1. Basics — Building Blocks
What is Statistics?
Statistics is the science of collecting, organising, presenting, analysing and interpreting numerical data. In Class IX, we focus on graphical representation — turning tables of numbers into visual pictures.
Data and Variables
A variable is the characteristic being measured (e.g., marks, weight, age). Data are the values of that variable for each individual. Data can be:
- Ungrouped (raw data) — individual values
- Grouped data — values arranged into class intervals with frequencies
Key Terms
| Term | Meaning | Example |
|---|---|---|
| Frequency | Number of times a value/class appears | 15 students in 40–45 kg |
| Class interval | Range of values in a group | 30.5 – 35.5 |
| Class size/width | Upper limit − lower limit | 35.5 − 30.5 = 5 |
| Class mark | (Upper + Lower) / 2 | (30.5 + 35.5)/2 = 33 |
| Frequency density | Frequency ÷ class width | Used in varying-width histograms |
Remember: "One picture is worth a thousand words." Graphs make it easy to compare data, spot trends, and communicate findings at a glance.
2. Core Concepts
2.1 Bar Graph
Definition
A bar graph is a pictorial representation using rectangular bars of equal width with equal spacing between them. Bars can be horizontal or vertical.
- X-axis: the variable (category)
- Y-axis: the frequency/value
- Gaps between bars — data is discrete/categorical
2.2 Histogram
Definition
A histogram is like a bar graph but for continuous class intervals. There are no gaps between bars because the data is continuous. The width of each bar equals the class size.
Key principle: In a histogram, Area of rectangle ∝ Frequency.
Demo: Weights of 36 students (uniform class size)
Each bar's height = frequency. Bars touch (no gaps). Kink on X-axis shows break from 0.
Uniform vs Varying Class Width
Uniform Width
When all class sizes are equal, bar height = frequency. Simple.
Varying Width — Use Frequency Density!
When class sizes differ, using frequency directly gives a misleading picture. Instead:
Length of bar = (Frequency / Class Width) × Minimum class size
This makes the area proportional to frequency, giving a correct picture.
2.3 Frequency Polygon
Definition
A frequency polygon is obtained by joining the mid-points (class-marks) of the tops of histogram bars with line segments. The polygon is "closed" by adding a class with zero frequency before the first and after the last class.
Class-mark = (Upper limit + Lower limit) / 2
Frequency polygons can be drawn without histograms — just plot (class-mark, frequency) points and join.
Comparison: Bar Graph vs Histogram
| Feature | Bar Graph | Histogram |
|---|---|---|
| Data type | Discrete / Categorical | Continuous (grouped) |
| Gaps between bars | Yes (equal gaps) | No gaps |
| Bar width | All equal (arbitrary) | = class size (significant) |
| Area meaning | Not significant | Proportional to frequency |
| Used for | Comparison of categories | Frequency distribution |
3. Examples — Step by Step
Example 1 (Textbook) — Reading a Bar Graph
40 students were asked their birth months. A bar graph was drawn. (i) How many born in November? (ii) In which month were maximum students born?
Solution
Read the height of the bar for November from the Y-axis: 4 students
Find the tallest bar: The bar for August is tallest (height = 6). Maximum students born in August.
Example 2 (Textbook) — Drawing a Bar Graph
A family's monthly expenditure: Grocery ₹4k, Rent ₹5k, Education ₹5k, Medicine ₹2k, Fuel ₹2k, Entertainment ₹1k, Miscellaneous ₹1k. Draw a bar graph.
Solution (Steps)
Draw X-axis for the Heads (categories), Y-axis for expenditure. Scale: 1 unit = ₹1000.
Draw bars of equal width for each head. Height = expenditure value. Maintain equal gaps between bars.
Label axes and give a title. Bars: Grocery(4), Rent(5), Education(5), Medicine(2), Fuel(2), Entertainment(1), Misc(1).
Observation: Education and Rent are the highest. Entertainment and Misc are lowest.
Example 3 (Textbook) — Histogram with Varying Widths
Marks of 90 students: 0-20: 7, 20-30: 10, 30-40: 10, 40-50: 20, 50-60: 20, 60-70: 15, 70-100: 8. Draw a histogram.
Solution
Note: Class sizes vary! (20, 10, 10, 10, 10, 10, 30). Minimum class size = 10.
