ML Aggarwal ICSE Solutions Class 10 Math 20th Chapter Measures of Central Tendency
class 10 chapter 21 Measures of Central TendencyExercise- 21.1
(1) Calculate the arithmetic mean of 5, 7, 6.6, 7.2, 9.3, 6.2
Solution:
Given observations are 5, 7, 6.6, 7.2, 9.3, 6.2
sum of observations = (5 + 7 + 6.6 + 7.2 + 9.3 + 6.2) = 35
Arithmetic mean = 35/5 = 7
A. mean = 7 is the required arithmetic mean.
(2) The marks obtained by is students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 24, 17, 03, 20 find
(i) The mean of their marks
(ii) The mean of their marks when the marks of each student are increased by 4.
(iii) The mean of their marks when 2 marks are deducted from the marks of each student.
Solution:
Given that,
15 students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20.
Sum of observations = (12 + 14 + 07 + 09 + 23 + 11 + 08 + 13 + 11 + 19 + 16 + 24 + 17 + 03 + 20)
= 207
(i) Mean = 207/15
mean = 13.8 is the required mean.
(ii) If mark of each student is increased by 4.
Then total marks increased = 4 × 15 = 60
Total sum of marks of all students = 207 + 60 = 267
Then, mean = 267/15 = 17.8
mean = 17.8 is the required mean when marks of each student are increased by 4.
(iii) If mark of each student is deducted by 2.
Then total marks deducted = 15 × 2 = 30
Total sum of marks of all students = 30 – 207 = 177
Then, mean = 177/15 = 11.8
mean = 11.8 is the required mean when marks of each student are deducted by 2.
(3) (a) The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.
(b) The mean of 9 variants is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15. Find the 9thvariate.
Solution:
(a) Given observations : 6, y, 7, x, 14
sum of observations = 6 + y + 7 + x + 14 = 27 + x + y
mean = 8
But, (27 + x + y)/5 = 8
27 + x + y = 40
y = 13 – x is the required answer.
(b) Mean of 9 variants = 11
sum of given 8 variants = (7 + 12 + 9 + 14 + 21 + 3 + 8 + 15) = 89
Mean = (8 variants sum + 9thvariate)/9 = 89 + 9th/9 = 11
99 = 89 + 9thvariate
=> 9thvariate = 99 – 89
9thvariate = 10 is the required answer.
(5) Find the mean of 25 given numbers when the mean of 10 of them is 13 and the mean of remaining numbers is 18.
Solution:
Given that,
Mean of 10 numbers = 13
Mean of remaining is numbers = 18
(sum of 10 no.) + (sum of 16 no.) = total sum
(10 × 13) + (18 × 15) = total sum
130 + 270 = total sum
=> total sum = 400
Then, mean of 25 numbers = 400/25 = 16
mean = 16 is the required answer.
(6) Find the mean of the following distribution:
Number | 5 | 10 | 15 | 20 | 25 | 30 | 35 |
Frequency | 1 | 2 | 5 | 6 | 3 | 2 | 1 |
Solution:
From given distribution we can make following table.
x | f | fx |
5 | 1 | 5 |
10 | 2 | 20 |
15 | 5 | 75 |
20 | 6 | 120 |
25 | 3 | 75 |
30 | 2 | 60 |
35 | 1 | 35 |
Total | 20 | 390 |
Thus,
mean = Σfx/Σf = 390/20 = 19.5
Mean = 19.5 is the required mean for given distribution.
(7) The contents of 100 match boxes were checked to determine the number of matches they contained
No. of matches | 35 | 36 | 37 | 38 | 39 | 40 | 41 |
No. of boxes | 6 | 10 | 18 | 25 | 21 | 12 | 8 |
(i) Calculate, correct to one decimal place, the mean of no. of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches.
Solution:
From given distribution we can make following table.
No. of matches (x) | No. of boxes (f) | fx |
35 | 6 | 210 |
36 | 10 | 360 |
37 | 18 | 666 |
38 | 25 | 950 |
39 | 21 | 819 |
40 | 12 | 480 |
41 | 8 | 328 |
Total | 100 | 3813 |
Thus,
mean = Σfx/Σf
= 3813/100
mean = 38.13 is the required mean.
