ML Aggarwal Solutions for Chapter 20 Statistics Class 9 Maths ICSE
Exercise 20.1
1. Find the mean of 8, 6, 10, 12, 1, 3, 4, 4.
Solution
Given data,
8, 6, 10, 12, 1, 3, 4, 4
Here, n = 8
∴ Mean (x̄) = Æ© x_{i}/n
= (8 + 6 + 10 + 12 + 1+ 3 + 4 + 4)/8
= 48/8 = 6
Therefore, mean of the given data is 6.
2. 5 people were asked about the time in a week they spend in doing social work in their community. They replied 10, 7, 13, 20 and 15 hours, respectively. Find the mean time in a week devoted by them for social work.
Solution
Given data,
10, 7, 13, 20, 15
Here, n = 5
∴ Mean (x̄) = Æ© x_{i}/n
= (10 + 7 + 13 + 20 + 15)/5
= 65/5 = 13
Therefore, the mean time in a week devoted by them for social work is 13 hours.
3. The enrollment of a school during six consecutive years was as follows:
1620, 2060, 2540, 3250, 3500, 3710.
Find the mean enrollment.
Solution
Given data,
1620, 2060, 2540, 3250, 3500, 3710
Here, n = 6
∴ Mean (x̄) = Æ© x_{i}/n
= (1620 + 2060 + 2540 + 3250 + 3500 + 3710)/5
= 16680/6 = 2780
Therefore, the mean enrollment is 2780.
4. Find the mean of the first twelve natural numbers.
Solution
The first twelve natural numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Here, n = 12
∴ Mean (x̄) = Æ© x_{i}/n
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/12
= 78/12 = 6.5
Therefore, the mean of the first twelve natural numbers is 6.5
5. (i) Find the mean of the first six prime numbers.
(ii) Find the mean of the first seven odd prime numbers.
Solution
(i) First 6 prime numbers are 2, 3, 5, 7, 11, 13
Here, n = 6
∴ Mean (x̄) = Æ© x_{i}/n
= (2 + 3 + 5 + 7 + 11 + 13)/6
= 41/6
Therefore, the mean of the first six prime numbers is 41/6.
(ii) First seven odd prime numbers are 3, 5, 7, 11, 13, 17, 19
Here, n = 7
∴ Mean (x̄) = Æ© x_{i}/n
= (3 + 5 + 7 + 11 + 13 + 17 + 19)/7
= 75/7
Therefore, the mean of the first six prime numbers is 75/7.
6. (i)The marks (out of 100) obtained by a group of students in a Mathematics test are 81, 72, 90, 90, 85, 86, 70, 93 and 71. Find the mean marks obtained by the group of students.
(ii) The mean of the age of three students Vijay, Rahul and Rakhi is 15 years. If their ages are in the ratio 4 : 5 : 6 respectively, then find their ages.
Solution
(i) The marks obtained by the group of students are:
81, 72, 90, 90, 85, 86, 70, 93, 71
Here, n = 9
∴ Mean (x̄) = Æ© x_{i}/n
= (81 + 72 + 90 + 90 + 85 + 86 + 70 + 93 + 71)/9
= 738/9 = 82
Therefore, the mean marks obtained by the group of students is 82.
(ii) Given, the mean of the age of three students Vijay, Rahul and Rakhi is 15 years.
So, n = 3
Now, the sum of ages of the 3 students = 15×3 = 45
Also given, ratio of their ages is 4 : 5 : 6
Sum of ratios = 4 + 5 + 6 = 15
Hence,
Vijay’s age = (45/15)×4 = 12 years
Rahul’s age = (45/15)×5 = 15 years
Rakhi’s age = (45/15)×6 = 18 years
7. The mean of 5 numbers is 20. If one number is excluded, mean of the remaining numbers becomes 23. Find the excluded number.
Solution
Given,
The mean of 5 numbers = 20
So, the total sum of the numbers = 20×5 = 100
After excluding one number,
The mean of the remaining 4 numbers = 23
So, the total sum of these numbers = 23×4 = 92
Hence, the excluded number is = 100 – 92 = 8.
8. The mean of 25 observations is 27. If one observation is included, the mean still remains 27. Find the included observation.
Solution
Given,
The mean of 25 observations is 27.
So, the total sum of all the 25 observations = 27×25 = 675
After one observation is included,
Now the mean of 26 (25 + 1) numbers = 27
So, the total sum of all the 26 observations = 27×26 = 702
Hence, the included observation = 702 – 675 = 27
9. The mean of 5 observations is 15. If the mean of first three observations is 14 and that of the last three is 17, find the third observation.
Solution
Given,
The mean of 5 observations = 15
So, total sum of the 5 observations = 15×5 = 75
Also given,
Mean of first 3 observations = 14
So, the sum of the 3 observations = 14×3 = 42
And, the mean of last 3 observations = 17
So, the sum of last 3 observations = 17×3 = 51
Thus, the total of 3 + 3 observations = 42 + 51 = 93
Hence, The third observation = 93 – 75 = 18.
10. The mean of 8 variate is 10.5. If seven of them are 3, 15, 7, 19, 2, 17 and 8, then find the 8^{th} variate.
Solution
Given,
Seven out of eight variates are: 3, 15, 7, 19, 2, 17 and 8
Mean of 8 variates = 10.5
So, the total of 8 variates = 10.5×8 = 84
Now,
Sum of seven variates = (3 + 15 + 7 + 19 + 2 + 17 + 8) = 71
Hence, the 8^{th} variate = 84  71 = 13.
11. The mean weight of 8 students is 45.5 kg. Two more students having weights 41.7 kg and 53.3 kg join the group. What is the new mean weight?
Solution
Given,
The mean weight of 8 students = 45.5 kg
So, the total weight of 8 students = 45.5 x 8 = 364 kg
Weight of two more students are 41.7 kg and 53.3 kg
Now,
The total weight of 10 (8 + 2) students = 364 + 41.7 + 53.3
= 364 + 95
= 459 kg