Calculate adjusted bar lengths (proportional to class size 10):
0-20: (7/20)×10 = 3.5 | 20-30: (10/10)×10 = 10 | 30-40: 10
40-50: 20 | 50-60: 20 | 60-70: 15 | 70-100: (8/30)×10 = 2.67
0-20: (7/20)×10 = 3.5 | 20-30: (10/10)×10 = 10 | 30-40: 10
40-50: 20 | 50-60: 20 | 60-70: 15 | 70-100: (8/30)×10 = 2.67
Draw rectangles with widths proportional to class size and heights as calculated above. Y-axis label: "Proportion of students per 10 marks interval".
This correctly represents frequency by area — the wide bar (70-100) has low height to compensate for its greater width.
Example 4 (Textbook) — Frequency Polygon from Histogram
51 students' marks (0-100) in 10 class intervals. Draw a frequency polygon.
Solution
Draw a histogram for the data.
Mark mid-points of the top of each rectangle: B(5,5), C(15,10), D(25,4), E(35,6), F(45,7), G(55,3), H(65,2), I(75,2), J(85,3), K(95,9).
Extend the X-axis to −10 (imaginary class before first). Plot A at midpoint (−5, 0). Join A → B.
Also plot L at (105, 0) — midpoint of imaginary class after last. Join K → L.
Connect all points in order: O → A → B → C → ... → K → L. This is the frequency polygon.
Example 5 (Textbook) — Frequency Polygon WITHOUT Histogram
Cost of living index (52 weeks): 140-150: 5, 150-160: 10, 160-170: 20, 170-180: 9, 180-190: 6, 190-200: 2. Draw frequency polygon directly.
Solution
Find class-marks: 145, 155, 165, 175, 185, 195.
Plot points: A(135,0), B(145,5), C(155,10), D(165,20), E(175,9), F(185,6), G(195,2), H(205,0).
Join all points in order with straight line segments: A-B-C-D-E-F-G-H. This is the frequency polygon. Note: A and H are the "closing" points at zero frequency.
4. Textbook Exercises — Solved
Exercise 12.1
Q1 — Women's Illness Data
Causes of illness: Reproductive 31.8%, Neuropsychiatric 25.4%, Injuries 12.4%, Cardiovascular 4.3%, Respiratory 4.1%, Other 22.0%. (i) Represent graphically. (ii) Major cause?
Solution
Draw a bar graph: X-axis = Causes, Y-axis = Female fatality rate (%). Scale: 1 unit = 5%.
Draw bars: Reproductive(31.8), Neuropsychiatric(25.4), Injuries(12.4), Cardiovascular(4.3), Respiratory(4.1), Other(22.0).
(ii) The tallest bar is Reproductive health conditions (31.8%) — the major cause.
(iii) Two factors (discuss): Poor maternal healthcare access, lack of family planning education.
Q2 — Girls per 1000 Boys
Draw a bar graph: SC:940, ST:970, Non SC/ST:920, Backward:950, Non-backward:920, Rural:930, Urban:910.
Solution
X-axis: Section categories. Y-axis: Number of girls per 1000 boys. Scale: Start Y from 900 (break shown), each unit = 10.
Draw bars of equal width for each section.
Conclusions: ST has the highest ratio (970); Urban has the lowest (910). Rural areas generally have better ratios than urban.
Q3 — Election Seats
Party A:75, B:55, C:37, D:29, E:10, F:37. (i) Bar graph. (ii) Max seats.
Solution
X-axis: Parties A-F. Y-axis: Seats won. Scale: 1 unit = 10 seats.
Draw bars with heights 75, 55, 37, 29, 10, 37.
(ii) Party A won the maximum seats (75).
Q4 — Leaf Lengths (Histogram)
Lengths (mm): 118-126:3, 127-135:5, 136-144:9, 145-153:12, 154-162:5, 163-171:4, 172-180:2. Draw histogram.
Solution
Hint: Class intervals are discontinuous (e.g., 118-126 then 127-135 — there's a gap). Make them continuous: 117.5-126.5, 126.5-135.5, 135.5-144.5, 144.5-153.5, 153.5-162.5, 162.5-171.5, 171.5-180.5. Each class width = 9.
Draw histogram with continuous bars. Height = frequency. All widths equal so no adjustment needed.
(ii) Yes — a frequency polygon is also suitable.
(iii) No — the maximum class is 145-153 mm, meaning the most leaves are between 145-153 mm, not all 153 mm.
Q5 — Neon Lamps
Life times: 300-400:14, 400-500:56, 500-600:60, 600-700:86, 700-800:74, 800-900:62, 900-1000:48. Histogram and lamps > 700 hrs.
Solution
Equal class width (100). Draw histogram directly with heights 14, 56, 60, 86, 74, 62, 48.
