📊 Data Visualization Question Paper
Bar Graphs, Histograms & Frequency Polygons
Total Marks: 80 | Time: 2 Hours
📋 INSTRUCTIONS
- All questions are compulsory
- Section A: 1-5 (1 mark each) = 5 marks
- Section B: 6-13 (2 marks each) = 16 marks
- Section C: 14-18 (4 marks each) = 20 marks
- Section D: 19-20 (19.5 marks each) = 39 marks
- Use of ruler and pencil for graphs is recommended
📈 Marks Distribution
- Very Short Answer (1 mark): 5 questions
- Short Answer (2 marks): 8 questions
- Medium Answer (4 marks): 5 questions
- Long Answer (19.5 marks each): 2 questions (with graphs/diagrams)
What is the main difference between a bar graph and a histogram?
Bar Graph: Has gaps between bars (represents separate/distinct categories)
Histogram: Has NO gaps between bars (represents continuous data)
Define a frequency polygon.
A frequency polygon is a line graph obtained by connecting the midpoints of the upper sides of the adjacent rectangles in a histogram with straight line segments.
What is a class-mark? Write its formula.
The class-mark is the midpoint of a class interval.
In a histogram with varying class widths, what determines the height of a rectangle?
The height is determined by frequency density, not just frequency.
This ensures the area of the rectangle is proportional to frequency.
Name one situation where frequency polygons are most useful.
Comparing two different sets of continuous data of the same nature.
Example: Comparing test scores of Section A vs Section B, or sales of two months.
A student drew a histogram for data with class intervals 0-10, 10-20, 20-30, but forgot to close the polygon at both ends. What steps should be taken to correct this?
Steps to correct:
- Find the class preceding the first class: (-10) - 0, with class-mark = -5
- Find the class succeeding the last class: 30 - 40, with class-mark = 35
- Plot points at (-5, 0) and (35, 0) on the graph
- Connect these points to close the frequency polygon
A frequency distribution has class intervals: 10-20 (frequency 5), 20-40 (frequency 10), 40-50 (frequency 8). Identify if this is suitable for a standard histogram and explain why or why not.
NOT suitable for a standard histogram without adjustment.
Reason: Class widths are varying (10, 20, 10). In a standard histogram, we assume uniform width.
What to do: Use frequency density method to adjust heights before drawing.
Calculate class-marks for the following class intervals: 50-60, 60-70, 70-80, 80-90.
| Class Interval | Class-Mark Calculation | Class-Mark |
|---|---|---|
| 50-60 | (50+60)÷2 | 55 |
| 60-70 | (60+70)÷2 | 65 |
| 70-80 | (70+80)÷2 | 75 |
| 80-90 | (80+90)÷2 | 85 |
Why is the width of bars important in a histogram but not in a bar graph?
In Bar Graph: Width doesn't matter because bars represent separate, distinct categories with no relationship between their widths and values.
In Histogram: Width matters because the AREA of the rectangle (not height) is proportional to frequency. Area = Height × Width, so width directly affects the representation.
Given frequency distribution for marks: 0-20 (7 students), 20-40 (15 students), 40-50 (20 students), 50-100 (8 students). Find the corrected heights for a histogram using minimum class width (10) as standard.
Step 1: Identify minimum class width = 10
| Marks | Frequency | Width | Corrected Height |
|---|---|---|---|
| 0-20 | 7 | 20 | 7÷20×10 = 3.5 |
| 20-40 | 15 | 20 | 15÷20×10 = 7.5 |
| 40-50 | 20 | 10 | 20÷10×10 = 20 |
| 50-100 | 8 | 50 | 8÷50×10 = 1.6 |
Differentiate between frequency distribution with uniform width and varying width class intervals.
Uniform Width: All class intervals have same width (e.g., 10-20, 20-30, 30-40). Height directly proportional to frequency.
Varying Width: Class intervals have different widths (e.g., 0-20, 20-30, 30-60). Need frequency density method. Height = Frequency ÷ Width.
What are the two methods to draw a frequency polygon? Name them briefly.
Method 1: Using Histogram - Draw histogram first, then join midpoints of upper rectangle sides with line segments.
Method 2: Direct Method (Using Class-Marks) - Calculate class-marks, plot points (class-mark, frequency), and join them with lines.
Can a histogram be drawn with continuous data that has non-continuous class intervals? Justify your answer.
Yes, but with correction. Histograms are for continuous data, so if class intervals are given as 10-20, 21-30, 31-40 (non-continuous), they must be converted to continuous form: 9.5-20.5, 20.5-30.5, 30.5-40.5.