Hence, the new mean weight of all the 10 students = 459/10 = 45.9 kg
12. Mean of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. Find the correct mean of the observations.
Solution
Given,
Mean of 9 observations = 35
So, the sum of all 9 observations = 35×9 = 315
Now, the difference due to misread = 81 – 18 = 63
Thus, the actual sum = 315 + 63 = 378
Hence, the actual mean = 378/ 9 = 42.
13. A student scored the following marks in 11 questions of a question paper:
7, 3, 4, 1, 5, 8, 2, 2, 5, 7, 6.
Find the median marks.
Solution
Given,
Marks scored in 11 questions of a question paper by the student are:
7, 3, 4, 1, 5, 8, 2, 2, 5, 7, 6
Arranging it in descending order, we have
1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8
Here, n = 11 which is odd
∴ Median = (n + 1)/2^{th} term
= (11 + 1)/2 = 12/2 = 6^{th} term i.e 5
Hence, the median mark is 5.
14. Calculate the mean and the median of the numbers:
2, 3, 4, 3, 0, 5, 1, 1, 3, 2.
Solution
First arrange the number in descending order
0, 1, 1, 2, 2, 3, 3, 3, 4, 5
So n = 10 which is even
Mean (x̄) = Æ© x_{i}/n
Substituting the values
= (0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5)/10
By further calculation
= 24/10
= 2.4
Median = ½ [10/2th + (10/2 + 1)th terms]
It can be written as
= ½ (5^{th} + 6^{th}) term
Substituting the values
= ½ (2 + 3)
= 5/2
= 2.5
15. A group of students was given a special test in Mathematics. The test was completed by the various students in the following time in (minutes):
24, 30, 28, 17, 22, 36, 30, 19, 32, 18, 20, 24.
Find the mean time and median time taken by the students to complete the test.
Solution
First arrange the data in descending order
17, 18, 19, 20, 22, 24, 24, 28, 30, 30, 32, 36
So n = 12 which is even
Mean (x̄) = Æ© x_{i}/n
Substituting the values
= (17 + 18 + 19 + 20 + 22 + 24 + 24 + 28 + 30 + 30 + 32 + 36)/2
So we get
= 300/12
= 25
Median = ½ [12/2th + (12/2 + 1)th terms]
It can be written as
= ½ (6^{th} + 7^{th}) terms
Substituting the values
= ½ (24 + 24)
= ½ (48)
= 24
16. In a Science test given to a group of students, the marks scored by them (out of 100) are
41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52.
Find the mean and median of this data.
Solution
On arranging the marks obtained by the students, we have
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Here, n = 15 which is odd
∴ Mean (x̄)
Æ© x_{i}/n = (39 + 40 + 40 + 41 + 42 + 46 + 48 + 52 + 52 + 52 + 54 + 60 + 62 + 96 + 98)/15
= 822/15 = 54.8
And,
Median = (15 + 1)/2^{th} term
= 16/2 = 8^{th} term i.e. 52
Therefore, for the given data mean = 54.8 and median = 52.
17. The points scored by a Kabaddi team in a series of matches are as follows:
7, 17, 2, 5, 27, 15, 8, 14, 10, 48, 10, 7, 24, 8, 28, 18.
Find the mean and the median of the points scored by the Kabaddi team.
Solution
Let’s arrange the given data in descending order:
2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48
Here, n = 16 when is even
∴ Mean (x̄)
Æ© x_{i}/ n = (2 + 5 + 7 + 7 + 8 + 8 + 10 + 10 + 14 + 15 + 17 + 18 + 24 + 27 + 28 + 48)/15
= 248/16 = 15.5 points
And,
Median = ½ [(16/2)^{th} term + (16/2 + 1)^{th} term]
= ½ (8^{th} term + 9^{th} term)
= ½ (10 + 14)
= ½ x 24 = 12 points
Therefore, the mean and the median of the points scored by the Kabaddi team are 15.5 and 12 respectively.
18. The following observations have been arranged in ascending order. If the median
the data is 47.5, find the value of x.
17, 21, 23, 29, 39, 40, x, 50, 51, 54, 59, 67, 91, 93.
Solution
Given data,
17, 21, 23, 29, 39, 40, x, 50, 51, 54, 59, 67, 91, 93
Here, n = 14 which is even
As the given data is arranged in descending order
Median = ½ [(14/2)^{th} term + (14/2 + 1)^{th} term]
= ½ (7^{th} term + 8^{th} term)
⇒ 47.5 = ½ (x + 50)
95 = x + 50
x = 95 – 50 = 45
Hence, the value of x is 45.
19. The following observations have been arranged in ascending order. If the median
the data is 13, find the value of x.
3, 6, 7, 10, x, x + 4, 19, 20, 25, 28.
Solution
Given observations in ascending order,
3, 6, 7, 10, x, x + 4, 19, 20, 25, 28
Here, n = 10 which is even and median = 13
So, Median = ½ [(10/2)^{th} term + (10/2 + 1)^{th} term]
= ½ (5^{th} term + 6^{th} term)
= ½ (x + x + 4)
= (2x + 4)/2
= x + 2
⇒ x + 2 = 13
x = 13 – 2 = 11
Hence, the value of x is 11.
Exercise 20.2
1. State which of the following variables are continuous and which are discrete:
(i) marks scored (out of 50) in a test.
(ii) daily temperature of your city.
(iii) sizes of shoes.
(iv) distance travelled by a man.
(v) time.
Solution
(i) Discrete
(ii) Continuous
(iii) Discrete
(iv) Continuous
(v) Continuous
2. Using class intervals 0 – 4, 5 – 9, 10 – 14, …… construct the frequency distribution for the following data:
13, 6, 10, 5, 11, 14, 2, 8, 15, 16, 9, 13, 17, 11, 19, 5, 7, 12, 20, 21, 18, 1, 8, 12, 18.
Solution
The frequency distribution for the following data is
Class 
Tally marks 
Frequency 
04 
II 
2 
59 