(ii) Given that, New mean = 39
Σfx = 39 × 100 = 3900
Then, no. of extra matches we have to add = 3900 – 3813 = 87
Thus, the no. of extra matches we have to add are found to be 87.
(8) Find the mean for the following distribution by short cut method.
Numbers | 60 | 61 | 62 | 63 | 64 | 65 | 66 |
Cumulative freq. | 8 | 18 | 33 | 40 | 49 | 55 | 60 |
Solution:
From given table we can make following frequency distribution table.
Numbers (x) | Cumulative freq. c.f | Frequency f | fx |
60 | 8 | 8 | 480 |
61 | 18 | 10 | 610 |
62 | 33 | 15 | 930 |
63 | 40 | 7 | 441 |
64 | 49 | 9 | 576 |
65 | 55 | 6 | 390 |
66 | 60 | 5 | 330 |
Total | 60 | 3757 |
Mean = Σfx/Σf = 3757/60 = 62.616 = 62.62
mean = 62.62 is the required mean for given distribution.
(10) If the mean of the following distribution is 7.5. Find the missing frequency f.
Variate | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Frequency | 20 | 17 | f | 10 | 8 | 6 | 7 | 6 |
Solution:
From given table we can make following distribution table.
Variate (x) | Frequency (f) | fx |
5 | 20 | 100 |
6 | 17 | 102 |
7 | f | 7f |
8 | 10 | 80 |
9 | 8 | 72 |
10 | 6 | 60 |
11 | 7 | 77 |
12 | 6 | 72 |
Total | Σf = 74 + f | 563 + 7f |
Thus, mean = Σfx/Σf
7.5 = 563 + 7f/74 + f
555 + 7.5f = 563 + 7f
0.5f = 8
f = 16 is the required missing frequency.
(11) Marks obtained by 40 students in a short assessment are given below. Where a and b are two missing data. If the mean of the distribution is 7.2, find a and b.
Marks | 5 | 6 | 7 | 8 | 9 |
No. of students | 6 | a | 16 | 13 | b |
Solution:
From given distribution table, we can make following frequency distribution table.
Marks (x) |
No. of students (f) |
fx |
5 | 6 | 30 |
6 | a | 6a |
7 | 16 | 112 |
8 | 13 | 104 |
9 | b | 9b |
35 + a + b = 40 | 246 + 6a + 9b |
Total no. of students = 40
Σf = 35 + a + b
40 = 35 + a + b
a = 5 – b …… (i)
Again, Mean = Σfx/Σf
7.2 = (246 + 6a + 3b)/40
=> 246 + 6a + 9b = 40 × 7.2
246 + 6a + 9b = 288
6a + 9b = 288 – 246
6a + 9b = 288 – 246
=> 2a + 3b = 14 ….. (ii)
from (i) & (ii)
2(5 – b) + 3b = 14
10 – 2b + 3b = 14
10 + b = 14
b = 4 put in (i)
=> a = 5 – b
a = 5 – 4
a = 1
Thus, required values of a and b found to be a = 1, b = 4.
(12) Calculate the mean of the following distribution:
Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Frequency | 8 | 5 | 12 | 35 | 24 | 16 |
Solution:
From given distribution table we can make following frequency distribution table.
Class interval | Frequency f1 | Class mark x1 | f1x1 |
0-10 | 8 | 5 | 40 |
10-20 | 5 | 15 | 75 |
20-30 | 12 | 25 | 300 |
30-40 | 35 | 35 | 1225 |
40-50 | 24 | 45 | 1080 |
50-60 | 16 | 55 | 880 |
Total | fi = 100 | fixi = 3600 |
Here, mean = Σfixi/Σfi = 3600/100
Mean = 36 is the required mean for given distribution.
(13) Calculate the mean of the following distribution using step deviation method.
