Class 9 Mathematics
Chapter 12: Statistics
Exercise 12.1 – Stepwise Solutions
Important Concepts Used
1. A bar graph is used to compare different categories of data.
2. In a histogram, class intervals must be continuous.
3. The height of each bar in a histogram represents the frequency.
4. The category with the greatest value is identified by the tallest bar.
Q1. A survey conducted by an organisation for the cause of illness and death among women between the ages 15–44 years worldwide found the following figures (in
Given data:
Reproductive health conditions = 31.8
Neuropsychiatric conditions = 25.4
Injuries = 12.4
Cardiovascular conditions = 4.3
Respiratory conditions = 4.1
Other causes = 22.0
(i) Represent the information graphically.
To represent this data graphically, draw a bar graph:
Horizontal axis (x-axis): Causes
Vertical axis (y-axis): Female fatality rate (
Take a suitable scale, for example:
1 unit = 5
Then draw bars of heights:
Reproductive health conditions → 31.8
Neuropsychiatric conditions → 25.4
Injuries → 12.4
Cardiovascular conditions → 4.3
Respiratory conditions → 4.1
Other causes → 22.0
(ii) Which condition is the major cause of women’s ill health and death worldwide?
The highest percentage is 31.8
So, the major cause is Reproductive health conditions.
(iii) Write any two factors which may play a major role in the cause in (ii) above being the major cause.
Two possible factors are:
1. Lack of proper medical facilities for women.
2. Poor nutrition and lack of awareness about reproductive health.
Q2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.
Given data:
Scheduled Caste (SC) = 940
Scheduled Tribe (ST) = 970
Non SC/ST = 920
Backward districts = 950
Non-backward districts = 920
Rural = 930
Urban = 910
(i) Represent the information above by a bar graph.
To draw the bar graph:
x-axis: Sections of Indian society
y-axis: Number of girls per thousand boys
Take a suitable scale, for example:
1 unit = 10 girls
Draw bars of heights:
SC → 940
ST → 970
Non SC/ST → 920
Backward districts → 950
Non-backward districts → 920
Rural → 930
Urban → 910
(ii) What conclusions can be arrived at from the graph?
From the graph, we conclude:
1. Scheduled Tribe (ST) has the highest number of girls per thousand boys, i.e. 970.
2. Urban section has the lowest number of girls per thousand boys, i.e. 910.
3. Backward districts have more girls per thousand boys than non-backward districts.
4. Rural areas have more girls per thousand boys than urban areas.
Q3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections.
Given data:
Party A = 75
Party B = 55
Party C = 37
Party D = 29
Party E = 10
Party F = 37
(i) Draw a bar graph to represent the polling results.
To draw the bar graph:
x-axis: Political Parties
y-axis: Number of seats won
Take a suitable scale, for example:
1 unit = 10 seats
Draw bars of heights:
A → 75
B → 55
C → 37
D → 29
E → 10
F → 37
(ii) Which political party won the maximum number of seats?
The maximum number of seats is 75.
So, Party A won the maximum number of seats.
Q4. The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table.
Given table:
118–126 → 3
127–135 → 5
136–144 → 9
145–153 → 12
154–162 → 5
163–171 → 4
172–180 → 2
(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
Since the lengths are measured to the nearest millimetre, we first convert the class intervals into continuous intervals:
117.5 – 126.5 → 3
126.5 – 135.5 → 5
135.5 – 144.5 → 9
144.5 – 153.5 → 12
153.5 – 162.5 → 5
162.5 – 171.5 → 4
171.5 – 180.5 → 2
Now draw the histogram:
x-axis: Length of leaves (in mm)
y-axis: Number of leaves
Draw adjacent rectangles with heights:
3, 5, 9, 12, 5, 4, 2
(ii) Is there any other suitable graphical representation for the same data?
Yes, another suitable graphical representation is a frequency polygon.
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
No, it is not correct.
The class interval 145–153 has the highest frequency, which means the maximum number of leaves have lengths lying between 145 mm and 153 mm.
It does not mean that all of them are exactly 153 mm long.
Class 9 Mathematics
Chapter 12: Statistics
Exercise 12.1 (Q5, Q6, Q7) – Stepwise Solutions
Q5. Life time of 400 neon lamps
Given data:
300–400 → 14
400–500 → 56
500–600 → 60
600–700 → 86
700–800 → 74
800–900 → 62
900–1000 → 48
(i) Represent the data using histogram
Steps:
1. All class intervals are equal (width = 100), so histogram can be drawn directly.
2. x-axis: Life time (in hours)
3. y-axis: Number of lamps
4. Draw rectangles with heights:
300–400 → 14
400–500 → 56
500–600 → 60
600–700 → 86
700–800 → 74
800–900 → 62
900–1000 → 48
(ii) How many lamps have life more than 700 hours?
