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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