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
No. of students | 10 | 9 | 25 | 30 | 16 | 10 |
Solution:
We can make following distribution table:
Marks | Mid values (xi) | No. of students (fi) | di = xi – A | ti = di/c | fiti |
0-10 | 5 | 10 | -20 | -2 | -20 |
10-20 | 15 | 9 | -10 | -1 | -9 |
20-30 | 25 | 25 = A | 0 | 0 | 0 |
30-40 | 35 | 30 | 10 | 1 | 30 |
40-50 | 45 | 16 | 20 | 2 | 32 |
50-60 | 55 | 10 | 30 | 3 | 30 |
Σfi = 100 | Σfiti = 63 |
Then, by step deviation method mean is given by
mean = x = A + h Σfiti/Σfi
= 25 + 10 × (63/100)
= 25 + 6.3
Mean = 31.3 is the required mean for given distribution by step deviation method.
(14) The data on the number of patients attending a hospital in a month are given below. Find the average (mean) no. of patients attending the hospital in a month by using the shortcut method.
Take assumed mean as 45. Give your answer correct to 2 decimal places.
No. of patients | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
No. of days | 5 | 2 | 7 | 9 | 2 | 5 |
Solution:
We can make following distribution table to find mean:
No. of patients | No. of days (fi) | Class marks | di = xi – a | fi di |
10-20 | 5 | 15 | -30 | -150 |
20-30 | 2 | 25 | -20 | -40 |
30-40 | 7 | 35 | -10 | -70 |
40-50 | 9 | a = 45 | 0 | 0 |
50-60 | 2 | 55 | 10 | 20 |
60-70 | 5 | 65 | 20 | 100 |
Σfi = 30 | Σfidi = -140 |
Then, Mean = x = a + fidi/fi
= 45 + (-140/30) = 45 – 4.67 = 40.33
Mean = 40.33 is the required mean for given distribution.
(16) Calculate the mean of the distribution given below using the shortcut method.
Marks | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 |
No. of students | 2 | 6 | 10 | 12 | 9 | 7 | 4 |
Solution:
From given distribution table we can make following table:
Marks | Frequency (f) | Mid value (x) | di = xi – A A = 45.5 |
ui = xi – A/10 | fxui |
11-20 | 2 | 15.5 | A – 30 | -3 | -6 |
21-30 | 6 | 25.5 | -20 | -2 | -12 |
31-40 | 10 | 35.5 | -10 | -1 | -10 |
41-50 | 12 | 45.5 = A | 0 | 0 | 0 |
51-60 | 9 | 55.5 | 10 | 1 | 9 |
61-70 | 7 | 65.5 | 20 | 2 | 14 |
71-80 | 4 | 75.5 | 30 | 3 | 12 |
Σf = 50 | Σfui = 7 |
Then, by short cut method mean is given by
mean = x = A + Σfidi/Σfi
= 45.5 + 70/50
= 45.5 + 1.4
Mean = 46.9 is the required mean for given distribution.
(17) A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean no. of days a students was absent.
No. of days | 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |
No. of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Solution:
From above table we can make following distribution table.
Mean = Σfixi/fi
No. of days | Mid value (x) | No. of students (f) | f × x |
0-6 | 3 | 11 | 33 |
6-10 | 8 | 10 | 80 |
10-14 | 12 | 7 | 84 |
14-20 | 17 | 4 | 68 |
20-28 | 24 | 4 | 96 |
28-38 | 33 | 3 | 99 |
38-40 | 39 | 1 | 39 |
Total | 40 | 499 |
Then, mean is given by
Mean = Σfixi/fi = 499/40
Mean = 12.475 is the required mean for given distribution.
(18) If the mean of the following distribution is 24. Find the value of ‘a’.
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
No. of students | 7 | a | 8 | 10 | 5 |
Solution:
From above table we can make following distribution table:
Marks | Class marks(xi) | No. of students (fi) | fi xi |
0-10 | 5 | 7 | 35 |
10-20 | 15 | a | 15a |
20-30 | 25 | 8 | 200 |
30-40 | 35 | 10 | 350 |
40-50 | 45 | 5 | 225 |
Total | 30 + a | 15a + 810 |
Mean = 24
∴ (15a + 810)/(30 + a) = 24
15a + 810 = 24(30 + a)
24a – 15a = 810 – 720
9a = 90
a = 10 is the required value of ‘a’ for given distribution.
