EDEXCEL GCSE MATHS · FOUNDATION & HIGHER
Averages and range
Interpreting and representing data · Lesson 5 of 6
Last Lesson and Before
Answer each one, then check.
1. Work out the mean of 2, 4 and 6.
(2 + 4 + 6)/3 = 4
2. Put 7, 3, 9, 1 in order.
1, 3, 7, 9
3. What is the middle value of 1, 3, 7?
3
4. Last lesson: what is interpolation?
Estimating inside the range of the data.
5. Round 12.46 to 1 decimal place.
12.5
Learning Objectives
1. Find the mean, median, mode and range of a set of data.
2. Choose the most suitable average.
3. Find averages and the range from a frequency table.
4. Estimate the mean of grouped data, and find the modal class and the class containing the median.
5. Solve problems with missing values and combined means.
Three Averages and a Spread
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Mean Add up all the values and divide by how many there are: mean = total/(number of values). |
Median The middle value when the data is in order. With an even number of values, the mean of the middle two. |
|
Mode The most common value. There can be more than one, or none. |
Range Largest value − smallest value. It measures spread, not average. |
All Four from a List
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Find the mean, median, mode and range of 4, 8, 3, 9, 8, 6, 12. |
1. Mean: add them and divide by 7
50/7 = 7.14 (2 d.p.)
2. Median: put them in order
3, 4, 6, 8, 8, 9, 12: the middle (4th) value is 8
3. Mode: the most common
8 appears twice
4. Range: largest − smallest
12 − 3 = 9
Answer: Mean 7.14, median 8, mode 8, range 9
Which Average?
|
MEAN |
MEDIAN AND MODE |
|
▸ Uses every value. ▸ Pulled up or down by outliers. ▸ Best when there are no extreme values. |
▸ The median is not affected by outliers - best for data like house prices or salaries. ▸ The mode is the only average for qualitative data, such as favourite colour. ▸ The mode can be unhelpful if every value is different. |
PART ONE
Frequency Tables
When each value comes up many times.
Pets Owned by 30 Students
Add a column for frequency × value: it gives the total number of pets.
|
Number of pets (x) |
Frequency (f) |
f × x |
|---|---|---|
|
0 |
8 |
0 |
|
1 |
12 |
12 |
|
2 |
6 |
12 |
|
3 |
3 |
9 |
|
4 |
1 |
4 |
|
Total |
30 |
37 |
Averages from a Frequency Table
|
Use the pets table to find the mean, median, mode and range. |
1. Mean = total of f × x ÷ total frequency
37/30 = 1.23 (2 d.p.)
2. Median: 30 values, so it is halfway between the 15th and 16th
The first 8 are 0; values 9 to 20 are 1
3. So the 15th and 16th values are both 1
Median = 1
4. Mode: the value with the highest frequency
1 (frequency 12)
5. Range: largest value − smallest value
4 − 0 = 4
Answer: Mean 1.23, median 1, mode 1, range 4
PART TWO
Grouped Data
When the exact values are unknown.
Journey Times of 30 Students
Each class is represented by its midpoint.
|
Time, t (minutes) |
Frequency (f) |
Midpoint |
f × midpoint |
|---|---|---|---|
|
0 < t ≤ 10 |
5 |
5 |
25 |
|
10 < t ≤ 20 |
12 |
15 |
180 |
|
20 < t ≤ 30 |
9 |
25 |
225 |
|
30 < t ≤ 40 |
4 |
35 |
140 |
|
Total |
30 |
|
570 |
Estimating the Mean of Grouped Data
|
Use the journey times table to estimate the mean, and find the modal class and the class containing the median. |
1. Use the midpoint of each class
5, 15, 25, 35
2. Multiply by the frequencies and add
25 + 180 + 225 + 140 = 570
3. Divide by the total frequency
570/30 = 19
4. Modal class: the class with the highest frequency
10 < t ≤ 20
5. Median: the 15.5th value. Running total: 5, then 5 + 12 = 17
so it is in 10 < t ≤ 20
Answer: Estimated mean 19 minutes; modal class 10 < t ≤ 20; the median is in 10 < t ≤ 20.
Why Only an Estimate?
With grouped data, you do not know the actual values.
▸ The midpoint stands in. We assume every value in a class is at its midpoint.
▸ Some are higher, some lower. The errors tend to cancel out, so the estimate is usually close.
▸ In an exam. "Explain why your answer is an estimate" wants: "the exact values are not known, so the midpoints were used".
Missing Values and Combined Means
Mean × number of values = total. That one fact solves most problems.
▸ Missing value. The mean of 5 numbers is 8, so they add up to 5 × 8 = 40. If four of them add up to 31, the fifth is 9.
▸ Adding a value. The mean of 4 numbers is 9 (total 36). After adding one, the mean of 5 is 10 (total 50). The new number is 50 − 36 = 14.
▸ Combined means. Class A: 20 students, mean 64 (total 1280). Class B: 25 students, mean 73 (total 1825). Mean of all 45: 3105/45 = 69.
▸ Never average the averages. (64 + 73)/2 = 68.5 is wrong: the classes are different sizes.
Key Terms
|
Mean The total of the values divided by the number of values. |
Median The middle value when the data is in order. |
|
Mode The most common value. |
Range The difference between the largest and smallest values. |
|
Modal class The class with the highest frequency in grouped data. |
Midpoint The value halfway through a class, used to estimate the mean. |
Your Task: Choose Your Average
12 minutes
|
For each data set, find the mean, median and mode, then decide which is the best average and why. (a) Salaries at a small firm: £22 000, £24 000, £24 000, £26 000, £28 000, £150 000. (b) Shoe sizes sold by a shop in one hour: 5, 6, 6, 7, 7, 7, 8, 9. (c) Favourite fruit of 10 people: apple 4, banana 3, grape 2, pear 1. 1. Work out all three averages. 2. Look for outliers and the type of data. 3. Choose and justify. |
A good answer shows: (a) Mean £45 667, median £25 000, mode £24 000 - the median, because the £150 000 outlier pulls the mean up. (b) Mean 6.875, median 7, mode 7 - the mode, because a shop orders whole shoe sizes. (c) Only the mode (apple) makes sense for qualitative data.
Can I...?
☐ Find the mean, median, mode and range of a list.
☐ Choose the best average.
☐ Find the mean from a frequency table.
☐ Find the median from a frequency table.
☐ Estimate the mean of grouped data.
☐ Find the modal class.
☐ Find the class containing the median.
☐ Solve missing value and combined mean problems.
Summary
✓ Mean = total ÷ number of values; median = middle; mode = most common; range = largest − smallest.
✓ Frequency table: mean = Σfx/Σf.
✓ Grouped data: use midpoints - the mean is only an estimate.
✓ Mean × number of values = total: the key to missing values and combined means.
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EXAM FOCUS Work out an estimate for the mean weight of the 40 parcels. (4 marks) For grouped data, add a midpoint column and an f × midpoint column to the table. Then divide by the total FREQUENCY, not by the number of classes. |