This note covers the three averages (mean, median and mode), the range as a measure of spread, and how to find them from lists, frequency tables and grouped frequency tables. Choosing and comparing averages is a basic skill in sport, science, business and everyday news reports.
Key ideas
An average is a single value that represents a whole set of data. There are three types, and each has strengths and weaknesses.
- The mean uses every value: add them all and divide by how many there are. It is affected by extreme values.
- The median is the middle value when the data are in order. It is not much affected by extreme values.
- The mode is the most common value. It is the only average that works for non-numerical data (such as favourite colour), but a set may have no mode, or more than one.
The range is the largest value minus the smallest value. It measures how spread out the data are. A small range means the values are close together and a large range means they are widely spread. The range is badly affected by a single extreme value.
Finding the median. Put the values in order first. If there are \(n\) values, the median is in position \(\dfrac{n + 1}{2}\). With an odd number of values, this is one of the data values. With an even number, the median is halfway between the two middle values.
Frequency tables. A frequency table shows each value and how many times it occurs. To work out the averages:
- Mode: the value with the highest frequency.
- Mean: add a column for value \(\times\) frequency. The mean is the total of this column divided by the total frequency.
- Median: find the total frequency \(n\), work out the position \(\dfrac{n+1}{2}\), then count along the cumulative frequencies until you reach that position.
- Range: largest value minus smallest value (with a non-zero frequency).
Grouped frequency tables. When data are grouped into classes (such as \(0 < t \le 10\)), you no longer know the exact values, so you can only estimate. Use the midpoint of each class as the value for every item in it.
- Estimate of the mean: total of (midpoint \(\times\) frequency) divided by total frequency.
- Modal class: the class with the highest frequency (when the classes have equal widths).
The answer is only an estimate because real values within each class could be anywhere in that class.
Choosing an average. Use the mean when the data are fairly even and you want to use all the values. Use the median when there are extreme values, for example house prices or salaries. Use the mode when you want the most popular item, such as the most common shoe size in a shop order. To compare two data sets, compare an average (which is typical) and the range (which is spread).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Mean \(= \dfrac{\text{sum of all values}}{\text{number of values}}\) | Average using every value. |
| Median | Middle value of ordered data, in position \(\dfrac{n+1}{2}\). |
| Mode | Most frequent value. |
| Range \(=\) largest \(-\) smallest | Measure of spread. |
| Mean from a table \(= \dfrac{\sum fx}{\sum f}\) | \(x\) is the value (or class midpoint) and \(f\) is the frequency. \(\sum\) means "the sum of". |
| Class midpoint | Halfway between the lower and upper limits of a class. |
| Modal class | Class with the highest frequency in a grouped table. |
Worked examples
Example 1 The number of goals scored by a team in eight matches was 6, 9, 4, 9, 12, 7, 9, 5. Find the mean, median, mode and range.
Order the values: 4, 5, 6, 7, 9, 9, 9, 12.
Mean: \(\dfrac{4 + 5 + 6 + 7 + 9 + 9 + 9 + 12}{8} = \dfrac{61}{8} = 7.625\).
Median: there are 8 values, so the median is halfway between the 4th and 5th values: \(\dfrac{7 + 9}{2} = 8\).
Mode: 9 occurs three times, more than any other value.
Range: \(12 - 4 = 8\).
Answer: mean 7.63 (3 s.f.), median 8, mode 9, range 8.
Example 2 A teacher records the number of errors in each of 24 short tests. Errors: 0, 1, 2, 3, 4 with frequencies 3, 5, 8, 6, 2. Find the mode, the median and the mean.
Mode: the highest frequency is 8, so the mode is 2 errors.
Median: total frequency is \(3 + 5 + 8 + 6 + 2 = 24\). The position is \(\dfrac{24 + 1}{2} = 12.5\), halfway between the 12th and 13th values. Cumulative frequencies are 3, 8, 16, 22, 24. Both the 12th and 13th values fall in the group with 2 errors, so the median is 2.
Mean: multiply each value by its frequency: \(0 \times 3 = 0\), \(1 \times 5 = 5\), \(2 \times 8 = 16\), \(3 \times 6 = 18\), \(4 \times 2 = 8\). The total is \(0 + 5 + 16 + 18 + 8 = 47\).
Mean \(= \dfrac{47}{24} = 1.958\ldots\)
Answer: mode 2, median 2, mean 1.96 errors.
Example 3 The table shows the time \(t\) minutes that 40 people waited at a clinic. Classes: \(0 \lt t \le 10\), \(10 \lt t \le 20\), \(20 \lt t \le 30\), \(30 \lt t \le 40\), \(40 \lt t \le 50\) with frequencies 4, 9, 15, 8, 4. Find the modal class and estimate the mean waiting time.
The highest frequency is 15, so the modal class is \(20 \lt t \le 30\).
Midpoints are 5, 15, 25, 35, 45.
\(fx\): \(5 \times 4 = 20\), \(15 \times 9 = 135\), \(25 \times 15 = 375\), \(35 \times 8 = 280\), \(45 \times 4 = 180\).
\(\sum fx = 20 + 135 + 375 + 280 + 180 = 990\) and \(\sum f = 40\).
Estimated mean \(= \dfrac{990}{40} = 24.75\).
Answer: modal class \(20 \lt t \le 30\); estimated mean 24.8 minutes (3 s.f.).
Common errors
- Common error: finding the median without putting the data in order first.
- Common error: from a frequency table, giving the highest frequency as the mode, instead of the value that has that frequency.
- Common error: dividing \(\sum fx\) by the number of rows in the table instead of by \(\sum f\).
- Common error: using the class boundaries or the class width instead of the midpoint when estimating a mean.
- Common error: stating the range as two numbers (such as "4 to 12"). The range is a single number: largest minus smallest.
Exam tip: When estimating a mean from grouped data, include a column for the midpoint and a column for midpoint \(\times\) frequency. Write the totals of both columns, and say that the answer is an estimate.
Check yourself
- Four numbers are 12, 15, 9 and 14. A fifth number is added and the mean of all five is 12. Find the fifth number.
Answer
The total of five numbers is \(5 \times 12 = 60\). The first four add to \(12 + 15 + 9 + 14 = 50\). The fifth number is \(60 - 50 = 10\). - Find the median of 4, 9, 6, 11, 5, 8.
Answer
In order: 4, 5, 6, 8, 9, 11. There are 6 values, so the median is halfway between the 3rd and 4th: \(\dfrac{6 + 8}{2} = 7\). - A shop sells shoes of sizes 5, 6, 7, 8 with frequencies 3, 10, 14, 5. Which average should the shop use to decide the most common size to order, and what is it?
Answer
The mode, because it shows the most popular size. The highest frequency is 14, so the mode is size 7. - Two groups of students have height ranges of 12 cm and 31 cm. What does this tell you?
Answer
The heights in the first group are closer together (less spread out) than in the second group.