7 
1014 

8 
1519 

6 
2024 
II 
2 
3. Given below are the marks obtained by 27 students in a test:
21, 3, 28, 38, 6, 40, 20, 26, 9, 8, 14, 18, 20, 16, 17, 10, 8, 5, 22, 27, 34, 2, 35, 31, 16, 28, 37.
(i) Using the class intervals 110, 1120 etc. construct a frequency table.
(ii) State the range of these marks.
(iii) State the class mark of the third class of your frequency table.
Solution
(i) The frequency table of the given data is
Class 
Tally marks 
Frequency 
110 

7 
1120 

8 
2130 

6 
3140 

6 
(ii) Range of these marks is 38.
(iii) The class mark of the third class of your frequency table = (21 + 30)/ 2 = 25.5
4. Explain the meaning of the following terms:
(i) variate
(ii) class size
(iii) class mark
(iv) class limits
(v) true class limits
(vi) frequency of a class
(vii) cumulative frequency of a class.
Solution
(i) Variant: A particular value of a variable is called variate.
(ii) Class size: The difference between the actual upper limit and the actual lower limit of a class is called its class size.
(iii) Class mark: The class mark of a class is the value midway between its actual lower limit and actual upper limit.
(iv) Class limits: In the frequency table the class interval is called class limits.
(v) True class limits: In a continuous distribution, the class limits are called true or actual class limits.
(vi) Frequency of a class: The number of tally marks opposite to a variate is its frequency and it is written in the next column opposite to tally marks of the variate.
(vii) Cumulative frequency of a class: The sum of frequency of all previous classes and that particular class is called the cumulative frequency of the class.
5. Fill in the blanks:
(i) The number of observations in a particular class is called …… of the class.
(ii) The difference between the class marks of two consecutive classes is the ….. of the class.
(iii) The range of the data 16, 19, 23, 13, 11, 25, 18 is …
(iv) The midpoint of the class interval is called its …
(v) The class mark of the class 4 – 9 is ….
Solution
(i) The number of observations in a particular class is called frequency of the class.
(ii) The difference between the class marks of two consecutive classes is the size of the class.
(iii) The range of the data 16, 19, 23, 13, 11, 25, 18 is 14.
(iv) The midpoint of the class interval is called its class marks.
(v) The class mark of the class 4 – 9 is 6.5. [Class mark = (4 + 9)/2 = 13/2 = 6.5]
6. The marks obtained (out of 50) by 40 students in a test are given below:
28, 31, 45, 03, 05, 18, 35, 46, 49, 17, 10, 28, 31, 36, 40, 44, 47, 13, 19, 25, 24, 31, 38, 32,
27, 19, 25, 28, 48, 15, 18, 31, 37, 46, 06, 01, 20, 10, 45, 02.
(i) Taking class intervals 1 10, 11 – 20, .., construct a tally chart and a frequency
distribution table.
(ii) Convert the above distribution to continuous distribution.
(iii) State the true class limits of the third class.
(iv) State the class mark of the fourth class.
Solution
(i) A tally chart and a frequency distribution of given data is
(ii) Converting the above distribution to continuous distribution.
(iii) The true class limits of the third class=lower limit = 20.5 and upper limit = 30.5
(iv) The class mark of the fourth class (31 + 40)/2 = 71/2 = 35.5
7. Use the adjoining table to find:
(i) upper and lower limits of fifth class.
(ii) true class limits of the fifth class.
(iii) class boundaries of the third class.
(iv) class mark of the fourth class.
(v) width of sixth class.
Class 
Frequency 
2832 
5 
3337 
8 
3842 
13 
4347 
9 
4852 
7 
5357 
5 
5862 
2 
Solution
(i) Upper and lower limits of fifth class are as follows.
Upper limit = 52 and lower limit = 48
(ii) True class limits of the fifth class
Upper limit = 52.5 and lower limit = 47.5
(iii) Class boundaries of the third class is 37.5 and 42.5.