Lamps with life > 700 hrs = 74 + 62 + 48 = 184 lamps
Q6 — Two Sections Frequency Polygon
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. Compare.
Solution
Class-marks: 5, 15, 25, 35, 45. Plot both polygons on same axes using different colours.
Section A polygon: A(−5,0)→B(5,3)→C(15,9)→D(25,17)→E(35,12)→F(45,9)→G(55,0)
Section B polygon: A(−5,0)→B(5,5)→C(15,19)→D(25,15)→E(35,10)→F(45,1)→G(55,0)
Comparison: Section B has more students in lower marks (10-20). Section A is more uniformly distributed. Section A performs slightly better overall (higher frequencies in upper ranges).
Q7 — Cricket Teams Frequency Polygon
Teams A and B — runs in balls 1-6 to 55-60. Represent on same graph.
Solution
Hint: Intervals 1-6, 7-12... are discontinuous. Make continuous: 0.5-6.5, 6.5-12.5, etc. Class width = 6.
Class-marks: 3.5, 9.5, 15.5, 21.5, 27.5, 33.5, 39.5, 45.5, 51.5, 57.5
Plot both polygons (Team A and Team B) on same graph. Use different line styles. Close the polygon with zero-frequency classes at both ends.
Q8 — Children in Park (Varying Width Histogram)
Age: 1-2:5, 2-3:3, 3-5:6, 5-7:12, 7-10:9, 10-15:10, 15-17:4. Draw histogram.
Solution
Minimum class size = 1 (from 1-2 and 2-3 intervals). Adjust lengths.
1-2(w=1): 5/1×1=5 | 2-3(w=1): 3 | 3-5(w=2): 6/2×1=3 | 5-7(w=2): 12/2=6
7-10(w=3): 9/3=3 | 10-15(w=5): 10/5=2 | 15-17(w=2): 4/2=2
Draw bars with these heights. Y-axis label: "Number of children per 1 year".
Q9 — Surnames Histogram
Letters: 1-4:6, 4-6:30, 6-8:44, 8-12:16, 12-20:4. Draw histogram. Class with max surnames.
Solution
Min class size = 2 (intervals 4-6 and 6-8). Adjust bar lengths.
1-4(w=3): 6/3×2=4 | 4-6(w=2): 30 | 6-8(w=2): 44 | 8-12(w=4): 16/4×2=8 | 12-20(w=8): 4/8×2=1
Draw histogram. (ii) Max surnames lie in class 6-8 (frequency = 44).
5. Tips & Tricks
Bar vs Histogram
Bar graph: gaps between bars, discrete/categorical data. Histogram: no gaps, continuous data.
Varying Width Histogram
Always adjust bar height = (frequency / class width) × min class size. Don't use raw frequency as height!
Class Mark Formula
Class-mark = (Upper limit + Lower limit) / 2. Used for frequency polygons and sometimes for calculations.
Closing the Polygon
Always add one imaginary class with frequency 0 before the first class and after the last class to "close" the frequency polygon.
Discontinuous Intervals
If intervals are like 10-20, 20-30 (sharing boundary), they're already continuous. If like 10-19, 20-29, subtract 0.5 from lower limits and add 0.5 to upper limits.
Reading Histograms
The class with the tallest bar has the highest frequency. For varying-width histograms, compare areas, not heights.
Frequency Polygon Uses
Best for comparing two datasets on the same graph — both polygons can be overlaid clearly.
Scale Selection
Choose scales so the graph fills the grid nicely. Start Y-axis with a kink/break if data doesn't start from 0.
6. Formula & Fact Sheet
Class Mark
(UL + LL) / 2
UL = upper limit, LL = lower limit
Class Size/Width
UL − LL
Must be equal for simple histogram
Adjusted Bar Height
(f / w) × w_min
f = frequency, w = class width, w_min = smallest class width
Frequency Density
f / w
Frequency per unit class width
Continuous Intervals
New LL = LL − 0.5
New UL = UL + 0.5
New UL = UL + 0.5
For discontinuous raw data
Closing Points (Polygon)
Add class before first
Add class after last
Both with f = 0
Add class after last
Both with f = 0
Frequency polygon closes to X-axis
Quick Facts
Area Rule: In a histogram, area of each bar ∝ frequency. This is the fundamental principle.
When class widths are equal: Height of bar ∝ frequency (simple case).
When class widths vary: Use frequency density (f/w) on Y-axis, not raw frequency.
Frequency polygon area = Histogram area (because we add zero-frequency classes at both ends).
