 2 }\)[5th term + 6th term]

= \(\frac { 1 }{ 2 }\) [11 + 13] = \(\frac { 24 }{ 2 }\) = 12

∴ Median = 12

Question 6.

15 students secured the following marks in a test in Statistics. Find the median marks : 35, 28, 13, 17, 20, 30, 19, 29, 11, 10, 29, 23,18,25,17

Solution:

No. of scores (n) = 15 which is odd

35, 28, 13, 17, 20, 30, 19, 29, 11, 10, 29, 23, 18, 25, 17

Arranging in ascending order

10, 11, 13, 17, 17, 18, 19, 20, 23, 25, 28, 29, 29, 30, 35

Median = \(\frac { n+1 }{ 2 }\)th term = \(\frac { 15+1 }{ 2 }\) = \(\frac { 16 }{ 2 }\) th

= 8 th term

Which is 20

∴ Median = 20 marks

Question 7.

Calculate the median from the following data:

Roll No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Marks | 25 | 55 | 5 | 45 | 15 | 35 | 60 |

Solution:

Here n = 7 which is odd

Arranging in descending order :

5, 15, 25, 35, 45, 55, 60

∴ Median = \(\frac { n+1 }{ 2 }\)th term = \(\frac { 7+1 }{ 2 }\) th term

= 4th term which is 35

∴ Median marks = 35

Question 8.

Out of total of 16 observations, arranged in ascending order, the 8th and 9th observations are 25 and 27. What is the median?

Solution:

No. of observations = 16 which is even

∴ Median = \(\frac{1}{2}\left[\frac{n}{2} \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }\right]\)

= \(\frac{1}{2}\left[\frac{16}{2} \text { th term }+\left(\frac{16}{2}+1\right) \text { th term }\right]\)

= \(\frac { 1 }{ 2 }\)[8th term + 9th term]

= \(\frac { 1 }{ 2 }\) (25 + 27) = \(\frac { 5 }{ 2 }\) = 26

∴ Median = 26

Question 9.

The median of the following observations arranged in ascending order 8, 9, 12, 18, (x + 2), (x + 4), 30, 31, 34, 39 is 24. Find x.

Solution:

The observations are given in ascending order 8, 9, 12, 18, (x + 2),(x + 4), 30, 31, 34, 39

Here n = 10 which is even

∴ Median = \(\frac{1}{2}\left[\frac{n}{2} \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }\right]\)

= \(\frac{1}{2}\left[\frac{10}{2} \text { th term }+\left(\frac{10}{2}+1\right) \text { th term }\right]\)

= \(\frac { 1 }{ 2 }\)[5th term + 6th term]

= \(\frac { 1 }{ 2 }\)[x + 2 + x + 4]

= \(\frac { 1 }{ 2 }\)[2x + 6] = x + 3

But median is given = 24

∴ x + 3 = 24 ⇒ x = 24 – 3 = 21

Hence x = 21

Question 10.

The following data have been in ascending order 18, 20, 25, 26, 30, *, 37, 38, 39, 48. If the median of the data is 35, find x. In the above data, if 48 is replaced by 28, find the new median.

Solution:

(i) The data is given ascending order

18,20, 25, 26, 30, x, 37, 38,39, 48

Which are 10 an even number

∴ Median = \(\frac{1}{2}\left[\frac{n}{2} \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }\right]\)

= \(\frac{1}{2}\left[\frac{10}{2} \text { th term }+\left(\frac{10}{2}+1\right) \text { th term }\right]\)

= \(\frac { 1 }{ 2 }\) [5th term + 6th term]

= \(\frac { 1 }{ 2 }\) [30 + x] = \(\frac { 30+x }{ 2 }\)

But median is given = 35

∴ \(\frac { 30+x }{ 2 }\) = 35 ⇒ 30 + x = 70

⇒ x = 70 – 30 = 40

(ii) By replacing 48 by 28, the terms will be

18, 20, 25, 26, 28, 30, .37, 38, 39, 40

New median = \(\frac { 1 }{ 2 }\) [5th term + 6th term]

= \(\frac { 1 }{ 2 }\) [28 + 30]

= \(\frac { 58 }{ 2 }\) = 29

Hence new median = 29