(20) The following table gives the life time in days of 100 electricity tubes of a certain make:
Find the mean lifetime of electricity tubes.
Lifetime in days |
No. of tubes |
less than 50 | 8 |
less than 100 | 23 |
less than 150 | 55 |
less than 200 | 81 |
less than 250 | 93 |
less than 300 | 100 |
Solution:
From the given table we can make following distribution table:
Lifetime (in days) (class intervals) |
c.f. | frequency (f) | class mark (x) | u = x – A/h | fu |
0-50 | 8 | 8 | 25 | -3 | -24 |
50-100 | 23 | 15 | 75 | -2 | -30 |
100-150 | 55 | 32 | 125 | -1 | -32 |
150-200 | 81 | 26 | 175 | 0 | 0 |
200-250 | 93 | 12 | 225 | 1 | 12 |
250-300 | 100 | 7 | 275 | 2 | 14 |
Total | 100 | -60 |
Here, class mark = (upper limit + lower limit)/2
let us consider (A) = 175, class size (h) = 50
Then, Mean = 175 + 50 × (0.6)
= 175 + 50 (-60/100)
= 175 – 30
Mean = 145 is the required mean of lifetime of electricity tubes.
(21) Using the information given that in the histogram, calculate the mean correct to one decimal place.
Solution:
From the given histogram we can make following distribution table.
Class interval | Frequency (f) | Class mark (x) | f x |
20-30 | 3 | 25 | 75 |
30-40 | 5 | 35 | 175 |
40-50 | 12 | 45 | 540 |
50-60 | 9 | 55 | 495 |
60-70 | 4 | 65 | 260 |
Total | 33 | 1545 |
Mean by short cut method is given by,
Mean = x = A + Σfidi/Σfi
= 45 + 60/33
= 45 + 1.81
Mean = 46.81 is the required mean for given distribution.
Exercise – 21.2
(1) A student scored the following marks in 11 questions of a question paper: 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7. Find the median marks.
Solution:
Initially, we will arrange given marks in ascending order as follows : 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8.
Thus, n = 11 -> odd number
Then, median (middle term) = n+1/2 = 11+1/2 = 12/2 = 6th term
Thus, the required median is 6th term = 5.
(2) For the following set of the number, find the median:
10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15
Solution:
Initially, We will arrange given numbers in ascending order as follows:
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81.
Here, n = 12 -> even number
Thus, Median = {(n/2)th term + (n/2+1)th term/2}
= {(12/2)th term + (12/2+1)th term/2}
= (6th term + 7th tern)/2
= (15 + 17)/2
= 32/2
= 16
Median = 16
(3) Calculate the mean and median of the numbers:
2, 1, 0, 3, 1, 2, 3, 4, 3, 5
Solution:
Initially, we will write given numbers in ascending order as follow:
0, 1, 1, 2, 2, 3, 3, 3, 4, 5
Here, n = 10 -> even number
Median = {(n/2)th term + (n/2 + 1)th term}/2
= {10/2th term + (10/2 + 1)th term}/2
= {5th term + 6th term}/2
= (2 + 3)/2
= 5/2
Median = 2.5 is the required median for given numbers
Then, mean = sum of all numbers/observations/No. of observations
= Σxi/n
= (0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5)/10
= 24/10
= 2.4
Mean = 2.4 is the required mean for given numbers.
(4) The median of the observations 11, 12, 14 (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of ‘x’ and hence find the mean.
Solution:
Given observations in ascending order as follows:
11, 12, 14, (x-2), (x-4), (x+9), 32, 38, 47 => n = 9
Median = (n + ½)th term = (9+1/2) = 10/2 = 5th term
But, 5th term = x + 4
=> Median = x + 4
24 = x + 4
x = 20
Now, sum of all observations = [11 + 12 + 14 + (x-2) + (x+4) + (x+9) + 32 + 38 + 47]
= 165 + 3x
Mean = 165+3x/9 = (165 + 3 × 20/9) = 225/9 = 25
Mean = 25 is the required mean for given observations.