A teacher collected data on the heights of 50 students in cm:
| Height (cm) | Number of Students |
|---|---|
| 140-145 | 5 |
| 145-150 | 10 |
| 150-155 | 18 |
| 155-160 | 12 |
| 160-165 | 5 |
(a) Is this data suitable for a bar graph or histogram? Why? (2 marks)
(b) Calculate class-marks for each interval. (2 marks)
(a) Suitable for Histogram (2 marks)
Why: Height is continuous data grouped into class intervals (ranges). Histogram is designed for such continuous, grouped data.
(b) Class-Marks (2 marks)
| Height Interval | Class-Mark |
|---|---|
| 140-145 | (140+145)÷2 = 142.5 |
| 145-150 | (145+150)÷2 = 147.5 |
| 150-155 | (150+155)÷2 = 152.5 |
| 155-160 | (155+160)÷2 = 157.5 |
| 160-165 | (160+165)÷2 = 162.5 |
Explain why a histogram with varying class widths needs correction. Give an example with calculations showing the need for correction.
Why Correction Needed: In a histogram, the AREA of rectangle must be proportional to frequency. If we only use frequency as height (ignoring width), larger class intervals will appear to have more frequency than they actually do, creating a misleading picture.
Example: Age distribution of 100 people
| Age Range | Frequency | Width | Height (if uncorrected) | Corrected Height (÷10) |
|---|---|---|---|---|
| 0-20 | 10 | 20 | 10 (misleading!) | 10÷20×10 = 5 |
| 20-30 | 30 | 10 | 30 | 30÷10×10 = 30 |
| 30-60 | 40 | 30 | 40 (misleading!) | 40÷30×10 = 13.33 |
Without correction, 0-20 (10 frequency) appears taller than 30-60 (40 frequency), which is wrong!
Draw a frequency polygon for the following data and list the points you would plot (assuming you're using the direct method with class-marks).
| Marks | 0-10 | 10-20 | 20-30 | 30-40 |
|---|---|---|---|---|
| Frequency | 4 | 6 | 8 | 2 |
Step 1: Calculate Class-Marks
| Marks | Class-Mark | Frequency |
|---|---|---|
| 0-10 | 5 | 4 |
| 10-20 | 15 | 6 |
| 20-30 | 25 | 8 |
| 30-40 | 35 | 2 |
Step 2: Points to Plot (including closing points)
- A(-5, 0) - Zero frequency before first class
- B(5, 4) - First interval
- C(15, 6) - Second interval
- D(25, 8) - Third interval (peak)
- E(35, 2) - Fourth interval
- F(45, 0) - Zero frequency after last class
Step 3: Plot these 6 points and join them with straight lines. The frequency polygon will look like a mountain with peak at (25, 8).
A student claims that frequency polygons cannot be drawn for data with non-uniform class widths. Is this claim correct? Justify with an example.
The claim is INCORRECT. Frequency polygons CAN be drawn for non-uniform class widths.
Key Point: Frequency polygon uses CLASS-MARKS (midpoints) and frequencies, not the class width. The width doesn't affect plotting.
Example:
| Marks Range | Width | Class-Mark | Frequency | Point to Plot |
|---|---|---|---|---|
| 0-20 | 20 | 10 | 5 | (10, 5) |
| 20-30 | 10 | 25 | 10 | (25, 10) |
| 30-50 | 20 | 40 | 8 | (40, 8) |
We can plot and join these points (10,5) → (25,10) → (40,8) even though widths differ!
Compare and contrast the uses of bar graphs, histograms, and frequency polygons. Create a comparison table with at least 4 features.
| Feature | Bar Graph | Histogram | Frequency Polygon |
|---|---|---|---|
| Data Type | Discrete/Categorical | Continuous (grouped) | Continuous (grouped) |
| Gaps Between Bars | YES (gaps present) | NO (no gaps) | N/A (line graph) |
| Width Importance | No (all same) | YES (affects area) | No (based on midpoints) |
| Best Use | Comparing items | Showing distribution | Comparing 2+ datasets |
| Examples | Favorite fruits, brands | Heights, weights, marks | Section A vs B scores |
The weights (in kg) of 40 students in a class are given below:
| Weight (kg) | Number of Students |
|---|---|
| 35-40 | 3 |
| 40-45 | 8 |
| 45-50 | 16 |
| 50-55 | 10 |
| 55-60 | 3 |
Requirements:
(a) Justify whether you would use a bar graph or histogram for this data. (2 marks)
(b) Draw a well-labeled histogram for this data. (5 marks)
(c) From the histogram, identify the class with maximum frequency. (2 marks)
(d) Using the class-marks, draw a frequency polygon on the same graph. (5 marks)
(e) Calculate and list all points (including closing points) needed to complete the frequency polygon. (5.5 marks)
(a) Justification for Histogram (2 marks):
This data should be represented as a HISTOGRAM because:
- Weight is continuous data
- Data is grouped into class intervals (ranges)
- Histogram is specifically designed for continuous, grouped data
- We want to show the distribution of weights across ranges
(b) Histogram (5 marks):
🎨 HISTOGRAM: Weight Distribution of 40 Students
(c) Class with Maximum Frequency (2 marks):
Looking at the histogram, the tallest bar is in the interval 45-50 kg with a frequency of 16 students.