(iv) Class mark of the fourth class = (43 + 47)/2 = 90/2 = 45
(v) Width of sixth class=57.5 – 52.5 = 5
8. The marks of 200 students in a test were recorded as follows:
Marks % 
1019 
2029 
3039 
4049 
5059 
6069 
7079 
8089 
No. of students 
7 
11 
20 
46 
57 
37 
15 
7 
Draw the cumulative frequency table.
Solution
The cumulative frequency table is as follows:
Marks % (Class) 
Frequency 
Cumulative Frequency 
1019 
7 
7 
2029 
11 
18 
3039 
20 
38 
4049 
46 
84 
5059 
57 
141 
6069 
37 
178 
7079 
15 
193 
8089 
7 
200 
9. Given below are the marks secured by 35 students in a test:
41, 32, 35, 21, 11, 47, 42, 00, 05, 18, 25, 24, 29, 38, 30, 04, 14, 24, 34, 44, 48, 33, 36, 38, 41, 46, 08, 34, 39, 11, 13, 27, 26, 43, 03.
Taking class intervals 010, 1020, 2030 …., construct frequency as well as cumulative frequency distribution table. Find the number of students obtaining below 20 marks.
Solution
The cumulative frequency distribution table is given below:
Class 
Tally Marks 
Frequency 
Cumulative Frequency 
010 

5 
5 
1020 

5 
10 
2030 

7 
17 
3040 

10 
27 
4050 

8 
35 
The number of students obtaining below 20 marks is 10.
10. The marks out of 100 of 50 students in a test are given below:
5 35 6 35 18 36 12 36 85 32
20 36 22 38 24 50 22 39 74 31
25 54 25 64 25 70 28 66 58 25
29 72 31 82 31 84 31 82 37 21
32 84 32 92 35 95 34 92 35 5
(i) Taking a class interval of size 10, construct a frequency as well as cumulative frequency table for the given data.
(ii) Which class has the largest frequency?
(iii) How many students score less than 40 marks?
(iv) How many students score first division (60% or more) marks?
Solution
(i) The cumulative frequency table for the given data
Class 
Tally marks 
Frequency 
Cumulative frequency 
010 
III 
3 
3 
1020 
II 
2 
5 
2030 

11 
16 
3040 

18 
34 
4050 

18 
34 
5060 
III 
3 
37 
6070 
II 
2 
39 
7080 
III 
3 
42 
8090 

5 
47 
90100 
III 
3 
50 
(ii) 3040 is the class which has the largest frequency.
(iii) 34 students score less than 40 marks.
(iv) 13 students score first division (60% or more) marks.
11. Construct the frequency distribution table from the following data:
Ages (in years) 
Below 4 
Below 7 
Below 10 
Below 13 
Below 16 
No. of children 
7 
38 
175 
248 
300 
State the number of children in the age group 1013.
Solution
The frequency distribution table from the given data:
Class 
Frequency 
04 
7 
47 
31 
710 
137 
1013 
73 
1316 
52 
Hence, the number of children in the age group 1013 is 73.
12. Rewrite the following cumulative frequency distribution into frequency distribution:
Less than or equal to 10 2
Less than or equal to 20 7
Less than or equal to 30 18
Less than or equal to 40 32
Less than or equal to 50 43
Less than or equal to 60 50
Solution
The given cumulative frequency distribution is rewritten into frequency distribution below:
Class 
Frequency 
010 
2 
1120 
5 
2130 
11 
3140 
14 
4150 
11 
5160 
7 
13. The water bills (in rupees) of 32 houses in a locality are given below. Construct a
frequency distribution table with a class size of 10.
80, 48, 52, 78, 103, 85, 37, 94, 72, 73, 66, 52, 92, 85, 78, 81, 64, 60, 75, 78, 108, 63, 71, 54,
59, 75, 100, 103, 35, 89, 95, 73.
Solution
A frequency distribution with a class size of 10 is follows:
14. The maximum temperatures (in degree Celsius) for Delhi for the month of April, 2014, as reported by the Meteorological Department, are given below:
27.4, 28.3, 23.9, 23.6, 25.4, 27.5, 28.1, 28.4, 30.5, 29.7, 30.6, 31.7, 32.2, 32.6, 33.4, 35.7, 36.1, 37.2, 38.4, 40.1, 40.2, 40.5, 41.1, 42.0, 42.1, 42.3, 42.4, 42.9, 43.1, 43.2.
Construct a frequency distribution table.
Solution
The frequency distribution table of the given data is as follows:
Class 
Tally marks 
Frequency 
23.527.5 
IIII 
4 
27.531.5 

