Selina Concise Class 10 Math Chapter 24 Exercise 24C Measures of Central Tendency Solutions

Selina Concise Class 10 Math Chapter 24 Exercise 24C Measures of Central Tendency Solutions

 

Exercise – 24C

(Q1) A student got the following marks in 9 questions of question paper.

3, 5, 7, 8, 0, 1, 4 and 6

Find the median of these marks

Solution:

Given number of marks 3, 5, 7, 3, 8, 0, 1, 4

These number of marks are arrange in increasing order 0, 1, 3, 4, 5, 6, 7, 8

{n (number of term) = odd}

So we have to use, median = (n+1/2)th term

and {n (Number of term = even) so, we have to use,

Median = 1/2 [(n/2)th + (n/2 + 1)th]

Given terms are in odd number, so we have to use

Median = (n+1/2)th term

= (9+1/2)

= 10/2

Median = 5th term

So, the 5th term is 4

∴ Median = 4

 

(Q2) The weights (in kg) of 10 students of a class are given below:

21, 28, 5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24

Find the median of their weights

Solution:

The given weights are, 21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24

Given terms are arrange in increasing order

20.5, 21, 22, 24, 24, 25.5, 27.5, 28, 28.5

Given terms are in even numbers so, we have to use,

Median = ½ [(n/2)th + (n/2 + 1)th] term

= 1/2 [(10/2)th + (10/2 + 1)th] term

= 1/2 [(5)th + (5+1)th] term

= 1/2 [(5)th + (6)th] term

= 1/2 [24 + 24]

The 5th and 6th terms are 24 and 24.

Median = 24+24/2

= 48/2

Median = 24

 

(Q3) The marks obtained by 19 students of a class are given below: 27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28

(i) Median

(ii) Lower quartile

(iii) Upper quartile

(iv) Interquartile

Solution:

(i) Median:

First we have arrange given numbers in increasing order.

22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37

Given are in odd numbers so, we have to use,

Median = (n+1/2)th term

= (19+1/2)th

= (20/2)th

= 10th term

∴ The 10th term is 29.

So, median = 29

(ii) Lower quartile:

if the number of terms are in odd number then we have to use –

Lower quartile = (n+1/4)th term

= (19+1/4)th

= (20/4)th

= 5th term

The 5th term is 26

The lower quartile is 26

(iii) Upper quartile:

If the number of terms are in odd number then we have to use –

Upper quartile = 3 (n+1/4) term

= 3 (19+1/4)th

= 3 (20/4)th

= 3 (5)th

= 15th term

∴ The 15th term is 35

∴ The upper quartile is 35

(iv) Interquartile:

We have to find interquartile, the difference between lower quartile and upper quartile.

Upper quartile – lower quartile = interquartile

35 – 26 = 9

∴ The interquartile = 9

 

(Q4) From the following data, find:

(i) Median

(ii) Upper quartile

(iii) Inter-quartile range

25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83

Solution:

First we have to arrange the given terms in increasing order.

0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95

(i) Median:

The given number of terms is in even we have to use

Median = 1/2 [(n/2)th + (n/2 + 1)th] term

∴ n = 16

Median = 1/2 [(16/2)th + (16/2 + 1)th]

= 1/2 [(8)th + (8 + 1)th] term

= 1/2 [8th + 9th] term

= 1/2 [40 + 45] term

∴ The 8th term = 40 and the 9th term is 45.

Median = 1/2 [85]

Median = 85/2

Median = 42.5

 

(ii) Upper quartile:

If the number of terms are in even then we have to use –

Upper quartile = 3 (n/4)th term

= 3 (16/4)th

= 3 (4)th

= 12th term

∴ The 12th term is 65

∴ The upper quartile is 65

 

(iii) Lower quartile

n = 16

Lower quartile = (n/4)th term

= (16/4)th

= 4th

∴ The 4th term is 18

∴ The lower quartile is 18

(iv) Inter – quartile:

The difference between upper quartile and lower quartile

Upper quartile – lower quartile = interquartile

65 – 18 = 47

 

 

Here is your solution of Selina Concise Class 10 Math Chapter 24 Exercise 24C Measures of Central Tendency

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