S4: Measures of Central Tendency
Interpret, analyse and compare distributions: median, mean, mode, modal class; range, quartiles, interquartile range (Higher)
๐ Key Concepts
Definition: Measures of central tendency identify the "average" or "typical" value in a set of data. The three main measures are mean, median, and mode.
Comparison of Averages
| Measure | What It Is | Best Used For |
|---|---|---|
| Mean | Sum of values รท number of values | Continuous data, no outliers |
| Median | Middle value when ordered | Data with outliers, skewed data |
| Mode | Most frequent value | Categorical data, identifying most common |
๐ The Mean
Formula: Mean = Sum of all values รท Number of values
Symbol: xฬ (x-bar) represents the mean
Symbol: xฬ (x-bar) represents the mean
Example 1
Find the mean of: 4, 7, 9, 12, 8
Solution:
Mean = (4 + 7 + 9 + 12 + 8) รท 5
Mean = 40 รท 5 = 8
Example 2
The mean of 6 numbers is 15. Five of the numbers are 12, 18, 14, 16, 20. Find the sixth number.
Solution:
Total sum = Mean ร Number of values = 15 ร 6 = 90
Sum of five numbers = 12 + 18 + 14 + 16 + 20 = 80
Sixth number = 90 - 80 = 10
Mean from a Frequency Table:
Mean = ฮฃ(f ร x) รท ฮฃf
where f = frequency and x = value
Mean = ฮฃ(f ร x) รท ฮฃf
where f = frequency and x = value
Example 3
Find the mean from this frequency table:
| Score (x) | Frequency (f) |
|---|---|
| 2 | 3 |
| 3 | 5 |
| 4 | 8 |
| 5 | 4 |
Solution:
| x | f | f ร x |
|---|---|---|
| 2 | 3 | 6 |
| 3 | 5 | 15 |
| 4 | 8 | 32 |
| 5 | 4 | 20 |
| Total | 20 | 73 |
Mean = 73 รท 20 = 3.65
๐ The Median
Definition: The middle value when data is arranged in order.
Position: For n values, median position = (n + 1) รท 2
Position: For n values, median position = (n + 1) รท 2
Example 4
Find the median of: 3, 7, 2, 9, 5
Solution:
Step 1: Order the data: 2, 3, 5, 7, 9
Step 2: Find middle position: (5 + 1) รท 2 = 3rd value
Median = 5
Example 5
Find the median of: 4, 8, 1, 6, 9, 3
Solution:
Step 1: Order the data: 1, 3, 4, 6, 8, 9
Step 2: 6 values, so median position = (6 + 1) รท 2 = 3.5
Step 3: Average of 3rd and 4th values: (4 + 6) รท 2 = 5
Median = 5
From a Frequency Table: Find the cumulative frequency, then locate the middle position.
Example 6
Find the median from:
| Score | Frequency | Cumulative Freq |
|---|---|---|
| 1 | 4 | 4 |
| 2 | 7 | 11 |
| 3 | 9 | 20 |
| 4 | 5 | 25 |
Solution:
Total = 25 values, median position = (25+1) รท 2 = 13th value
The 13th value is in the group where cumulative frequency reaches 20
Median = 3
๐ The Mode
Definition: The value that occurs most frequently.
- Data can have one mode (unimodal), two modes (bimodal), or no mode
- Only average that can be used for categorical data
Example 7
Find the mode of: 5, 3, 7, 5, 2, 5, 8, 3
Solution:
5 appears 3 times (most frequent)
Mode = 5
Example 8
Find the mode: Red, Blue, Red, Green, Blue, Blue, Red
Solution:
Red: 3 times, Blue: 3 times, Green: 1 time
Bimodal: Mode = Red and Blue
Modal Class: For grouped data, the class with the highest frequency is the modal class.
Example 9
Find the modal class:
| Height (cm) | Frequency |
|---|---|
| 150-160 | 8 |
| 160-170 | 15 |
| 170-180 | 12 |
Solution:
Highest frequency = 15 in class 160-170
Modal class = 160-170 cm
๐ The Range
Definition: Range = Highest value - Lowest value
Measures the spread of data. A larger range means more variation.
Measures the spread of data. A larger range means more variation.
Example 10
Find the range of: 12, 45, 23, 67, 34, 89, 56
Solution:
Highest = 89, Lowest = 12
Range = 89 - 12 = 77
๐ Quartiles and Interquartile Range (Higher)
Quartiles divide data into four equal parts:
- Q1 (Lower Quartile): 25% of data below
- Q2 (Median): 50% of data below
- Q3 (Upper Quartile): 75% of data below
Interquartile Range (IQR): IQR = Q3 - Q1
Measures the spread of the middle 50% of data. Less affected by outliers than the range.
Measures the spread of the middle 50% of data. Less affected by outliers than the range.
Example 11
Find the quartiles and IQR: 2, 5, 7, 8, 10, 12, 15, 18, 22
Solution:
Data already ordered. n = 9
Q2 (Median): Position 5 = 10
Q1: Median of lower half (2, 5, 7, 8) = (5+7)รท2 = 6
Q3: Median of upper half (12, 15, 18, 22) = (15+18)รท2 = 16.5
IQR = Q3 - Q1 = 16.5 - 6 = 10.5
๐ Comparing Distributions
When comparing distributions, discuss:
- A measure of average (mean or median)
- A measure of spread (range or IQR)
- Compare using phrases like "higher than", "more varied than"
Example 12
Class A: Mean = 72, Range = 25
Class B: Mean = 68, Range = 40
Compare the test scores.
