What are Mean, Mode, Median & Range?
Mean, Mode, and Median are types of 'averages' or 'measures of central tendency.' They help us summarize a set of data by giving us a typical or central value. The Range is a measure of spread or consistency.
- Mean: Often called the 'average'. It's calculated by adding up all the values in a data set and then dividing by the number of values. For grouped data or frequency tables, it involves multiplying values (or midpoints) by their frequencies.
- Mode: The value that appears most frequently in a data set. A data set can have one mode, more than one mode (bimodal, trimodal, etc.), or no mode at all. For grouped data, it's the group with the highest frequency (modal class).
- Median: The middle value in a data set that has been arranged in order from least to greatest. If there is an even number of values, the median is the mean of the two middle values.
- Range: A measure of how spread out or consistent the data is. It's calculated by subtracting the smallest value in the data set from the largest value. A smaller range means the data is more consistent.
Calculating Mean, Mode, Median & Range: Examples
Calculating the Mean (from a list)
Problem: Find the mean of the data set: 4, 7, 7, 8, 9.
- Add up all the values:
4 + 7 + 7 + 8 + 9 = 35. - Count the number of values: There are 5 values.
- Divide the sum by the number of values:
35 ÷ 5 = 7.
Answer: The Mean is 7.
Calculating the Mean from a Frequency Table (Discrete Data)
Problem: The number of pets owned by students in a class is shown below. Calculate the mean number of pets per student.
| Number of Pets (x) | Frequency (f) | Pets × Frequency (fx) |
|---|---|---|
| 0 | 5 | 0 × 5 = 0 |
| 1 | 12 | 1 × 12 = 12 |
| 2 | 8 | 2 × 8 = 16 |
| 3 | 3 | 3 × 3 = 9 |
| 4 | 2 | 4 × 2 = 8 |
| Total | Σf = 30 | Σfx = 45 |
- Multiply the number of pets (x) by its frequency (f) for each row to get the
fxcolumn. - Sum the frequencies (Σf) to find the total number of students:
5 + 12 + 8 + 3 + 2 = 30. - Sum the
fxvalues (Σfx) to find the total number of pets:0 + 12 + 16 + 9 + 8 = 45. - Calculate the mean: Mean = Σfx / Σf =
45 ÷ 30 = 1.5.
Answer: The Mean number of pets is 1.5.
Estimating the Mean from a Grouped Frequency Table (using Midpoints)
Problem: The time taken (in minutes) for 25 people to complete a puzzle is recorded below. Estimate the mean time.
| Time (minutes) | Midpoint (x) | Frequency (f) | Midpoint × Frequency (fx) |
|---|---|---|---|
| 0 < t ≤ 10 | 5 | 4 | 5 × 4 = 20 |
| 10 < t ≤ 20 | 15 | 10 | 15 × 10 = 150 |
| 20 < t ≤ 30 | 25 | 7 | 25 × 7 = 175 |
| 30 < t ≤ 40 | 35 | 4 | 35 × 4 = 140 |
| Total | Σf = 25 | Σfx = 485 |
- Find the midpoint (x) for each group. (e.g., for 0 < t ≤ 10, midpoint is (0+10)/2 = 5).
- Multiply the midpoint (x) by its frequency (f) for each row to get the
fxcolumn. - Sum the frequencies (Σf) to find the total number of people:
4 + 10 + 7 + 4 = 25. - Sum the
fxvalues (Σfx) to find the total (midpoint × frequency):20 + 150 + 175 + 140 = 485. - Estimate the mean: Mean ≈ Σfx / Σf =
485 ÷ 25 = 19.4.
Answer: The Estimated Mean time is 19.4 minutes.
Note: This is an estimate because we use midpoints, assuming data is evenly spread within each group.
Finding the Mode
Problem 1: Find the mode of: 2, 3, 5, 5, 6, 7, 5.
The number 5 appears most often (3 times).
Answer: The Mode is 5.
Problem 2: Find the mode of: 1, 2, 3, 4, 5.
All values appear only once.
Answer: There is No Mode.
Problem 3: Find the mode of: 1, 1, 2, 3, 3, 4.
The numbers 1 and 3 both appear twice (most often).
Answer: The Modes are 1 and 3 (this is a bimodal set).
Finding the Median
Problem 1 (Odd number of values): Find the median of: 9, 2, 7, 3, 5.
- Arrange the data in order: 2, 3, 5, 7, 9.