7 
31.535.5 
IIII 
4 
35.539.5 
IIII 
4 
39.543.5 

11 
15. (i) The class marks of a distribution are 94, 104, 114, 124, 134, 144 and 154. Determine the class size and the class limits of the fourth class.
(ii) The class marks of a distribution are 9.5, 16.5, 23.5, 30.5, 37.5 and 44.5. Determine the class size and the class limits of the third class.
Solution
(i) We know that
Class size is the difference between two successive class marks
Class size = 104 – 94 = 10
Class limits of the fourth class
Lower limit = 119 and upper limit = 129
(ii) We know that
Class size is the difference between two successive class marks
Class size = 16.5 – 9.5 = 7
Class limits of the third class
Lower limit = 20 and upper limit = 27.
Exercise 20.3
1. The area under wheat cultivation last year in the following states, correct to the nearest lacs hectares was:
State 
Punjab 
Haryana 
U.P. 
M.P. 
Maharashtra 
Rajasthan 
Cultivated area 
220 
120 
100 
40 
80 
30 
Represent the above information by a bar graph.
Solution
The required bar graph is given below:
2. The number of books sold by a shopkeeper in a certain week was as follows:
Day 
Monday 
Tuesday 
Wednesday 
Thursday 
Friday 
Saturday 
No. of books 
420 
180 
230 
340 
160 
120 
Draw a graph for the above data.
Solution
The required bar graph is given below:
3. Given below is the data of percentage of passes of a certain school in the ICSE for consecutive years:
Year 
2000 
2001 
2002 
2003 
2004 
2005 
2006 
% of passes 
92 
80 
70 
86 
54 
78 
94 
Draw a bar graph to represent the above data.
Solution
The required bar graph is given below:
4. Birth rate per thousand of different countries over a certain period is:
Country 
India 
Pakistan 
China 
U.S.A. 
France 
Birth rate 
36 
45 
12 
18 
20 
Draw a horizontal bar graph to represent the above data.
Solution
The required bar graph is given below:
5. Given below is the data of number of students (boys and girls) in class IX of a certain school:
Class 
IX A 
IX B 
IX C 
IX D 
Boys 
28 
22 
40 
15 
Girls 
18 
34 
12 
25 
Draw a bar graph to represent the above data.
Solution
The required bar graph is given below:
6. Draw a histogram to represent the following data:
Marks obtained 
010 
1020 
2030 
3040 
4050 
5060 
No. of students 
4 
10 
6 
8 
5 
9 
Solution
The required histogram is given below:
7. Draw a histogram to represent the following frequency distribution of monthly wages of 255 workers of a factory.
Monthly wages (in rupees) 
850950 
9501050 
10501150 
11501250 
12501350 
No. of workers 
35 
45 
75 
60 
40 
Solution
The required histogram is given below:
8. Draw a histogram for the following data:
Class marks 
12.5 
17.5 
22.5 
27.5 
32.5 
37.5 
Frequency 
7 
12 
20 
28 
8 
11 
Solution
The required histogram is given below:
9. Draw a histogram for the following frequency distribution:
Age (in years) 
Below 2 
Below 4 
Below 6 
Below 8 
Below 10 
Below 12 
No. of children 
12 
15 
36 
45 
72 
90 
Solution
First convert the given cumulative frequency into frequency distribution table:
Age (in years) 
c.f. 
f. 
02 
12 
12 
24 
15 
3 
46 
36 
21 
68 
45 
9 
810 
72 
27 
1012 
90 
18 
So represent age on xaxis and number of children on yaxis and draw the histogram as given below:
10. Draw a histogram for the following data:
Classes 
5965 
6672 
7379 
8086 
8793 
94100 
Frequency 
10 
5 
25 
15 
30 
10 
Solution
First write the given data in continuous classes:
Classes 
Classes after adjustment 
Frequency 
5956 
58.565.5 
10 
6672 
65.572.5 
5 
7379 
72.579.5 
25 
8086 
79.586.5 
15 
8793 
86.593.5 
30 
94100 
93.5100.5 
10 
Represent classes on xaxis and frequency on yaxis and draw a histogram as given below:
11. Draw a frequency polygon for the following data:
Class intervals 
4050 
5060 
6070 
7080 
8090 
90100 
Frequency 
15 
28 
45 
32 
41 
18 
Solution
Take class intervals on xaxis and frequency on yaxis.
Construct a histogram and then taking the midpoint of each class join them with xaxis to get the frequency polygon.
12. In a class of 60 students, the marks obtained in a monthly test were as under:
Marks 
1020 
2030 
3040 
4050 
5060 
Students 
10 
25 
12 
08 
05 
Draw a frequency polygon to represent the above data.
Solution
Consider marks on xaxis and number of students on yaxis.
Construct a histogram and then by joining the midpoints with xaxis we set a frequency polygon as given below.
13. In a class of 90 students, the marks obtained in a weekly test were as under:
Marks 
1620 
2125 
2630 
3125 
3640 
4145 
4650 
No. of students 
4 
12 
18 
26 
14 
10 
6 
Draw a frequency polygon for the above data.
Solution
Now write the classes as continuous classes:
Marks 
Classes after adjustment 
Class Mark 
No. of students (f) 
1620 
15.520.5 
18.0 
4 
2125 
20.525.5 
23.0 
12 
2630 
25.530.5 
28.0 
18 
3135 
30.535.5 
33.0 
26 
3640 
35.540.5 
38.0 
14 
4145 
40.545.5 
43.0 
10 
4650 
45.550.5 
48.0 
6 
Represent marks on xaxis and frequency on yaxis and then draw frequency polygon as shown below:
14. In a city, the weekly observations made in a study on the cost of living index are given in the following table:
Cost of living index 
140150 
150160 
160170 
170180 
180190 
190200 
Number of weeks 
5 
10 
20 
9 
6 
2 
Draw a frequency polygon for the data given above.
Solution
Cost of living index 
No. of weeks 
Midpoints 
140150 
5 
145 
150160 
10 
155 
160170 
20 
165 
170180 
9 
175 
180190 
6 
185 
190200 
2 
195 
Mark the midpoints of the cost of living index on xaxis and number of weeks on the yaxis.
Plot the midpoints and join them to form a frequency polygon as shown in the figure.
15. Construct a combined histogram and frequency polygon for the following data:
Weekly earnings (in rupees) 
150165 
165180 
180195 
195210 
210225 
225240 
No. of workers 
8 
14 
22 
12 
15 
6 
Solution
Take weekly earnings on xaxis and number of workers on yaxis.
Draw histogram with given data and then by joining the midpoints of each class with xaxis we find frequency polygon as given below.
16. In a study of diabetic patients, the following data was obtained:
Age (in years) 
1020 
2030 
3040 
4050 
5060 
6070 
7080 
No. of patients 
3 
8 
30 
36 
27 
15 
6 
Represent the above data by a histogram and a frequency polygon.
Solution
Take age (in years) on xaxis and number of patients on yaxis.
Draw a histogram with the given data and then by joining the midpoints of the classes with xaxis, we get a frequency polygon as given below.
17. The water bills (in rupees) of 32 houses in a locality are given below:
30, 48, 52, 78, 103, 85, 37, 94, 72, 73, 66, 52, 92, 65, 78, 81, 64, 60, 75, 78, 108, 63, 71, 54, 59, 75, 100, 103, 35, 89, 95, 73.
Taking class intervals 3040, 4050, 5060, ….. ,form frequency distribution table. Construct a combined histogram and frequency polygon.
Solution
First represent the given data in the form of a frequency distribution table.
Class Intervals 
Tally Marks 
Frequency 
3040 
III 
3 
4050 
I 
1 
5060 
IIII 
4 
6070 