Lamps with life > 700 hours:
700–800 → 74
800–900 → 62
900–1000 → 48
Total = 74 + 62 + 48
= 184
Answer: 184 lamps
Q6. Distribution of students (Section A and B)
Section A:
0–10 → 3
10–20 → 9
20–30 → 17
30–40 → 12
40–50 → 9
Section B:
0–10 → 5
10–20 → 19
20–30 → 15
30–40 → 10
40–50 → 1
Step 1: Find class marks
0–10 → 5
10–20 → 15
20–30 → 25
30–40 → 35
40–50 → 45
(i) Draw frequency polygons
Plot points for Section A:
(5,3), (15,9), (25,17), (35,12), (45,9)
Plot points for Section B:
(5,5), (15,19), (25,15), (35,10), (45,1)
Join the points with straight lines to form two frequency polygons on the same graph.
(ii) Compare performance
1. Section A has more students in higher marks range (20–40).
2. Section B has more students in lower marks range (10–20).
Conclusion: Section A performed better overall.
Q7. Runs scored by Team A and Team B
Given data:
Team A:
1–6 → 2
7–12 → 1
13–18 → 8
19–24 → 9
25–30 → 4
31–36 → 5
37–42 → 6
43–48 → 10
49–54 → 6
55–60 → 2
Team B:
1–6 → 5
7–12 → 6
13–18 → 2
19–24 → 10
25–30 → 5
31–36 → 6
37–42 → 3
43–48 → 4
49–54 → 8
55–60 → 10
Step 1: Make class intervals continuous
0.5–6.5, 6.5–12.5, 12.5–18.5, 18.5–24.5, 24.5–30.5, 30.5–36.5, 36.5–42.5, 42.5–48.5, 48.5–54.5, 54.5–60.5
Step 2: Find class marks
Class marks:
3.5, 9.5, 15.5, 21.5, 27.5, 33.5, 39.5, 45.5, 51.5, 57.5
Step 3: Plot frequency polygons
Team A points:
(3.5,2), (9.5,1), (15.5,8), (21.5,9), (27.5,4), (33.5,5), (39.5,6), (45.5,10), (51.5,6), (57.5,2)
Team B points:
(3.5,5), (9.5,6), (15.5,2), (21.5,10), (27.5,5), (33.5,6), (39.5,3), (45.5,4), (51.5,8), (57.5,10)
Join the respective points to form two frequency polygons.
Conclusion:
1. Team A shows steady performance in middle intervals.
2. Team B shows higher variation with strong performance in the final overs.
Class 9 Mathematics
Chapter 12: Statistics
Solutions (Question 8 and Question 9)
Q8. A random survey of the number of children of various age groups playing in a park was found as follows. Draw a histogram to represent the data above.
Given data:
1–2 → 5
2–3 → 3
3–5 → 6
5–7 → 12
7–10 → 9
10–15 → 10
15–17 → 4
Step 1: Check class intervals
The class intervals are unequal.
Widths are:
1–2 → 1
2–3 → 1
3–5 → 2
5–7 → 2
7–10 → 3
10–15 → 5
15–17 → 2
Step 2: Find frequency density / adjusted height
Since class intervals are unequal, histogram is drawn using:
Adjusted frequency = Frequency / Class width
So,
1–2 → 5/1 = 5
2–3 → 3/1 = 3
3–5 → 6/2 = 3
5–7 → 12/2 = 6
7–10 → 9/3 = 3
10–15 → 10/5 = 2
15–17 → 4/2 = 2
Step 3: Draw histogram
x-axis: Age (in years)
y-axis: Adjusted frequency
Draw bars with heights:
1–2 → 5
2–3 → 3
3–5 → 3
5–7 → 6
7–10 → 3
10–15 → 2
15–17 → 2
Answer: The histogram should be drawn using the above adjusted frequencies.
Q9. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the surnames was found as follows:
Given data:
1–4 → 6
4–6 → 30
6–8 → 44
8–12 → 16
12–20 → 4
(i) Draw a histogram to depict the given information.
Step 1: Check class intervals
The class intervals are unequal.
Widths are:
1–4 → 3
4–6 → 2
6–8 → 2
8–12 → 4
12–20 → 8
Step 2: Find adjusted frequencies
Adjusted frequency = Frequency / Class width
1–4 → 6/3 = 2
4–6 → 30/2 = 15
6–8 → 44/2 = 22
8–12 → 16/4 = 4
12–20 → 4/8 = 0.5
Step 3: Draw histogram
x-axis: Number of letters
y-axis: Adjusted frequency
Draw bars with heights:
1–4 → 2
4–6 → 15
6–8 → 22
8–12 → 4
12–20 → 0.5
(ii) Write the class interval in which the maximum number of surnames lie.
From the original frequencies:
1–4 → 6
4–6 → 30
6–8 → 44
8–12 → 16
12–20 → 4
The maximum frequency is 44.
So, the class interval in which the maximum number of surnames lie is:
6–8