When class widths are equal: Height of bar ∝ frequency (simple case).
When class widths vary: Use frequency density (f/w) on Y-axis, not raw frequency.
Frequency polygon area = Histogram area (because we add zero-frequency classes at both ends).
7. 50 Practice Questions
Covering MCQ · 1-mark · 2-mark · 3-mark · 5-mark types. Click Show Answer to reveal.
MCQ (1 mark each)
MCQ
Q1
In a histogram, the area of each rectangle is proportional to:
Show Answer
Frequency. (B)
MCQ
Q2
The class mark of the class interval 40 – 50 is:
Show Answer
(40+50)/2 = 45. (C)
MCQ
Q3
In a bar graph, the bars:
Show Answer
Have equal gaps between them. (B)
MCQ
Q4
A histogram is used for:
Show Answer
Continuous grouped data. (B)
MCQ
Q5
A frequency polygon is drawn by joining the:
Show Answer
Mid-points (class-marks). (B)
MCQ
Q6
For a class interval 20 – 30 in a histogram of varying widths where minimum class size is 5, the bar length for frequency 10 is:
Show Answer
(10/10)×5 = 5. (B)
MCQ
Q7
Frequency polygons are particularly useful for:
Show Answer
Comparing two or more datasets. (B)
MCQ
Q8
In a histogram with continuous class intervals, the bars:
Show Answer
Touch each other (no gaps). (B)
MCQ
Q9
To make the intervals 10-19, 20-29, 30-39 continuous, adjust to:
Show Answer
Subtract 0.5 from lower, add 0.5 to upper: 9.5-19.5, 19.5-29.5, 29.5-39.5. (B)
MCQ
Q10
Which of these is NOT a graphical representation covered in this chapter?
Show Answer
Pie chart is not covered in this chapter. (C)
1-Mark Questions
1 Mark
Q11
Write the formula for class mark.
Show Answer
Class-mark = (Upper limit + Lower limit) / 2
1 Mark
Q12
Find the class mark of 25 – 35.
Show Answer
(25+35)/2 = 30
1 Mark
Q13
What does "kink" on the X-axis of a histogram represent?
Show Answer
A break/gap in the axis, indicating the axis does not start from 0 but from the first class boundary.
1 Mark
Q14
Name the graphical representation where the width of bars plays a significant role.
Show Answer
Histogram
1 Mark
Q15
The class width of the interval 140 – 150 is ______.
Show Answer
10
1 Mark
Q16
In Exercise Q5 (neon lamps), how many lamps have life time more than 700 hours?
Show Answer
74 + 62 + 48 = 184 lamps
1 Mark
Q17
State the main difference between a bar graph and a histogram.
Show Answer
Bar graphs have gaps between bars (discrete/categorical data); histograms have no gaps (continuous data) and area ∝ frequency.
1 Mark
Q18
For a frequency polygon, what is plotted on the X-axis?
Show Answer
Class marks (mid-points of class intervals)
2-Mark Questions
2 Mark
Q19
Find the class marks of: 10-20, 20-30, 30-40, 40-50, 50-60.
Show Answer
15, 25, 35, 45, 55
2 Mark
Q20
A class interval 0-20 has frequency 14 and minimum class size in the dataset is 10. What is the adjusted bar height for the histogram?
Show Answer
Adjusted height = (14/20) × 10 = 7
2 Mark
Q21
Why does a frequency polygon need imaginary classes at the beginning and end?
Show Answer
To "close" the polygon by bringing it back to the X-axis, and to ensure the area of the frequency polygon equals the area of the histogram (by properties of congruent triangles).
2 Mark
Q22
The scores of 30 students are given in the interval 0-10: 5, 10-20: 10, 20-30: 8, 30-40: 7. Find total students and class marks.
Show Answer
Total = 5+10+8+7 = 30. Class marks: 5, 15, 25, 35.
2 Mark
Q23
Convert the following discontinuous intervals to continuous: 5-14, 15-24, 25-34, 35-44.
Show Answer
4.5-14.5, 14.5-24.5, 24.5-34.5, 34.5-44.5
2 Mark
Q24
In a bar graph of election results, Party A got 75 seats. The scale is 1 cm = 10 seats. How tall is Party A's bar?
Show Answer
75 ÷ 10 = 7.5 cm
2 Mark
Q25
What is wrong with drawing a histogram for marks 0-20, 20-30, ..., 70-100 using frequency directly as bar height?
Show Answer
The class widths are unequal (20, 10, ..., 30). Using frequency directly as height makes wider bars look like they have more data than they do. Area would not be proportional to frequency, giving a misleading picture.