5 
7080 

9 
8090 
III 
3 
90100 
III 
3 
100110 
IIII 
4 
Take class intervals on xaxis and frequency on yaxis and draw a histogram with the given data.
Join the midpoints of each class with xaxis, we get a frequency polygon as given below.
18. The number of matchsticks in 40 boxes on counting was found as given below:
44, 41, 42, 43, 47, 50, 51, 49, 43, 42, 40, 42, 44, 45, 49, 42, 46, 49, 45, 49, 45, 47, 48, 43, 43, 44, 48, 43, 46, 50, 43, 52, 46, 49, 52, 51, 47, 43, 43, 45.
Taking classes 4042, 4244 ….., construct the frequency distribution table for the above data. Also draw a combined histogram and frequency polygon to represent the distribution.
Solution
First represent the given data in a frequency distribution table as shown below:
Class Intervals 
Tally Marks 
Frequency 
4042 
II 
2 
4244 

12 
4446 

7 
4648 

6 
4850 

7 
5052 
IIII 
4 
5254 
II 
2 
Total 
40 
Take class on xaxis and frequency on yaxis and draw histogram with the given data.
Join the midpoint of each class with the xaxis, we get a frequency polygon as given below.
19. The histogram showing the weekly wages (in rupees) of workers in a factory is given alongside.
Answer the following about the frequency distribution:
(i) What is the frequency of the class 400425?
(ii) What is the class having minimum frequency?
(iii) What is the cumulative frequency of the class 425 – 450?
(iv) Construct a frequency and cumulative frequency table for the given distribution.
Solution
From the figure given in the question, we get
(i) The frequency of the class 400425 is 18.
(ii) The class having minimum frequency is 475500.
(iii) The cumulative frequency of the class 425450 is (6 + 18 + 10) = 34.
(iv) The frequency table for the given distribution is:
Classes 
Frequency 
Cumulative Frequency 
375400 
6 
6 
400425 
18 
24 
425450 
10 
34 
450475 
20 
54 
475500 
4 
58 
20. The runs scored by two teams A and B on the first 42 balls in a cricket match are given below:
No. of balls 
16 
712 
1318 
1924 
2530 
3136 
3742 
Runs scored by Team A 
2 
1 
8 
9 
4 
5 
6 
Runs scored by Team B 
5 
6 
2 
10 
5 
6 
3 
Draw their frequency polygons on the same graph.
Solution
No. of balls 
Class marks 
Team A 
Team B 
16 
3.5 
2 
5 
712 
9.5 
1 
6 
1318 
15.5 
8 
2 
1924 
21.5 
9 
10 
2530 
27.5 
4 
5 
3136 
33.5 
5 
6 
3742 
39.5 
6 
3 
Frequency polygons for both the teams is given on the same graph below:
Chapter test
1. Find the mean and the median of the following set of numbers:
8, 0, 5, 3, 2, 9, 1, 5, 4, 7, 2, 5.
Solution
By arranging in descending order
0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9
n = 12 which is even
Mean (x̄) = Æ© x_{i}/n
Substituting the values
= (0 + 1 + 2 + 2 + 3 + 4 + 5 + 5 + 5 + 7 + 8 + 9)/12
By further calculation
= 51/12
= 17/4
= 4.25
Median = ½ [12/2th + (12/2 + 1)th terms]
= ½ [6^{th} + 7^{th} terms]
Substituting the values
= ½ (4 + 5)
= 9/2
= 4.5
2. Find the mean and the median of all the (positive) factors of 48.
Solution
We know that
Positive factors of 48 are
1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Here N = 10 which is even
Mean (x̄) = Æ© x_{i}/ n
Substituting the values
= (1 + 2 + 3 + 4 + 6 + 8 + 12 + 16 + 24 + 48)/10
So we get
= 124/10
= 12.4
Median = ½ [10/2 th + (10/2 + 1)th terms]
= ½ [5^{th} + 6^{th} terms]
Substituting the values
= ½ (6 + 8)
= 14/2
= 7
3. The mean weight of 60 students of a class is 52.75 kg. If the mean weight of 35 of them is 54 kg, find the mean weight of the remaining students.
Solution
It is given that
Mean weight of 60 students of a class=52.75 kg
So the total weight of 60 students = 52.75 × 60 = 3165 kg
Mean weight of 35 students among them = 54 kg
So the total weight of 35 students = 54 × 35 = 1890 kg
Remaining students = 60 – 35 = 25
Total weight of 25 students = 3165 – 1890 = 1275 kg
So the mean weight of 25 students = 1275 ÷ 25 = 51 kg
Hence, the mean weight of the remaining students is 51 kg.
4. The mean age of 18 students of a class is 14.5 years. Two more students of age 15 years and 16 years join the class. What is the new mean age?
Solution
It is given that