Solution:
Class A performed better on average (72 > 68).
Class B's scores were more varied (range 40 > 25).
Class A's results were more consistent.
โ Practice Questions
Q1: Find the mean, median, and mode of: 4, 6, 8, 6, 5, 6, 9
Q2: The mean of 5 numbers is 12. Four of the numbers are 10, 15, 8, 14. Find the fifth number.
Q3: Find the range: 23, 45, 12, 67, 34, 89, 56
Q4: A data set has Q1 = 15, Q2 = 28, Q3 = 42. Find the interquartile range.
Q5: Which average would you use for favourite colours? Explain why.
โ Answers
- Mean = 44รท7 โ 6.29; Ordered: 4,5,6,6,6,8,9; Median = 6; Mode = 6
- Total = 5ร12 = 60; Fifth = 60 - (10+15+8+14) = 60 - 47 = 13
- Range = 89 - 12 = 77
- IQR = Q3 - Q1 = 42 - 15 = 27
- Mode - because colour is categorical data (you cannot calculate mean or median for colours).
๐ฏ Exam Tips
- Always order data before finding the median
- Check if there's more than one mode
- For grouped data, estimate the mean using midpoints
- Use median for data with outliers (less affected by extreme values)
- When comparing, comment on both average and spread
- Higher tier: Know how to find quartiles and IQR from cumulative frequency graphs
๐ง Problem-Solving Strategies
Problem-Solving
For averages and spread problems: (1) Always order data before finding the median, (2) For frequency tables, use ฮฃ(fรx)/ฮฃf for the mean, (3) When a value is missing and the mean is given, use: missing value = (mean ร n) โ sum of known values, (4) To compare distributions, always comment on BOTH average (mean/median) AND spread (range/IQR), (5) Use the median for data with outliers โ it's not affected by extreme values.
Multi-Step Problem
Seven numbers have a mean of 12. Six of the numbers are: 8, 15, 10, 14, 16, 9. (a) Find the seventh number. (b) Find the median of all seven numbers. (c) If an eighth number of 50 is added, find the new mean and median, and explain why they change differently.
Solution: (a) Total = 12 ร 7 = 84. Known sum = 72. Seventh = 84 โ 72 = 12. (b) Ordered: 8, 9, 10, 12, 14, 15, 16. Median = 4th value = 12. (c) New mean = (84+50)/8 = 16.75. New ordered: 8, 9, 10, 12, 14, 15, 16, 50. Median = (12+14)/2 = 13. The mean increased a lot (from 12 to 16.75) because 50 is an outlier. The median only changed slightly (12 to 13) because it's resistant to outliers.
โ ๏ธ Common Errors
Watch Out!
1. Wrong: Finding the median without ordering the data first Correct: Always arrange data in ascending order before identifying the middle value
2. Wrong: Confusing mean, median and mode โ e.g. saying "the most common value is the mean" Correct: Mode = most frequent, Median = middle when ordered, Mean = sum รท count โ they are different measures
3. Wrong: Using the mean for data with extreme outliers and claiming it represents a "typical" value Correct: For data with outliers (like salaries, house prices), the median better represents a typical value since the mean is pulled towards the outlier
โ๏ธ 6-Mark Exam Question
Extended Answer
6 marks: Two classes take the same test. Class A: 25 students, mean = 62, range = 35. Class B: 30 students, mean = 58, range = 50. (a) Calculate the combined mean for all 55 students. (b) Compare the performance of the two classes, commenting on both average and spread. (c) A new student joins Class A and scores 95. Explain the effect on the mean and median of Class A.
(a) Total for A = 25 ร 62 = 1550. Total for B = 30 ร 58 = 1740. Combined total = 3290. Combined mean = 3290/55 = 59.8 โ 60.
(b) Class A performed better on average (mean 62 > 58). Class A's results were more consistent (range 35 < 50). Class B had more variation in scores, with some students performing much better or worse than others.
(c) New mean = (1550+95)/26 = 1645/26 โ 63.3 (increased). The new score of 95 is well above the mean, pulling it up. The median will increase slightly because 95 is higher than most scores, shifting the middle value upward, but the change will be much smaller than the change in mean since median is resistant to outliers.
Mark scheme: M1 for class totals, A1 for combined mean โ60, M1 for comparing averages, M1 for comparing spread, A1 for clear comparison statement, M1 for explaining mean increase, A1 for explaining median is less affected
๐ AO3: Reason & Interpret
Reasoning and Interpretation
A company reports: "The mean salary is ยฃ42,000. The median salary is ยฃ28,000."
(a) What does the difference between mean and median tell you about the distribution of salaries?
(b) Which average should a journalist use to describe a "typical" salary? Justify your choice.
(c) The company removes the CEO's salary of ยฃ250,000 and recalculates. Would the mean or median change more? Explain.
Answers: (a) Mean > Median indicates right-skewed (positive skew). A few very high salaries pull the mean up above the median. (b) The median (ยฃ28,000) โ it better represents a typical salary because the mean is inflated by a small number of very high earners. (c) The mean would change more โ removing the CEO's salary significantly reduces the total sum, bringing the mean down. The median would barely change because it depends on the middle value, which shifts by at most one position.
๐ Exam Questions by Topic
๐ฌ Video Resources
Ready to ace your GCSE Mathematics exams?
Get the best revision books and guides to boost your grades.