Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
How to Use This Page
Work through each exercise in the Samacheer Kalvi textbook on your own first, then use the solutions below to verify your steps and answers. All questions from Chapter 8 (Statistics) are solved with full working — arrange data, apply the formula, and confirm your result.
Key Concepts (Quick Recap)
Measures of Central Tendency
- Arithmetic Mean: Sum of all observations / Number of observations
- For ungrouped data: Mean = (x₁ + x₂ + ... + xₙ) / n
- For grouped data (direct method): Mean = Σ(fᵢ × xᵢ) / Σfᵢ, where xᵢ = class mark
- Class mark (midpoint) = (upper limit + lower limit) / 2
- Median: The middle value when data is arranged in order
- For n odd: Median = ((n+1)/2)th observation
- For n even: Median = average of (n/2)th and (n/2+1)th observations
- Mode: The value that occurs most frequently
- Empirical relationship: Mode ≈ 3 Median − 2 Mean
Data Presentation
- Bar graph: Represents data using rectangular bars of equal width with heights proportional to values
- Histogram: Like a bar graph but for continuous grouped data with no gaps between bars
- Frequency polygon: Line graph connecting midpoints of class intervals
- Cumulative frequency: Running total of frequencies, used for ogive curves
Exercise 8.1 — Collection and Organisation of Data
Q1. Define the following terms:
(i) Primary data (ii) Secondary data
(ii) Secondary data: Data that has been collected by someone else and is obtained from published or unpublished sources. Examples: data from newspapers, government reports, websites.
Q2. Classify the following as primary or secondary data:
(i) Data collected by a researcher through a questionnaire (ii) Data obtained from a government census report (iii) Data from the RBI website on inflation rates (iv) Data collected by interviewing people in a village
Q3. What are the different methods of collecting primary data?
Q4. What is the difference between discrete and continuous data?
- Discrete data: Data that can take only specific values (usually integers). Examples: number of students in a class (cannot be 30.5), number of books on a shelf, number of goals in a match.
- Continuous data: Data that can take any value within a range (including decimals). Examples: height of students (can be 150.5 cm), weight, temperature, time.
Q5. The marks scored by 20 students in a test are:
12, 15, 18, 12, 20, 15, 18, 12, 15, 20, 18, 12, 15, 18, 20, 12, 15, 18, 15, 12
Organise the data in a frequency distribution table.
| Marks | Tally | Frequency | |||||||
|---|---|---|---|---|---|---|---|---|---|
| 12 | 5 | ||||||||
| 15 | 6 | ||||||||
| 18 | 7 | ||||||||
| 20 | 2 | ||||||||
| Total | 20 |
Verification: 5 + 6 + 7 + 2 = 20 ✓
Exercise 8.2 — Arithmetic Mean
Q1. Find the arithmetic mean of the following data:
5, 8, 12, 15, 20
Q2. Find the mean of the first 10 natural numbers.
Q3. Find the mean of the first five even numbers.
Q4. The marks of 8 students are 25, 30, 22, 35, 28, 32, 27, 21. Find the mean marks.
Q5. If the mean of 6 observations is 15, find the sum of the observations.
Q6. The mean of 5 numbers is 18. If one number is excluded, the mean of the remaining 4 numbers becomes 16. Find the excluded number.
Q7. The mean of 10 numbers is 20. If each number is increased by 5, what is the new mean?
Alternatively: If every observation increases by 5, the mean also increases by 5. New mean = 20 + 5 = 25
Q8. The mean of 15 numbers is 40. If each number is multiplied by 3, what is the new mean?
Q9. Find the mean of the following frequency distribution:
| x | 5 | 10 | 15 | 20 | 25 |
|---|---|---|---|---|---|
| f | 3 | 5 | 6 | 4 | 2 |
| x | f | f × x |
|---|---|---|
| 5 | 3 | 15 |
| 10 | 5 | 50 |
| 15 | 6 | 90 |
| 20 | 4 | 80 |
| 25 | 2 | 50 |
| Total | 20 | 285 |
Mean = Σ(f × x) / Σf = 285 / 20 = 14.25
Q10. The mean weight of 8 students is 45 kg. Two more students of weights 40 kg and 42 kg join the group. Find the new mean weight.