- The middle value is 5.
Answer: The Median is 5.
Problem 2 (Even number of values): Find the median of: 8, 3, 10, 4, 6, 1.
- Arrange the data in order: 1, 3, 4, 6, 8, 10.
- There are two middle values: 4 and 6.
- Find the mean of these two middle values:
(4 + 6) ÷ 2 = 10 ÷ 2 = 5.
Answer: The Median is 5.
Calculating the Range & Understanding Consistency
The range tells us how spread out our data is. A smaller range indicates the data is more consistent, while a larger range means it's less consistent.
Problem: Find the range of the data set: 15, 8, 22, 12, 25, 8.
- Identify the largest value in the data set: 25.
- Identify the smallest value in the data set: 8.
- Subtract the smallest value from the largest value:
25 - 8 = 17.
Answer: The Range is 17.
Comparing Consistency Using the Range
Problem: Two bowlers, Alice and Bob, record their scores over five games.
Alice's scores: 150, 155, 160, 152, 158
Bob's scores: 120, 180, 190, 130, 145
Who is the more consistent bowler?
Workings:
1. Calculate Alice's Range:
- Largest score: 160
- Smallest score: 150
- Alice's Range =
160 - 150 = 10
2. Calculate Bob's Range:
- Largest score: 190
- Smallest score: 120
- Bob's Range =
190 - 120 = 70
3. Compare the Ranges:
- Alice's range (10) is much smaller than Bob's range (70).
- A smaller range means greater consistency.
Answer: Alice is the more consistent bowler because her range of scores is smaller.
Test Your Skills!
1. Find the mean of: 10, 12, 15, 13, 10.
Answer: Mean = 12
Explanation: (10+12+15+13+10) = 60. Then 60 ÷ 5 = 12.
2. Find the mode of: 7, 8, 9, 8, 7, 8, 10.
Answer: Mode = 8
Explanation: 8 appears most frequently (3 times).
3. Find the median of: 15, 11, 19, 13, 17.
Answer: Median = 15
Explanation: Ordered: 11, 13, 15, 17, 19. The middle value is 15.
4. Find the range of: 3, 9, 5, 2, 10, 5.
Answer: Range = 8
Explanation: Largest (10) - Smallest (2) = 8.
Exam-Style Averages Problems
1. The scores of 7 students in a maths test were: 65, 72, 58, 81, 72, 65, 70.
Calculate: a) the mean score, b) the mode score, c) the median score, d) the range of scores.
Answers:
a) Mean: 69
b) Mode: 65 and 72 (bimodal)
c) Median: 70
d) Range: 23
Explanation:
a) Mean: (65+72+58+81+72+65+70) = 483. Then 483 ÷ 7 = 69.
b) Mode: 65 appears twice, 72 appears twice. Both are modes.
c) Median: Ordered scores: 58, 65, 65, 70, 72, 72, 81. The middle value is 70.
d) Range: Largest (81) - Smallest (58) = 23.
2. A shop records the number of items bought by customers in an hour. The data is shown in the frequency table below:
| Items Bought (x) | Frequency (f) |
|---|---|
| 1 | 8 |
| 2 | 12 |
| 3 | 7 |
| 4 | 5 |
| 5 | 3 |
Calculate: a) the mean number of items bought, b) the modal number of items bought, c) the median number of items bought.
Answers:
a) Mean: 2.51 (approx to 2dp)
b) Mode: 2 items
c) Median: 2 items
Explanation:
a) Mean:
Create an fx column: (1×8=8), (2×12=24), (3×7=21), (4×5=20), (5×3=15).
Sum of f (Σf) = 8+12+7+5+3 = 35 customers.
Sum of fx (Σfx) = 8+24+21+20+15 = 88 items.
Mean = Σfx / Σf = 88 ÷ 35 ≈ 2.51428... ≈ 2.51 (to 2 decimal places).
b) Mode: The highest frequency is 12, which corresponds to 2 items bought. So, the mode is 2 items.
c) Median:
Total frequency (Σf) = 35. Median position = (35+1)/2 = 18th value.
Cumulative frequencies:
1 item: 8 (up to 8th customer)
2 items: 8+12 = 20 (up to 20th customer)
The 18th value falls into the '2 items' category. So, the median is 2 items.
