How to Use This Page
1. Solve each exercise problem on your own first — attempt every step before looking here. 2. Compare your method with the step-by-step solution shown. 3. If your answer differs, rework from the step where you diverged. 4. Pay special attention to the Common Mistakes section — these are the errors that cost marks in exams. 5. Use the Exam Checklist before your test to ensure you have covered every topic.
Key Concepts (Quick Recap)
Measures of Central Tendency
- Mean (Average): Sum of all observations / Number of observations
- For ungrouped data: Mean = (x1 + x2 + ... + xn) / n
- For frequency distribution: Mean = Sum(fi × xi) / Sum(fi), where xi is the value and fi is its frequency
- For grouped data: Mean = Sum(fi × xi) / Sum(fi), where xi = class mark = (upper + lower) / 2
- Median: Middle value of sorted data
- For odd n: Median = (n+1)/2 th observation
- For even n: Median = average of (n/2)th and (n/2 + 1)th observations
- Mode: Most frequently occurring value
Frequency Distribution
- Class interval: Range of values grouped together (e.g., 0-10, 10-20)
- Class mark (midpoint): (Upper limit + Lower limit) / 2
- Class width: Upper limit - Lower limit
- Frequency: Number of observations in a class interval
- Cumulative frequency: Running total of frequencies
Important Formulas
- Range = Maximum value - Minimum value
- Mean using assumed mean method: Mean = A + Sum(fi × di) / Sum(fi), where A = assumed mean, di = xi - A
- Empirical relationship: Mode = 3 × Median - 2 × Mean
Graphical Representation
- Histogram: Bar graph for continuous data (no gaps between bars)
- Frequency polygon: Line graph connecting midpoints of histogram bars
- Ogive: Graph of cumulative frequency (S-curve)
Exercise 6.1 — Mean, Median, Mode for Ungrouped Data
Q1. Find the mean of: 4, 6, 8, 10, 12.
Mean = Sum of observations / Number of observations
= (4 + 6 + 8 + 10 + 12) / 5
= 40 / 5
= 8
Answer: Mean = 8
Q2. Find the mean of: 15, 18, 21, 24, 27, 30.
Mean = (15 + 18 + 21 + 24 + 27 + 30) / 6
= 135 / 6
= 22.5
Answer: Mean = 22.5
Q3. Find the mean of the first five natural numbers.
First five natural numbers: 1, 2, 3, 4, 5
Mean = (1 + 2 + 3 + 4 + 5) / 5
= 15 / 5
= 3
Answer: Mean = 3
Q4. The marks of 8 students are 45, 52, 48, 55, 60, 43, 50, 47. Find the mean marks.
Mean = (45 + 52 + 48 + 55 + 60 + 43 + 50 + 47) / 8
= 400 / 8
= 50
Answer: Mean = 50
Q5. If the mean of 6 numbers is 15, find their sum.
Mean = Sum / Number of observations
15 = Sum / 6
Sum = 15 × 6 = 90
Answer: Sum = 90
Q6. The mean of 5 numbers is 18. If one number is excluded, the mean of the remaining 4 numbers is 16. Find the excluded number.
Sum of 5 numbers = 5 × 18 = 90
Sum of 4 numbers = 4 × 16 = 64
Excluded number = 90 - 64 = 26
Answer: 26
Q7. Find the median of: 3, 7, 5, 9, 1, 8, 6.
Step 1: Arrange in ascending order: 1, 3, 5, 6, 7, 8, 9
Step 2: Number of observations n = 7 (odd)
Median = (n+1)/2 th observation = (7+1)/2 = 4th observation
Median = 6
Answer: Median = 6
Q8. Find the median of: 12, 15, 18, 20, 22, 25.
Already arranged in ascending order.
n = 6 (even)
Median = average of (n/2)th and (n/2 + 1)th observations
= average of 3rd and 4th observations
= (18 + 20) / 2
= 38 / 2
= 19
Answer: Median = 19
Q9. Find the median of: 25, 18, 32, 10, 15, 28, 20, 22.
Step 1: Arrange in ascending order: 10, 15, 18, 20, 22, 25, 28, 32
Step 2: n = 8 (even)
Median = average of 4th and 5th observations
= (20 + 22) / 2
= 42 / 2
= 21
Answer: Median = 21
Q10. Find the mode of: 2, 3, 5, 3, 7, 3, 5, 2, 3.
Arrange the data: 2, 2, 3, 3, 3, 3, 5, 5, 7
Frequency of each value:
2 occurs 2 times
3 occurs 4 times
5 occurs 2 times
7 occurs 1 time
Mode = 3 (occurs most frequently, 4 times)
Answer: Mode = 3
Q11. Find the mode of: 5, 7, 9, 7, 3, 5, 7, 5, 7.
Frequency:
3 occurs 1 time
5 occurs 3 times
7 occurs 4 times
9 occurs 1 time
Mode = 7 (occurs 4 times)
Answer: Mode = 7
Q12. The ages (in years) of 10 students are: 12, 14, 13, 12, 15, 14, 13, 12, 16, 12. Find the mean, median, and mode.