2 Mark
Q26
From the cost-of-living index data (Table 12.6), find the class marks for all 6 classes.
Show Answer
140-150: 145; 150-160: 155; 160-170: 165; 170-180: 175; 180-190: 185; 190-200: 195
2 Mark
Q27
For the frequency polygon of cost-of-living data, write the coordinates of the two closing points.
Show Answer
A(135, 0) — before first class; H(205, 0) — after last class.
2 Mark
Q28
The ages of 25 people are: 1-10: 4, 10-20: 7, 20-30: 9, 30-40: 5. Which age group is largest?
Show Answer
Group 20-30 with 9 people is the largest.
3-Mark Questions
3 Mark
Q29
Draw a frequency polygon without histogram for the data: 0-10: 8, 10-20: 12, 20-30: 15, 30-40: 6, 40-50: 4. (Show all coordinates.)
Show Answer
Class marks: 5, 15, 25, 35, 45.
Closing points: A(−5, 0) and F(55, 0).
Points to plot: A(−5,0) → B(5,8) → C(15,12) → D(25,15) → E(35,6) → F(45,4) → G(55,0).
Join all points with straight lines.
Closing points: A(−5, 0) and F(55, 0).
Points to plot: A(−5,0) → B(5,8) → C(15,12) → D(25,15) → E(35,6) → F(45,4) → G(55,0).
Join all points with straight lines.
3 Mark
Q30
The following data gives lifetime of 400 neon lamps. Compute adjusted bar heights for a histogram (minimum class size = 100). Data: 300-400:14, 400-500:56, 500-600:60, 600-700:86, 700-800:74, 800-900:62, 900-1000:48.
Show Answer
All class sizes = 100 = minimum. So adjusted height = frequency itself.
Heights: 14, 56, 60, 86, 74, 62, 48. (No adjustment needed since equal widths.)
Heights: 14, 56, 60, 86, 74, 62, 48. (No adjustment needed since equal widths.)
3 Mark
Q31
For the children-in-park data (Q8), compute the adjusted bar heights and state what the Y-axis should be labelled.
Show Answer
Min class size = 1. Adjusted heights:
1-2(w=1): 5 | 2-3(w=1): 3 | 3-5(w=2): 3 | 5-7(w=2): 6 | 7-10(w=3): 3 | 10-15(w=5): 2 | 15-17(w=2): 2
Y-axis: "Number of children per 1 year interval"
1-2(w=1): 5 | 2-3(w=1): 3 | 3-5(w=2): 3 | 5-7(w=2): 6 | 7-10(w=3): 3 | 10-15(w=5): 2 | 15-17(w=2): 2
Y-axis: "Number of children per 1 year interval"
3 Mark
Q32
Compare bar graphs and histograms under four criteria: data type, bar gaps, Y-axis meaning, and shape of graph.
Show Answer
Data type: Bar graph → discrete/categorical; Histogram → continuous grouped.
Bar gaps: Bar graph → equal gaps; Histogram → no gaps.
Y-axis: Bar graph → value/frequency; Histogram → frequency or frequency density.
Shape: Bar graph → isolated bars; Histogram → solid figure (bars touch).
Bar gaps: Bar graph → equal gaps; Histogram → no gaps.
Y-axis: Bar graph → value/frequency; Histogram → frequency or frequency density.
Shape: Bar graph → isolated bars; Histogram → solid figure (bars touch).
3 Mark
Q33
A survey of 50 families shows monthly income (₹'000): 0-10:2, 10-20:8, 20-30:14, 30-40:18, 40-50:8. Draw a frequency polygon (give coordinates).
Show Answer
Class marks: 5, 15, 25, 35, 45.
Points: A(−5,0) B(5,2) C(15,8) D(25,14) E(35,18) F(45,8) G(55,0).
Join with straight lines. Highest frequency in 30-40 range.
Points: A(−5,0) B(5,2) C(15,8) D(25,14) E(35,18) F(45,8) G(55,0).
Join with straight lines. Highest frequency in 30-40 range.
3 Mark
Q34
For Section A: 0-10:3, 10-20:9, 20-30:17, 30-40:12, 40-50:9 and Section B: 0-10:5, 10-20:19, 20-30:15, 30-40:10, 40-50:1 — compare performance of both sections using the frequency polygon approach.
Show Answer
Section B peaks at 10-20 (high frequency 19), meaning more students scored lower marks.
Section A peaks at 20-30 (frequency 17), suggesting better performance.