(5) The mean of the numbers 1, 7, 5, 3, 4, 4, is m. The numbers 3, 2, 4, 2, 3, 3, p have mean (m – 1) and median q.
Find (i) p
(ii) q
(iii) the mean of p and q.
Solution:
(i) Mean of given numbers 1, 7, 5, 3, 4, 4 is m.
Here, n = 6
Then, mean = m = (1 + 7 + 5 + 3 + 4 + 4)/6
m = 24/6
m = 4 is the required mean.
Again, the numbers 3, 2, 4, 2, 3, 3, p have m can (m – 1) and median q.
Thus, (m-1) = (3 + 2 + 4 + 2 + 3 + 3 + p)/7
(m – 1) = (17 + p)/7
(4 – 1) = (17 + p)/7 ∵ m = 4
3 × 7 = 17 + p
p = 21 – 17
=> p = 4 is the required value of p.
(ii) Also, given that the no. 3, 2, 4, 2, 3, 3, 4 has median q. We arrange given no. m ascending order as follows.
2, 2, 3, 3, 3, 4, 4 and n = 7 -> odd no.
Then, median = (n + 1)/2th term
q = (7 + 1)/2th term = 8/2th term
q = 4th term
q = 3 is the required value of q.
(iii) Mean of p and q is given by
mean = (p + q)/2 = (4 + 3)/2 = 7/2
mean = 3.5 is the required mean.
(6) Find the median for the following distribution
Wages per day (Rs) | 38 | 45 | 48 | 55 | 62 | 65 |
No. of workers | 14 | 8 | 7 | 10 | 6 | 2 |
Solution:
From above distribution table we will prepare following cumulative frequency table.
Wages per day (in Rs) | No. of workers | c.f. |
38 | 14 | 14 |
45 | 8 | 22 |
48 | 7 | 29 |
55 | 10 | 39 |
62 | 6 | 45 |
65 | 2 | 47 |
Here, n = 47 -> odd
median = (n+1)/2th term
= (48/2)th term
= 24th term
Median = 48 is the required median for given distribution.
(7) Marks obtained by 70 students are given below:
Marks | 20 | 70 | 50 | 60 | 75 | 90 | 40 |
No. of students | 8 | 12 | 18 | 18 | 9 | 5 | 12 |
Calculate the median marks.
Solution:
From the table we will make c.f. table as given below.
Marks | No. of students (f) | c.f. |
20 | 8 | 8 |
40 | 12 | 20 |
50 | 18 | 38 |
60 | 6 | 44 |
70 | 12 | 56 |
75 | 9 | 65 |
90 | 5 | 70 |
Here, n = 70
Median = [n/2th term + (n/2 + 1)th term]/2
= (70/2th term + (70/2 + 1)th term)/2
= (35th term + 36th term)/2
= (50 + 50)/2
= 50
Thus, Median = 50 is the required median for given distribution.
(8) Calculate the mean and median of the following distribution:
Number | 5 | 10 | 15 | 20 | 25 | 30 | 35 |
Frequency | 1 | 2 | 5 | 6 | 3 | 2 | 1 |
Solution:
From given distribution table we will make c.f. table as below.
Number (x) | frequency (f) | c.f. | fx |
5 | 1 | 1 | 5 |
10 | 2 | 3 | 20 |
15 | 5 | 8 | 75 |
20 | 6 | 14 | 120 |
25 | 3 | 17 | 75 |
30 | 2 | 19 | 60 |
35 | 1 | 20 | 35 |
Total | 20 | 390 |
Then, mean is found to be
mean = Σfx/Σf
= 390/20
Mean = 19.5 is the required mean for given distribution.
Here, n = 20 -> even
Then, median is given by
median = [nth/2 term + (n/2 + 1)th term]/2
= [20/2th term + (20/2 + 1)th term]/2
= (10th term + 11th term)/2
= (20 + 20)/2
Median = 20 is the required median for the given distribution.