Arranged in order: 12, 12, 12, 12, 13, 13, 14, 14, 15, 16
Mean = (12 + 12 + 12 + 12 + 13 + 13 + 14 + 14 + 15 + 16) / 10
= 133 / 10
= 13.3
Median: n = 10 (even)
Median = average of 5th and 6th observations
= (13 + 13) / 2
= 13
Mode = 12 (occurs 4 times)
Answer: Mean = 13.3, Median = 13, Mode = 12
Q13. If the mean of 10 observations is 20 and one observation 15 is replaced by 25, find the new mean.
Original sum = 10 × 20 = 200
Change in sum = 25 - 15 = 10
New sum = 200 + 10 = 210
New mean = 210 / 10 = 21
Answer: New mean = 21
Q14. The mean of 20 numbers is 30. If each number is increased by 5, find the new mean.
Original sum = 20 × 30 = 600
Increase in each number = 5
Total increase = 20 × 5 = 100
New sum = 600 + 100 = 700
New mean = 700 / 20 = 35
Alternative method: If each number is increased by k, the new mean = old mean + k
New mean = 30 + 5 = 35
Answer: 35
Q15. Find the mean of the first 10 even numbers.
First 10 even numbers: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Mean = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20) / 10
= 110 / 10
= 11
Answer: Mean = 11
Exercise 6.2 — Frequency Distribution and Grouped Data
Q1. The following data shows the marks obtained by 20 students in a test. Prepare a frequency distribution table with class intervals 0-10, 10-20, etc. Data: 5, 12, 18, 23, 7, 15, 25, 9, 14, 20, 3, 16, 22, 11, 8, 19, 27, 13, 6, 17
Frequency Distribution Table:
| Class Interval | Tally | Frequency |
|----------------|---------------|-----------|
| 0 - 10 | |||| | 5 |
| 10 - 20 | |||| |||| || | 12 |
| 20 - 30 | ||| | 3 |
| Total | | 20 |
Answer: Frequencies are 5, 12, 3 for the three class intervals.
Q2. Find the mean from the following frequency distribution:
Q3. Find the mean from the following grouped frequency distribution:
Q4. Find the mean using the assumed mean method for the data:
Q5. The weights (in kg) of 25 students are given below. Find the mean. 38, 42, 35, 40, 44, 38, 42, 36, 40, 39, 41, 37, 43, 38, 42, 36, 40, 38, 42, 40, 35, 43, 37, 41, 39
Frequency Table:
| Weight (xi) | fi | fi × xi |
|-------------|-----|---------|
| 35 | 2 | 70 |
| 36 | 2 | 72 |
| 37 | 2 | 74 |
| 38 | 4 | 152 |
| 39 | 2 | 78 |
| 40 | 4 | 160 |
| 41 | 2 | 82 |
| 42 | 4 | 168 |
| 43 | 2 | 86 |
| 44 | 1 | 44 |
| Total | 25 | 986 |
Mean = 986 / 25 = 39.44
Answer: Mean = 39.44 kg
Q6. Find the mean of the following distribution using the direct method:
Q7. The daily wages of 50 workers are given below. Find the mean daily wage.
Q8. Find the median from the following data: 13, 16, 12, 14, 19, 12, 14, 13, 14
Step 1: Arrange in ascending order: 12, 12, 13, 13, 14, 14, 14, 16, 19
Step 2: n = 9 (odd)
Median = (n+1)/2 th observation = (9+1)/2 = 5th observation
Median = 14
Answer: Median = 14
Q9. Find the median for the following frequency distribution:
Q10. Find the mode from the following frequency distribution:
Q11. The following data shows the number of children in 30 families. Find the mode.
Q12. Find the range of the data: 25, 18, 20, 22, 28, 35, 30, 15
Maximum value = 35
Minimum value = 15
Range = Maximum - Minimum = 35 - 15 = 20
Answer: Range = 20
Q13. Find the mean, median, and mode for: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
Mean = (6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24) / 10
= 150 / 10
= 15
Median: n = 10 (even)
Median = average of 5th and 6th observations
= (14 + 16) / 2
= 15
Mode: Each value appears exactly once. No mode.
Verify empirical formula: Mode = 3 × Median - 2 × Mean
No mode exists, so the empirical formula does not apply here.
Answer: Mean = 15, Median = 15, No mode
Q14. If the mean of a distribution is 25 and the mode is 15, find the median using the empirical formula.
Empirical formula: Mode = 3 × Median - 2 × Mean
15 = 3 × Median - 2 × 25
15 = 3 × Median - 50
3 × Median = 15 + 50
3 × Median = 65
Median = 65/3 = 21.67
Answer: Median = 21.67 (approximately)
Q15. If the mean is 20 and the median is 22, find the mode using the empirical formula.
Empirical formula: Mode = 3 × Median - 2 × Mean
Mode = 3 × 22 - 2 × 20
Mode = 66 - 40
Mode = 26
Answer: Mode = 26
Exercise 6.3 — Cumulative Frequency and Ogive
Q1. From the following frequency distribution, prepare a cumulative frequency table:
Q2. Find the median from the following grouped frequency distribution:
Q3. Find the median for the following distribution:
Q4. Find the mode for the following grouped frequency distribution:
Q5. The heights of 50 students are given below. Find the median height.