Section A has 9 students in 40-50 vs only 1 in Section B → Section A performs better in higher marks.
Conclusion: Section A performs better overall.
Section A peaks at 20-30 (frequency 17), suggesting better performance.
Section A has 9 students in 40-50 vs only 1 in Section B → Section A performs better in higher marks.
Conclusion: Section A performs better overall.
3 Mark
Q35
The marks data 0-20:7, 20-30:10, 30-40:10, 40-50:20, 50-60:20, 60-70:15, 70-100:8 has varying widths. Calculate adjusted heights and explain why it gives a correct picture.
Show Answer
Min class size = 10. Adjusted heights:
0-20: (7/20)×10=3.5 | 20-30:10 | 30-40:10 | 40-50:20 | 50-60:20 | 60-70:15 | 70-100:(8/30)×10=2.67
Why correct: Area = width × adjusted height = class_width × (f/class_width × 10) = f × 10. Since ×10 is constant, area ∝ frequency. ✓
0-20: (7/20)×10=3.5 | 20-30:10 | 30-40:10 | 40-50:20 | 50-60:20 | 60-70:15 | 70-100:(8/30)×10=2.67
Why correct: Area = width × adjusted height = class_width × (f/class_width × 10) = f × 10. Since ×10 is constant, area ∝ frequency. ✓
5-Mark Questions
5 Mark
Q36
The following data shows runs scored by two teams in 60 balls. 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. Represent both on the same graph and compare.
Show Answer
Make continuous: 0.5-6.5, 6.5-12.5, ..., 54.5-60.5. Class marks: 3.5, 9.5, ..., 57.5.
Team A points: (−2.5,0)→(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)→(63.5,0)
Team B points: (−2.5,0)→(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)→(63.5,0)
Comparison: Team B scores more in early overs (1-12) and late overs (49-60). Team A is better in middle overs (13-24, 43-48). Team B ends with a stronger finish.
Team A points: (−2.5,0)→(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)→(63.5,0)
Team B points: (−2.5,0)→(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)→(63.5,0)
Comparison: Team B scores more in early overs (1-12) and late overs (49-60). Team A is better in middle overs (13-24, 43-48). Team B ends with a stronger finish.
5 Mark
Q37
For the surnames data (Q9): 1-4:6, 4-6:30, 6-8:44, 8-12:16, 12-20:4. Calculate adjusted bar heights, draw the histogram description, and identify (i) class with max surnames (ii) class with min surnames.
Show Answer
Min class size = 2 (intervals 4-6 and 6-8).
1-4 (w=3): (6/3)×2 = 4
4-6 (w=2): (30/2)×2 = 30
6-8 (w=2): (44/2)×2 = 44
8-12 (w=4): (16/4)×2 = 8
12-20 (w=8): (4/8)×2 = 1
Y-axis: "Number of surnames per 2-letter interval"
(i) Max surnames: class 6-8 (height 44, frequency 44)
(ii) Min surnames: class 12-20 (frequency 4)
1-4 (w=3): (6/3)×2 = 4
4-6 (w=2): (30/2)×2 = 30
6-8 (w=2): (44/2)×2 = 44
8-12 (w=4): (16/4)×2 = 8
12-20 (w=8): (4/8)×2 = 1
Y-axis: "Number of surnames per 2-letter interval"
(i) Max surnames: class 6-8 (height 44, frequency 44)
(ii) Min surnames: class 12-20 (frequency 4)
5 Mark
Q38
Explain with a full example why a frequency polygon has the same area as the corresponding histogram. Use the concept of congruent triangles.
Show Answer
Proof sketch:
When we draw a frequency polygon on a histogram:
1. At each pair of adjacent rectangles, the polygon cuts off a triangle on the right side of one bar and adds a triangle on the left of the next bar.
2. These two triangles are congruent (same base = half-class-width, same height difference).
3. Therefore, the area cut off = area added, for every transition point.
4. The imaginary classes at the start and end have zero frequency. The polygon forms triangles here too — the area lost at one end equals the area gained by extending to that zero class.
5. Hence, total area of frequency polygon = total area of histogram = total frequency.
Conclusion: Both represent the same total data.
When we draw a frequency polygon on a histogram:
1. At each pair of adjacent rectangles, the polygon cuts off a triangle on the right side of one bar and adds a triangle on the left of the next bar.
2. These two triangles are congruent (same base = half-class-width, same height difference).
3. Therefore, the area cut off = area added, for every transition point.
4. The imaginary classes at the start and end have zero frequency. The polygon forms triangles here too — the area lost at one end equals the area gained by extending to that zero class.