Exercise 8.3 — Mean for Grouped Data
Q1. Find the mean of the following grouped data using the direct method:
| Class Interval | Frequency (f) |
|---|---|
| 0 – 10 | 4 |
| 10 – 20 | 6 |
| 20 – 30 | 8 |
| 30 – 40 | 5 |
| 40 – 50 | 2 |
| Class Interval | f | Class Mark (x) | f × x |
|---|---|---|---|
| 0 – 10 | 4 | 5 | 20 |
| 10 – 20 | 6 | 15 | 90 |
| 20 – 30 | 8 | 25 | 200 |
| 30 – 40 | 5 | 35 | 175 |
| 40 – 50 | 2 | 45 | 90 |
| Total | 25 | 575 |
Class mark = (upper limit + lower limit) / 2
Mean = Σ(f × x) / Σf = 575 / 25 = 23
Q2. Find the mean of the following distribution:
| Class Interval | Frequency |
|---|---|
| 10 – 20 | 5 |
| 20 – 30 | 8 |
| 30 – 40 | 12 |
| 40 – 50 | 10 |
| 50 – 60 | 5 |
| Class Interval | f | Class Mark (x) | f × x |
|---|---|---|---|
| 10 – 20 | 5 | 15 | 75 |
| 20 – 30 | 8 | 25 | 200 |
| 30 – 40 | 12 | 35 | 420 |
| 40 – 50 | 10 | 45 | 450 |
| 50 – 60 | 5 | 55 | 275 |
| Total | 40 | 1420 |
Mean = Σ(f × x) / Σf = 1420 / 40 = 35.5
Q3. The following data gives the daily income of 50 workers in a factory. Find the mean daily income.
| Daily Income (Rs) | Number of Workers |
|---|---|
| 100 – 120 | 12 |
| 120 – 140 | 14 |
| 140 – 160 | 8 |
| 160 – 180 | 6 |
| 180 – 200 | 10 |
| Daily Income | f | Class Mark (x) | f × x |
|---|---|---|---|
| 100 – 120 | 12 | 110 | 1320 |
| 120 – 140 | 14 | 130 | 1820 |
| 140 – 160 | 8 | 150 | 1200 |
| 160 – 180 | 6 | 170 | 1020 |
| 180 – 200 | 10 | 190 | 1900 |
| Total | 50 | 7260 |
Mean = Σ(f × x) / Σf = 7260 / 50 = 145.2
Q4. Find the mean of the following frequency distribution:
| Marks | Number of Students |
|---|---|
| 0 – 4 | 4 |
| 4 – 8 | 6 |
| 8 – 12 | 8 |
| 12 – 16 | 5 |
| 16 – 20 | 2 |
| Marks | f | Class Mark (x) | f × x |
|---|---|---|---|
| 0 – 4 | 4 | 2 | 8 |
| 4 – 8 | 6 | 6 | 36 |
| 8 – 12 | 8 | 10 | 80 |
| 12 – 16 | 5 | 14 | 70 |
| 16 – 20 | 2 | 18 | 36 |
| Total | 25 | 230 |
Mean = 230 / 25 = 9.2
Exercise 8.4 — Median
Q1. Find the median of the following data:
3, 7, 5, 9, 12, 8, 6
Q2. Find the median of the following data:
15, 12, 18, 10, 20, 16, 8, 14
Q3. Find the median of the following data:
25, 18, 20, 22, 16, 24, 30, 28, 14
Q4. The marks obtained by 11 students in a test are:
35, 28, 42, 30, 25, 38, 45, 32, 40, 27, 33
Find the median marks.