Mean age of 18 students = 14.5 years
Total age = 14.5 × 18 = 261 years
Total age of 2 more students = 15 + 16 = 31 years
Total age of 18 + 2 = 20 students = 261 + 31 = 292 years
Mean age = 292/20 = 14.6 years
Hence, the new mean age is 14.6 years.
5. If the mean of the five observations x + 1, x + 3, x + 5, 2x + 2, 3x + 3 is 14, find the mean of first three observations.
Solution
We know that
Mean of the five observations
x + 1, x + 3, x + 5, 2x + 2, 3x + 3 is 14
Mean = (x + 1 + x + 3 + x + 5 + 2x + 2 + 3x + 3)/5
By further calculation
= (8x + 14)/5
Equating to mean
(8x + 14)/5 =14
By cross multiplication
8x + 14 = 70
So we get
8x = 70 – 14 = 56
By division
x = 56/8 = 7
Mean of x + 1 + x + 3 + x + 5
We get
Mean = (x + 1 + x + 3 + x + 5)/3
By further calculation
= (3x + 9)/3
= x + 3
Substituting the value of x
= 7 + 3
= 10
Hence, the mean of first three observations is 10.
6. The mean height of 36 students of a class is 150.5 cm. Later on, it was detected that the height of one student was wrongly copied as 165 cm instead of 156 cm. Find the correct mean height.
Solution
It is given that
Mean height of 36 students of a class=150.5 cm
Total height = 150.5 × 36 = 5418 cm
Difference in height which was wrongly copied = 165 + 56 = 9 cm
Actual height = 5418 – 9 = 5409 cm
Actual mean height = 5409/36 = 150.25 cm
Hence, the correct mean height is 150.25 cm.
7. The mean of 40 items is 35. Later on, it was discovered that two items were misread as 36 and 29 instead of 63 and 22. Find the correct mean.
Solution
It is given that
Mean of 40 items = 35
Total of 40 items = 35 × 40 = 1400
Difference between two items which were wrongly read = (63 + 22) – (36 + 29)
By further calculation
= 85 – 65
= 20
Here,
Actual total = 1400 + 20 = 1420
Correct mean = 1420/40 = 35.5
Hence, the correct mean is 35.5.
8. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 75, 87, 91.
Solution
We know that
N = 10 which is even
Median = ½ [n/2 th + (n/2 + 1)th term]
It can be written as
63 = ½ [10/2th + (10/2 + 1)th term]
63 = ½ (5^{th} + 6^{th} terms)
Substituting the values
63 = ½ (x + x + 2)
63 × 2 = 2x + 2
By further calculation
2x = 126 – 2 = 124
By division
x = 124/2 = 62
Therefore, the value of x is 62.
9. Draw a histogram showing marks obtained by the students of a school in a Mathematics paper carrying 60 marks.
Marks 
010 
1020 
2030 
3040 
4050 
5060 
Students 
4 
5 
10 
8 
30 
40 
Solution
Represent marks on xaxis and number of students on yaxis and draw a histogram as shown below:
10. In a class of 60 students, the marks obtained in a surprise test were as under:
Marks 
1420 
2026 
2632 
3238 
3844 
4450 
5056 
5662 
No. of students 
4 
10 
9 
15 
12 
5 
3 
2 
Represent the above data by a histogram and a frequency polygon.
Solution
Take marks on xaxis and number of students on the yaxis. Draw a histogram from the given data and then by joining the midpoint of each class with xaxis we obtain a frequency polygon as given below.
11. Construct a combined histogram and frequency polygon for the following distribution:
Classes 
91100 
101110 
111120 
121130 
131140 
141150 
151160 
Frequency 
16 
28 
44 
20 
32 
12 
4 
Solution
Write the classes in continuous frequency classes:
Classes 
Classes after adjustment 
Frequency 
91100 
90.5100.5 
16 
101110 
100.5110.5 
28 
111120 
110.5120.5 
44 
121130 
120.5130.5 
20 
131140 
130.5140.5 
32 
141150 
140.5150.5 
12 
151160 
150.5160.5 
4 
Represent classes on xaxis and frequency on yaxis and draw first histogram and from it we can draw a frequency polygon as given below:
12. The electricity bills (in rupees) of 40 houses in a locality are given below:
78 87 81 52 59 65 101 108 115 95
98 65 62 121 128 63 76 84 89 91
65 101 95 81 87 105 129 92 75 105
78 72 107 116 127 100 80 82 61 118
Form a frequency distribution table with a class size of 10. Also represent the above data with a histogram and frequency polygon.
Solution
We know that
Least term = 52
Greatest term = 129
Range = 129 – 52 = 77
Construct a frequency distribution table:
Class Interval 
Tally Numbers 
Frequency 
5060 
II 
2 
6070 