3. The heights (in cm) of 30 plants were measured and recorded in the grouped frequency table below:
| Height (cm) | Frequency (f) |
|---|---|
| 10 < h ≤ 20 | 5 |
| 20 < h ≤ 30 | 12 |
| 30 < h ≤ 40 | 8 |
| 40 < h ≤ 50 | 5 |
a) Estimate the mean height of the plants.
b) State the modal class.
c) Find the class interval in which the median height lies.
Answers:
a) Estimated Mean: 29.33 cm (approx to 2dp)
b) Modal Class: 20 < h ≤ 30 cm
c) Median Class: 20 < h ≤ 30 cm
Explanation:
a) Estimated Mean:
Midpoints (x): (10+20)/2=15; (20+30)/2=25; (30+40)/2=35; (40+50)/2=45.
fx values: (15×5=75); (25×12=300); (35×8=280); (45×5=225).
Sum of f (Σf) = 5+12+8+5 = 30 plants.
Sum of fx (Σfx) = 75+300+280+225 = 880.
Estimated Mean = Σfx / Σf = 880 ÷ 30 = 29.333... ≈ 29.33 cm (to 2 decimal places).
b) Modal Class: The class with the highest frequency (12) is 20 < h ≤ 30 cm.
c) Median Class:
Total frequency (Σf) = 30. Median position = (30+1)/2 = 15.5th value. This means we look for where the 15th and 16th values lie.
Cumulative frequencies:
10 < h ≤ 20: 5
20 < h ≤ 30: 5+12 = 17
The 15th and 16th values both fall into the 20 < h ≤ 30 cm class.
4. Maria is a delivery service supervisor. She compares delivery times for two services: "Standard" and "Express".
For the Standard service, based on 10 deliveries, the median delivery time was 25.5 minutes. The range of delivery times was 15 minutes.
The results below show the delivery times (in minutes) for 10 Express service deliveries: 10, 28, 35, 8, 15, 55, 12, 18, 13, 22
Maria thinks that with the Express service:
- the median delivery time decreases.
- the Express service delivery times are more consistent.
Is Maria correct? Show why you think this, giving reasons with your answers.
Workings:
Express Service Data Analysis:
1. Order the Express service data: 8, 10, 12, 13, 15, 18, 22, 28, 35, 55.
2. Calculate Express Service Median:
There are 10 data points. The median is the average of the 5th and 6th values.
5th value = 15, 6th value = 18.
Express Median = (15 + 18) / 2 = 33 / 2 = 16.5 minutes.
3. Calculate Express Service Range:
Range = Maximum value - Minimum value.
Express Range = 55 - 8 = 47 minutes.
Comparison with Standard Service Data:
- Standard Median = 25.5 minutes.
- Standard Range = 15 minutes.
Evaluate Maria's Claims:
Claim 1: The median delivery time decreases with Express service.
- Express Median (16.5 minutes) vs Standard Median (25.5 minutes).
- Since 16.5 < 25.5, the median delivery time has decreased with the Express service.
- Maria is correct on this claim.
Claim 2: The Express service delivery times are more consistent.
- Consistency is often measured by the spread of data. A smaller range indicates more consistency.
- Express Range (47 minutes) vs Standard Range (15 minutes).
- Since 47 > 15, the range of delivery times is larger for the Express service, meaning the service is less consistent.
- Maria is incorrect on this claim.
Conclusion: Maria is correct that the median delivery time decreases. However, she is incorrect that the Express service is more consistent; it is less consistent as shown by the larger range.
Interactive Calculator (from a simple list of numbers)
Enter a list of numbers (separated by commas or spaces) to calculate the Mean, Mode, Median, and Range.
Mean: ?
Mode(s): ?
Median: ?
Range: ?
Interactive Calculator (from a Frequency Table)
Enter data as value (x) and frequency (f) pairs. For grouped data, input the group range and the calculator will use midpoints.
Mean (or Estimated Mean): ?
Mode(s) (Value/Midpoint with highest frequency): ?
Median (Value/Midpoint): ?
Range (of Values/Midpoints): ?
Key Points to Remember
- The Mean can be affected by very high or very low values (outliers). When data is grouped, the mean is an estimate.
- The Mode is the only average that can be used for non-numerical data. For grouped data, we identify the modal class.
- The Median is often a good choice when a data set has extreme outliers.
- Always arrange your data in order (or use cumulative frequency for tables) before finding the Median.
- The Range shows the spread of data. A smaller range means more consistent data, while a larger range means less consistent data. It can be heavily influenced by a single very high or very low value.