Q5. Find the median of the following frequency distribution:
| x | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| f | 4 | 6 | 8 | 5 | 2 |
Expanded data (in ascending order): 10, 10, 10, 10, 20, 20, 20, 20, 20, 20, 30, 30, 30, 30, 30, 30, 30, 30, 40, 40, 40, 40, 40, 50, 50
n = 25 (odd) Median = ((25+1)/2)th = 13th observation
Counting: 4 values of 10 (positions 1-4), 6 values of 20 (positions 5-10), 8 values of 30 (positions 11-18) The 13th value falls in the 30 group. Median = 30
Exercise 8.5 — Mode
Q1. Find the mode of the following data:
5, 7, 3, 5, 8, 5, 9, 3, 7, 5
- 3 appears 2 times
- 5 appears 4 times
- 7 appears 2 times
- 8 appears 1 time
- 9 appears 1 time
Mode = 5 (highest frequency = 4)
Q2. Find the mode of the following data:
2, 3, 5, 3, 2, 4, 5, 3, 4, 2, 5, 3
- 2 appears 3 times
- 3 appears 4 times
- 4 appears 2 times
- 5 appears 3 times
Mode = 3 (highest frequency = 4)
Q3. Find the mode of the following data:
1, 2, 3, 4, 5
Q4. Find the mode of the following distribution:
| x | 5 | 10 | 15 | 20 | 25 |
|---|---|---|---|---|---|
| f | 3 | 7 | 5 | 7 | 2 |
Q5. Find the mode of the following grouped frequency distribution:
| Class Interval | Frequency |
|---|---|
| 0 – 10 | 5 |
| 10 – 20 | 8 |
| 20 – 30 | 12 |
| 30 – 40 | 7 |
| 40 – 50 | 3 |
Using the formula: Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
Where:
- l = lower limit of modal class = 20
- f₁ = frequency of modal class = 12
- f₀ = frequency of class before modal class = 8
- f₂ = frequency of class after modal class = 7
- h = class width = 10
Mode = 20 + [(12 − 8) / (2×12 − 8 − 7)] × 10 = 20 + [4 / (24 − 8 − 7)] × 10 = 20 + [4 / 9] × 10 = 20 + 40/9 = 20 + 4.44 = 24.44
Q6. Find the mode of the following distribution:
| Class Interval | Frequency |
|---|---|
| 10 – 20 | 3 |
| 20 – 30 | 5 |
| 30 – 40 | 9 |
| 40 – 50 | 6 |
| 50 – 60 | 2 |
Using the formula: Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
Where:
- l = 30
- f₁ = 9
- f₀ = 5
- f₂ = 6
- h = 10
Mode = 30 + [(9 − 5) / (2×9 − 5 − 6)] × 10 = 30 + [4 / (18 − 5 − 6)] × 10 = 30 + [4 / 7] × 10 = 30 + 40/7 = 30 + 5.71 = 35.71
Exercise 8.6 — Relationship Between Mean, Median, and Mode
Q1. The mean and median of a distribution are 25 and 27 respectively. Find the mode using the empirical relationship.
Q2. The mode and mean of a distribution are 24 and 28 respectively. Find the median using the empirical relationship.
Q3. The mean and mode of a distribution are 20 and 14 respectively. Find the median.
Q4. If the median is 35 and the mean is 30, find the mode.
Q5. For a distribution, mean = 50, median = 48, and mode = 44. Verify the empirical relationship between mean, median, and mode.
The given mode is 44, and the calculated mode using the empirical relationship is also 44. Hence verified.
Exercise 8.7 — Bar Graphs and Histograms
Q1. Draw a bar graph for the following data showing the number of students in different classes:
| Class | Number of Students |
|---|---|
| I | 40 |
| II | 45 |
| III | 35 |
| IV | 50 |
| V | 30 |
6. Label the axes and give the graph a title.
- Class I: 40 students → height = 4 units
- Class II: 45 students → height = 4.5 units
- Class III: 35 students → height = 3.5 units
- Class IV: 50 students → height = 5 units
- Class V: 30 students → height = 3 units
Note: In an exam, draw the bar graph on graph paper using a ruler and pencil. Ensure all bars have equal width and are equally spaced.
Q2. Draw a histogram for the following data showing the marks of students:
| Marks | Number of Students |
|---|---|
| 0 – 10 | 5 |
| 10 – 20 | 8 |
| 20 – 30 | 12 |
| 30 – 40 | 10 |
| 40 – 50 | 7 |
5. The width of each rectangle equals the class width (10 units). 6. Label the axes and give the histogram a title.
- 0–10: height proportional to 5
- 10–20: height proportional to 8
- 20–30: height proportional to 12
- 30–40: height proportional to 10
- 40–50: height proportional to 7
Key difference from a bar graph: In a histogram, there are no gaps between the bars because the data is continuous.