6 
7080 

5 
8090 

8 
90100 

5 
100110 

7 
110120 
III 
3 
120130 
IIII 
4 
Total 
40 
Take class intervals on xaxis and frequency on yaxis
Histogram and frequency polygon are shown below:
13. The data given below represent the marks obtained by 35 students:
21 26 21 20 23 24 22 19 24
26 25 23 26 29 21 24 19 25
26 25 22 23 23 27 26 24 25
30 25 23 28 28 24 28 28
Taking class intervals 1920, 2122 etc., make a frequency distribution for the above data.
Construct a combined histogram and frequency polygon for the distribution.
Solution
It is given that
Least mark = 19
Greatest marks = 30
Range = 30 – 19 = 11
Construct the frequency distribution table
Class Interval 
Actual Intervals 
Frequency 
1920 
18.520.5 
3 
2122 
20.522.5 
5 
2324 
22.524.5 
10 
2526 
24.526.5 
10 
2728 
26.528.5 
5 
2930 
28.530.5 
2 
Total 
35 
Take class interval on xaxis and frequency on the yaxis
Histogram and frequency polygon are show below.
14. The given histogram and frequency polygon shows the ages of teachers in a school. Answer the following:
(i) What is the class size of each class?
(ii) What is the class whose class mark is 48?
(iii) What is the class whose frequency is maximum?
(iv) Construct a frequency table for the given distribution.
Solution
(i) The class size of each class is 6.
(ii) The class whose class mark is 48 is 45 – 51
It can be written as
= (45 + 51)/2
= 96/2
= 48
(iii) The class 5157 has the maximum frequency i.e., 20.
(iv) Frequency table for the given distribution:
Classes 
2733 
3339 
3945 
4551 
5157 
5763 
Frequency 
4 
12 
18 
6 
20 
8 