This represents the weight range where most students fall, making it the modal class.
(d) Frequency Polygon (5 marks):
The frequency polygon is drawn by:
- Finding midpoints (class-marks) of each rectangle top
- Joining these midpoints with straight line segments
- Extending to zero frequency at both ends
The polygon is shown as the red line in the diagram above, overlapping the histogram.
(e) All Points for Frequency Polygon (5.5 marks):
| Class Interval | Class-Mark | Frequency | Point (x, y) |
|---|---|---|---|
| (preceding) 30-35 | 32.5 | 0 | (32.5, 0) ← Closing point |
| 35-40 | 37.5 | 3 | (37.5, 3) |
| 40-45 | 42.5 | 8 | (42.5, 8) |
| 45-50 | 47.5 | 16 | (47.5, 16) ← Peak |
| 50-55 | 52.5 | 10 | (52.5, 10) |
| 55-60 | 57.5 | 3 | (57.5, 3) |
| (succeeding) 60-65 | 62.5 | 0 | (62.5, 0) ← Closing point |
The histogram clearly shows data distribution with peak at 45-50. The frequency polygon smoothly represents the same data and makes it easy to identify the trend. Total = 19.5 marks
A mathematics test was conducted for two sections (A and B) with 30 students each. The frequency distributions are:
| Marks Range | Section A (Frequency) | Section B (Frequency) |
|---|---|---|
| 0-10 | 2 | 4 |
| 10-20 | 5 | 8 |
| 20-30 | 8 | 10 |
| 30-40 | 10 | 6 |
| 40-50 | 5 | 2 |
Requirements:
(a) Calculate class-marks for each interval and create a table with class-marks and frequencies. (4 marks)
(b) Draw frequency polygons for BOTH sections on the SAME graph with proper labeling and legend. (8 marks)
(c) Using the overlapping frequency polygons, compare the performance of both sections. (4 marks)
(d) Which section performed better overall? Justify your answer using observations from the graph. (3.5 marks)
(a) Class-Marks Table (4 marks):
| Marks Interval | Class-Mark | Section A (Freq) | Section B (Freq) |
|---|---|---|---|
| 0-10 | (0+10)÷2 = 5 | 2 | 4 |
| 10-20 | (10+20)÷2 = 15 | 5 | 8 |
| 20-30 | (20+30)÷2 = 25 | 8 | 10 |
| 30-40 | (30+40)÷2 = 35 | 10 | 6 |
| 40-50 | (40+50)÷2 = 45 | 5 | 2 |
(b) Frequency Polygons for Both Sections (8 marks):
📊 COMPARISON: Section A vs Section B Performance
(c) Comparison of Sections A and B (4 marks):
Key Observations:
- Section A Peak: Occurs at marks 30-40 (frequency 10) - most students score in this range
- Section B Peak: Occurs at marks 20-30 (frequency 10) - peak is at lower marks compared to Section A
- Section A Distribution: Skewed towards higher marks, with more students in 30-40 and 40-50 ranges
- Section B Distribution: Skewed towards lower marks, with more concentration in 10-30 range
- Overlap: Both sections have similar frequencies in the 20-30 mark range
(d) Which Section Performed Better? (3.5 marks):
Answer: SECTION A performed better overall.
Justification:
- Section A's polygon is shifted to the RIGHT compared to Section B, indicating higher marks
- Higher marks: Section A has 15 students (10+5) scoring 30+ marks vs. Section B's 8 students (6+2)
- Lower marks: Section B has 12 students (4+8) scoring below 20, vs. Section A's 7 students (2+5)
- Peak position: Section A's peak at 30-40 is higher than Section B's peak at 20-30
- Overall trend: The rightward shift of Section A's polygon indicates generally better performance
When comparing two frequency polygons on the same graph, the polygon that is positioned towards the right (higher values) indicates better performance for positive metrics like test scores. This visual comparison is the PRIMARY advantage of frequency polygons! Total = 19.5 marks
✅ SUMMARY FOR STUDENTS
- Total Questions: 20 questions (covering all three topics)
- Total Marks: 80 marks (5+16+20+39)
- Key Concepts Tested: Definitions, applications, calculations, graph drawing, comparisons, and analysis
- Preparation Tips: Practice drawing graphs with accurate scales, memorize formulas, understand when to use each graph type
- Scoring Strategy: Attempt Q1-5 first (easy 5 marks), then Q6-13 (moderate difficulty), then Q14-18 (application), finally Q19-20 (application with graphs)