Q3. Construct a frequency polygon for the following data:
| Class Interval | Frequency |
|---|---|
| 10 – 20 | 4 |
| 20 – 30 | 6 |
| 30 – 40 | 9 |
| 40 – 50 | 5 |
| 50 – 60 | 3 |
| Class Interval | Class Mark (x) | Frequency (f) |
|---|---|---|
| 10 – 20 | 15 | 4 |
| 20 – 30 | 25 | 6 |
| 30 – 40 | 35 | 9 |
| 40 – 50 | 45 | 5 |
| 50 – 60 | 55 | 3 |
2. Add imaginary class intervals at both ends:
- Before 10–20: 0–10 with class mark 5, frequency 0
- After 50–60: 60–70 with class mark 65, frequency 0
3. Plot the points (5, 0), (15, 4), (25, 6), (35, 9), (45, 5), (55, 3), (65, 0) on the graph. 4. Connect all the points with straight line segments. 5. The resulting closed figure is the frequency polygon.
Exercise 8.8 — Combined and Word Problems
Q1. The mean of 25 observations is 60. If the mean of the first 15 observations is 55 and the mean of the last 10 observations is 65, verify the overall mean.
Total sum = 825 + 650 = 1475
But the total should be 1500. Let me recheck. The sum of first 15 is 825 and last 10 is 650, giving 1475, not 1500. This indicates that the given values are inconsistent unless we recalculate:
Overall mean = (825 + 650) / 25 = 1475 / 25 = 59
So the overall mean is 59, not 60. The given overall mean of 60 would require a total of 1500, but the subgroups sum to only 1475. There is an inconsistency in the stated values.
Q2. The mean of 20 numbers is 45. If one number is excluded, the mean of the remaining 19 numbers becomes 43. Find the excluded number.
Q3. The average weight of 10 students is 50 kg. If one more student joins, the average weight becomes 52 kg. Find the weight of the new student.
Q4. The mean of 5 numbers is 18. If one number is included, the mean becomes 20. Find the included number.
Q5. Find the mean, median, and mode of the following data:
12, 15, 11, 15, 18, 14, 15, 10, 16, 20
Mean: Sum = 12 + 15 + 11 + 15 + 18 + 14 + 15 + 10 + 16 + 20 = 146 Mean = 146 / 10 = 14.6
Median: Arrange in ascending order: 10, 11, 12, 14, 15, 15, 15, 16, 18, 20 n = 10 (even) Median = average of 5th and 6th observations = (15 + 15) / 2 = 15
Mode: 15 appears 3 times (most frequent) Mode = 15
Verification using empirical relationship: Mode ≈ 3 × Median − 2 × Mean = 3 × 15 − 2 × 14.6 = 45 − 29.2 = 15.8 The actual mode is 15. The approximation gives 15.8, which is close.
Q6. Find the mean, median, and mode of the following frequency distribution:
| x | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|
| f | 5 | 8 | 10 | 7 | 5 |
Mean:
| x | f | f × x |
|---|---|---|
| 4 | 5 | 20 |
| 6 | 8 | 48 |
| 8 | 10 | 80 |
| 10 | 7 | 70 |
| 12 | 5 | 60 |
| Total | 35 | 278 |
Mean = 278 / 35 = 7.94
Median: Total frequency = 35 (odd) Median position = ((35+1)/2) = 18th observation
Cumulative frequencies: 5, 13, 23, 30, 35 The 18th observation falls in the cumulative frequency of 23 (x = 8). Median = 8
Mode: Highest frequency = 10 (for x = 8) Mode = 8
Q7. The following table shows the ages of 30 residents of a locality. Find the mean age.
| Age (years) | Number of Residents |
|---|---|
| 10 – 20 | 4 |
| 20 – 30 | 6 |
| 30 – 40 | 8 |
| 40 – 50 | 7 |
| 50 – 60 | 5 |
| Age | f | Class Mark (x) | f × x |
|---|---|---|---|
| 10–20 | 4 | 15 | 60 |
| 20–30 | 6 | 25 | 150 |
| 30–40 | 8 | 35 | 280 |
| 40–50 | 7 | 45 | 315 |
| 50–60 | 5 | 55 | 275 |
| Total | 30 | 1080 |
Mean = 1080 / 30 = 36
Q8. The mean of 6, 10, 12, x, 8 is 9. Find x.