Q6. The marks obtained by 60 students are given below. Find the mean marks using the direct method.
Q7. Draw a frequency distribution table for the following data and find the mean: 15, 22, 18, 25, 12, 20, 22, 15, 18, 25, 20, 22, 15, 18, 20
Frequency Table:
| xi | Tally | fi |
|----|-------|----|
| 12 | | | 1 |
| 15 | ||| | 3 |
| 18 | ||| | 3 |
| 20 | ||| | 3 |
| 22 | ||| | 3 |
| 25 | || | 2 |
| Total | | 15 |
| xi | fi | fi × xi |
|----|----|---------|
| 12 | 1 | 12 |
| 15 | 3 | 45 |
| 18 | 3 | 54 |
| 20 | 3 | 60 |
| 22 | 3 | 66 |
| 25 | 2 | 50 |
| Total | 15 | 287 |
Mean = 287 / 15 = 19.13
Answer: Mean = 19.13 (approximately)
Q8. The following data gives the number of books issued from a library on different days. Find the median. 25, 20, 30, 35, 15, 18, 22, 28, 32, 26, 24
Step 1: Arrange in ascending order: 15, 18, 20, 22, 24, 25, 26, 28, 30, 32, 35
Step 2: n = 11 (odd)
Median = (11+1)/2 = 6th observation
Median = 25
Answer: Median = 25
Q9. Find the mean, median, and mode for: 4, 6, 7, 8, 8, 9, 10
Arranged: 4, 6, 7, 8, 8, 9, 10
Mean = (4 + 6 + 7 + 8 + 8 + 9 + 10) / 7 = 52 / 7 = 7.43
Median: n = 7 (odd), Median = 4th observation = 8
Mode = 8 (occurs twice, most frequent)
Verify: Mode = 3 × Median - 2 × Mean = 3(8) - 2(7.43) = 24 - 14.86 = 9.14
(The empirical formula is approximate for small datasets.)
Answer: Mean = 7.43, Median = 8, Mode = 8
Q10. From the cumulative frequency table below, find the median:
Common Mistakes
1. Confusing mean, median, and mode — Mean is the average (sum/count), median is the middle value (must sort first), and mode is the most frequent value. These are three different measures and can give very different results for the same dataset.
2. Forgetting to sort data before finding median — Median requires data in ascending or descending order. Finding the middle value of unsorted data gives the wrong answer. Always sort first, then find the middle position.
3. Wrong class mark calculation — For grouped data, class mark = (upper limit + lower limit) / 2. A common error is using just the upper limit or just the lower limit. For the class 10-20, the class mark is (10+20)/2 = 15, not 10 or 20.
4. Incorrect cumulative frequency for median — When finding the median class, use the cumulative frequency JUST GREATER than N/2, not the class with frequency equal to N/2. Also, cf in the formula is the cumulative frequency BEFORE the median class, not the cumulative frequency of the median class.
5. Wrong median formula application — For grouped data: Median = L + [(N/2 - cf) / f] × h. Students often confuse cf (cumulative frequency before median class) with f (frequency of median class). Remember: cf is the cumulative frequency of the class BEFORE the median class.
6. Omitting class boundaries for continuous data — When data values like 10, 20, 30 appear, adjust class intervals. If classes are 0-10, 10-20, etc., and data includes 10, it belongs to 10-20, not 0-10. Use class boundaries: 0-10 becomes -0.5 to 9.5, or redefine classes as 0-10, 10-20 with the convention that 10 goes in 10-20.
7. Misapplying the empirical formula — Mode = 3 × Median - 2 × Mean is an approximate relationship for moderately skewed data. It does NOT work for symmetric or heavily skewed data. Do not use it to find exact values when the actual mode can be directly determined.
8. Not simplifying final answers — Always express fractions in lowest terms and round decimals appropriately. An answer of 22/44 should be written as 1/2, and decimal answers should be rounded to 2 decimal places unless stated otherwise.
Exam Checklist
- [ ] I can calculate the mean for ungrouped data (sum of values divided by count)
- [ ] I can calculate the mean for a frequency distribution using Sum(fi × xi) / Sum(fi)
- [ ] I can calculate the mean for grouped data using class marks
- [ ] I can use the assumed mean method for calculating mean
- [ ] I can find the median for ungrouped data (odd and even n)
- [ ] I can find the median for grouped data using the cumulative frequency method
- [ ] I can find the mode for ungrouped data and frequency distributions
- [ ] I can find the mode for grouped data using the modal class formula
- [ ] I can create frequency distribution tables with tally marks
- [ ] I can create cumulative frequency tables (less than type)
- [ ] I can calculate range, class intervals, and class marks
- [ ] I can use the empirical formula Mode = 3 × Median - 2 × Mean
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Frequently Asked Questions
For grouped data, Mean = Sum of (fi × xi) / Sum of fi, where fi is the frequency of each class and xi is the class mark (midpoint) of each class interval. Class mark = (upper limit + lower limit)/2.