5. Hence, total area of frequency polygon = total area of histogram = total frequency.
Conclusion: Both represent the same total data.
5 Mark
Q39
A student draws a histogram for varying-width data and uses raw frequency as bar height. Show with a specific example how this gives a wrong picture and explain the correct method.
Show Answer
Example: Class 0-20 has frequency 7; class 20-30 has frequency 10.
Wrong method: Bar for 0-20 has height 7, bar for 20-30 has height 10. The bar for 20-30 looks taller, suggesting more data there.
But: Area of 0-20 bar = 20 × 7 = 140; Area of 20-30 bar = 10 × 10 = 100.
The 0-20 interval actually has a larger area, meaning more students — but the graph shows it as shorter! This is misleading.
Correct method: Adjusted heights: 0-20 → (7/20)×10 = 3.5; 20-30 → 10.
Areas: 0-20 → 20×3.5=70; 20-30 → 10×10=100. Wait — 100 ≠ 70. But wait: 70 = 7×10 and 100 = 10×10; both = frequency × 10. So ratio 70:100 = 7:10 = frequency ratio. ✓ Area is now proportional to frequency.
Wrong method: Bar for 0-20 has height 7, bar for 20-30 has height 10. The bar for 20-30 looks taller, suggesting more data there.
But: Area of 0-20 bar = 20 × 7 = 140; Area of 20-30 bar = 10 × 10 = 100.
The 0-20 interval actually has a larger area, meaning more students — but the graph shows it as shorter! This is misleading.
Correct method: Adjusted heights: 0-20 → (7/20)×10 = 3.5; 20-30 → 10.
Areas: 0-20 → 20×3.5=70; 20-30 → 10×10=100. Wait — 100 ≠ 70. But wait: 70 = 7×10 and 100 = 10×10; both = frequency × 10. So ratio 70:100 = 7:10 = frequency ratio. ✓ Area is now proportional to frequency.
5 Mark
Q40
The height of 50 students is given: 130-135:5, 135-140:8, 140-145:12, 145-150:10, 150-155:9, 155-160:6. (a) Draw a frequency polygon. (b) Find the class marks. (c) Which height range has the most students?
Show Answer
(b) Class marks: 132.5, 137.5, 142.5, 147.5, 152.5, 157.5
(a) Points: A(127.5, 0) → B(132.5, 5) → C(137.5, 8) → D(142.5, 12) → E(147.5, 10) → F(152.5, 9) → G(157.5, 6) → H(162.5, 0)
Join all points. The polygon rises to a peak at D(142.5, 12) then falls.
(c) 140-145 cm has the most students (12).
(a) Points: A(127.5, 0) → B(132.5, 5) → C(137.5, 8) → D(142.5, 12) → E(147.5, 10) → F(152.5, 9) → G(157.5, 6) → H(162.5, 0)
Join all points. The polygon rises to a peak at D(142.5, 12) then falls.
(c) 140-145 cm has the most students (12).
Additional Mixed Practice
2 Mark
Q41
In the women's illness data, what percentage of fatalities are due to non-communicable causes (Cardiovascular + Respiratory)?
Show Answer
4.3 + 4.1 = 8.4%
2 Mark
Q42
For the election data (A:75, B:55, C:37, D:29, E:10, F:37), by how many seats did Party A beat Party B?
Show Answer
75 − 55 = 20 seats
3 Mark
Q43
Define frequency density and explain when it must be used instead of frequency on the Y-axis of a histogram.
Show Answer
Frequency density = Frequency ÷ Class width.
It represents frequency per unit class width.
When to use: When class intervals have unequal widths, using raw frequency on the Y-axis makes wider bars appear to contain more data even if they don't. Using frequency density (or adjusted heights) on the Y-axis ensures that the area of each bar is proportional to the frequency, giving an accurate visual representation.
It represents frequency per unit class width.
When to use: When class intervals have unequal widths, using raw frequency on the Y-axis makes wider bars appear to contain more data even if they don't. Using frequency density (or adjusted heights) on the Y-axis ensures that the area of each bar is proportional to the frequency, giving an accurate visual representation.
2 Mark
Q44
Total 400 neon lamps: how many have lifetime 500-700 hours?
Show Answer
60 + 86 = 146 lamps
2 Mark
Q45
The class mark of a class is 25 and class width is 10. Find the class interval.
Show Answer
Class mark = (UL+LL)/2 = 25 and UL−LL = 10 → UL = 30, LL = 20. Class: 20-30
3 Mark
Q46
Why is a frequency polygon better than a histogram for comparing two sets of data?