Q9. The following data shows the heights of 15 students (in cm):
148, 152, 155, 148, 150, 152, 155, 158, 150, 152, 155, 148, 152, 155, 150
Find the mean, median, and mode.
Arranging data in ascending order: 148, 148, 148, 150, 150, 150, 152, 152, 152, 152, 155, 155, 155, 155, 158
Mean: Sum = 3×148 + 3×150 + 4×152 + 4×155 + 158 = 444 + 450 + 608 + 620 + 158 = 2280 Mean = 2280 / 15 = 152
Median: n = 15 (odd) Median = ((15+1)/2)th = 8th observation The 8th value in the sorted list is 152. Median = 152
Mode: 152 appears 4 times (highest frequency) Mode = 152
Note: Since Mean = Median = Mode = 152, this is a symmetric distribution.
Q10. A group of students scored the following marks in a test:
12, 17, 16, 15, 18, 17, 14, 16, 17, 15
(i) Find the mean (ii) Find the median (iii) Find the mode (iv) Verify the empirical relationship
(i) Mean: Sum = 12 + 17 + 16 + 15 + 18 + 17 + 14 + 16 + 17 + 15 = 157 Mean = 157 / 10 = 15.7
(ii) Median: Arrange in ascending order: 12, 14, 15, 15, 16, 16, 17, 17, 17, 18 n = 10 (even) Median = average of 5th and 6th observations = (16 + 16) / 2 = 16
(iii) Mode: 17 appears 3 times (most frequent) Mode = 17
(iv) Verify empirical relationship: 3 × Median − 2 × Mean = 3 × 16 − 2 × 15.7 = 48 − 31.4 = 16.6 Actual Mode = 17 The approximation gives 16.6, which is close to 17.
Common Mistakes
1. Mistake: Not arranging data before finding the median — Always arrange data in ascending (or descending) order before identifying the middle value. The median is the middle value of the sorted data, not the original data.
2. Mistake: Confusing mean, median, and mode formulas — Mean is the average (sum/n), median is the middle value, mode is the most frequent value. Don't mix them up. Also, the median position for odd n is ((n+1)/2)th, not (n/2)th.
3. Mistake: Using class boundaries instead of class marks for grouped data — For the mean of grouped data, use the class mark (midpoint) = (upper limit + lower limit)/2, not the class boundaries. Students often use the upper or lower limit by mistake.
4. Mistake: Wrong formula for mode of grouped data — The correct formula is Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h. Students sometimes mix up f₀ and f₂ (the frequencies of the classes before and after the modal class).
5. Mistake: Forgetting that data can have no mode or more than one mode — If all values occur with the same frequency, there is no mode. If two values have the same highest frequency, the data is bimodal. Don't assume every data set has exactly one mode.
6. Mistake: Not adding imaginary class intervals for frequency polygons — A frequency polygon must be closed. Add class intervals with frequency 0 at both ends (before the first class and after the last class) to close the polygon.
7. Mistake: Confusing histograms with bar graphs — Histograms have no gaps between bars (continuous data), while bar graphs have gaps (discrete data). The x-axis of a histogram shows class intervals, not categories.
Exam Checklist
- [ ] I can distinguish between primary and secondary data
- [ ] I can organise raw data into a frequency distribution table
- [ ] I can calculate arithmetic mean for ungrouped data
- [ ] I can calculate mean for grouped data using the direct method (class marks)
- [ ] I can find the median for ungrouped data (both odd and even number of observations)
- [ ] I can find the mode for ungrouped and grouped data
- [ ] I can use the empirical formula: Mode ≈ 3 Median − 2 Mean
- [ ] I can solve word problems involving changes in mean when observations are added or removed
- [ ] I can draw bar graphs, histograms, and frequency polygons
- [ ] I know when to use mean, median, or mode (mean for normal data, median for skewed data, mode for categorical data)
- [ ] I can handle bimodal and no-mode situations
Internal Links
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- [Class 9 Maths Subject Guide](/school/tamil-nadu-state-board/class/9/maths)