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

__Exercise – 24C__

__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 = 5^{th} term

So, the 5^{th} 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 5^{th} and 6^{th} 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}

= 10^{th} term

∴ The 10^{th} 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 5^{th} 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}

= 15^{th} term

∴ The 15^{th} 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 [8^{th} + 9^{th}] term

= 1/2 [40 + 45] term

∴ The 8^{th} term = 40 and the 9^{th} 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}

= 12^{th} term

∴ The 12^{th} term is 65

∴ The upper quartile is 65

(iii) Lower quartile

n = 16

Lower quartile = **(n/4) ^{th}** term

= (16/4)^{th}

= 4^{th}

∴ The 4^{th} 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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