Show Answer
1. A frequency polygon is a single line, making it easy to draw two polygons on the same graph without overlap or confusion.
2. Two histograms on the same graph would have overlapping bars, making it very hard to read.
3. The shape of the polygon immediately shows differences in distribution — peaks, spread, and skewness.
4. Frequency polygons are particularly useful when data is continuous and very large.
2. Two histograms on the same graph would have overlapping bars, making it very hard to read.
3. The shape of the polygon immediately shows differences in distribution — peaks, spread, and skewness.
4. Frequency polygons are particularly useful when data is continuous and very large.
2 Mark
Q47
The girls-per-1000-boys data shows Urban = 910 and ST = 970. What is the difference, and which is better?
Show Answer
Difference = 970 − 910 = 60. ST has a better ratio (more girls per 1000 boys), which is socially more equitable.
3 Mark
Q48
A data has class intervals 0-10, 10-25, 25-35, 35-60. The minimum class size is 10. Frequencies are 5, 15, 8, 12. Calculate adjusted bar heights.
Show Answer
0-10 (w=10): (5/10)×10 = 5
10-25 (w=15): (15/15)×10 = 10
25-35 (w=10): (8/10)×10 = 8
35-60 (w=25): (12/25)×10 = 4.8
10-25 (w=15): (15/15)×10 = 10
25-35 (w=10): (8/10)×10 = 8
35-60 (w=25): (12/25)×10 = 4.8
5 Mark
Q49
For the leaf-length data (Q4), (i) make intervals continuous, (ii) compute class marks, (iii) write all coordinates for the frequency polygon, (iv) identify the class with maximum leaves.
Show Answer
(i) Continuous intervals: 117.5-126.5, 126.5-135.5, 135.5-144.5, 144.5-153.5, 153.5-162.5, 162.5-171.5, 171.5-180.5
(ii) Class marks: 122, 131, 140, 149, 158, 167, 176
(iii) Add zero classes: before (108.5-117.5, cm=113) and after (180.5-189.5, cm=185).
Points: A(113,0) → B(122,3) → C(131,5) → D(140,9) → E(149,12) → F(158,5) → G(167,4) → H(176,2) → I(185,0)
(iv) Class 145-153 mm (continuous: 144.5-153.5) has maximum leaves (12).
(ii) Class marks: 122, 131, 140, 149, 158, 167, 176
(iii) Add zero classes: before (108.5-117.5, cm=113) and after (180.5-189.5, cm=185).
Points: A(113,0) → B(122,3) → C(131,5) → D(140,9) → E(149,12) → F(158,5) → G(167,4) → H(176,2) → I(185,0)
(iv) Class 145-153 mm (continuous: 144.5-153.5) has maximum leaves (12).
5 Mark
Q50
Write a complete comparison of bar graphs, histograms, and frequency polygons in terms of: (a) type of data, (b) how they are drawn, (c) what they show, (d) when to use them, (e) their relationship to each other.
Show Answer
(a) Type of data:
Bar graph → categorical/discrete. Histogram → continuous grouped. Frequency polygon → continuous grouped.
(b) How drawn:
Bar graph: bars with equal gaps. Histogram: touching bars, area ∝ frequency. Frequency polygon: line joining class-mark tops.
(c) What they show:
Bar graph: comparison between categories. Histogram: frequency distribution shape. Frequency polygon: distribution curve shape.
(d) When to use:
Bar graph: comparing different groups. Histogram: showing distribution of continuous data. Frequency polygon: comparing two distributions on same graph.
(e) Relationship:
A frequency polygon is derived from a histogram by joining mid-points of bar tops. It can also be drawn independently using class marks. Both histogram and frequency polygon cover the same area, representing the same total frequency.
Bar graph → categorical/discrete. Histogram → continuous grouped. Frequency polygon → continuous grouped.
(b) How drawn:
Bar graph: bars with equal gaps. Histogram: touching bars, area ∝ frequency. Frequency polygon: line joining class-mark tops.
(c) What they show:
Bar graph: comparison between categories. Histogram: frequency distribution shape. Frequency polygon: distribution curve shape.
(d) When to use:
Bar graph: comparing different groups. Histogram: showing distribution of continuous data. Frequency polygon: comparing two distributions on same graph.
(e) Relationship:
A frequency polygon is derived from a histogram by joining mid-points of bar tops. It can also be drawn independently using class marks. Both histogram and frequency polygon cover the same area, representing the same total frequency.