Averages summarise a data set with one typical value, and the range and interquartile range describe how spread out it is. This note covers the mean, median, mode, range and quartiles for lists, frequency tables and grouped data, and reading and drawing common statistical diagrams. These skills are used whenever data must be summarised and compared.
Key ideas
Averages. The three averages answer "what is typical?" in different ways.
- Mean = total of all the values \(\div\) number of values. It uses every value, so one extreme value pulls it up or down.
- Median = the middle value when the data are in order. With \(n\) values it is in position \(\tfrac{n+1}{2}\). If there are two middle values, the median is halfway between them.
- Mode = the most common value. A set can have no mode, one mode or more than one. For grouped data we give the modal class, the class with the highest frequency (when the class widths are equal).
Spread. The range = largest value \(-\) smallest value. It depends only on the two extremes. The quartiles split ordered data into four equal parts: \(Q_1\) is the lower quartile, \(Q_2\) is the median and \(Q_3\) is the upper quartile. The interquartile range \(IQR=Q_3-Q_1\) measures the spread of the middle half of the data and is not affected by extreme values.
Frequency tables. For a table of values \(x\) with frequencies \(f\), the total frequency is \(\sum f\) and the mean is \(\dfrac{\sum fx}{\sum f}\). Add a column \(fx\) to your table. For the median, find the cumulative frequency and locate the \(\tfrac{n+1}{2}\)th value.
Grouped data. When the data are in classes, we do not know the exact values. To estimate the mean, use the midpoint of each class as its representative value: mean \(\approx \dfrac{\sum f\times\text{midpoint}}{\sum f}\). The answer is an estimate, and you should say so. You can only name the class that contains the median.
Statistical diagrams.
- Pictograms and bar charts compare categories. Check the key on a pictogram, because one symbol may stand for several items. Bars have gaps between them, and equal widths.
- Pie charts show proportions. The angle for a category is \(\dfrac{\text{frequency}}{\text{total}}\times360^\circ\).
- Stem and leaf diagrams keep the actual values. Put the leaves in order, write a key such as \(2\,|\,3\) means 23, and read the median by counting along the leaves.
- Frequency polygons are drawn by plotting each frequency against the midpoint of its class and joining the points with straight lines.
- Scatter graphs show the relationship between two variables. The pattern is called correlation: positive (both increase), negative (one increases as the other decreases) or none. A line of best fit passes close to the points, with roughly equal numbers on each side, and can be used to estimate one value from the other. Estimates within the data range (interpolation) are more reliable than those outside it (extrapolation). Correlation does not prove that one variable causes the other.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Mean \(=\dfrac{\sum x}{n}\) | List of \(n\) values |
| Mean \(=\dfrac{\sum fx}{\sum f}\) | Frequency table |
| Estimated mean \(=\dfrac{\sum f\times\text{midpoint}}{\sum f}\) | Grouped data |
| Median position \(\tfrac{n+1}{2}\) | Find the middle of ordered data |
| Range \(=\) largest \(-\) smallest | Simple measure of spread |
| \(IQR=Q_3-Q_1\) | Spread of the middle 50% of values |
| Pie chart angle \(=\dfrac{f}{\text{total}}\times360^\circ\) | Drawing and reading pie charts |
Worked examples
Example 1 A football club records the number of goals scored in each of 20 matches: 0 goals (3 matches), 1 goal (5), 2 goals (6), 3 goals (4), 4 goals (2). Find the mean, median, mode and range.
Total frequency \(=3+5+6+4+2=20\).
\(\sum fx=0\times3+1\times5+2\times6+3\times4+4\times2=0+5+12+12+8=37\).
Mean \(=37\div20=1.85\).
The median is halfway between the 10th and 11th values. The cumulative frequencies are 3, 8, 14, 18, 20. The 9th to 14th values are all 2, so the 10th and 11th are both 2, and the median is 2.
The mode is the value with the highest frequency, which is 2 (frequency 6). Range \(=4-0=4\).
Answer: mean 1.85 goals, median 2, mode 2, range 4.
Example 2 The waiting times, \(t\) minutes, of 50 patients at a clinic are: \(0\lt t\le10\): 8, \(10\lt t\le20\): 14, \(20\lt t\le30\): 18, \(30\lt t\le40\): 7, \(40\lt t\le50\): 3. Estimate the mean waiting time, and state the modal class and the class containing the median.
The midpoints are 5, 15, 25, 35 and 45.
\(\sum f\times\text{midpoint}=8\times5+14\times15+18\times25+7\times35+3\times45=40+210+450+245+135=1080\).
Estimated mean \(=1080\div50=21.6\) minutes.
The highest frequency is 18, so the modal class is \(20\lt t\le30\).
The median is between the 25th and 26th values. The cumulative frequencies are 8, 22, 40, 47, 50. The 25th and 26th values are in the third class.
Answer: estimated mean 21.6 minutes, modal class \(20\lt t\le30\), median in \(20\lt t\le30\).
Example 3 Nine delivery times in minutes are: 4, 7, 8, 10, 11, 13, 15, 18, 40. Find the mean, median, range and interquartile range. Then find the mean and median again without the value 40.
The data are already in order. Total \(=126\), so mean \(=126\div9=14\).
The median is the 5th value, which is 11. Range \(=40-4=36\).
Quartiles: the lower half (without the median) is 4, 7, 8, 10, with median \(\tfrac{7+8}{2}=7.5\), so \(Q_1=7.5\). The upper half is 13, 15, 18, 40, with median \(\tfrac{15+18}{2}=16.5\), so \(Q_3=16.5\). \(IQR=16.5-7.5=9\).
Without 40, there are 8 values with total 86. The mean is \(86\div8=10.75\). The median is halfway between the 4th and 5th values: \(\tfrac{10+11}{2}=10.5\).
Answer: mean 14, median 11, range 36, IQR 9. Without 40: mean 10.75 (3 s.f. 10.8), median 10.5. The mean changed a lot and the median hardly changed.
Common errors
- Finding the median of a list that is not in order. Always sort the data first.
- For a frequency table, taking the median or mode to be a frequency instead of a value. The mode is the value, not how often it occurs.
- Using \(\sum f\) as the number of different values (the number of rows) when finding the mean. The divisor is the total frequency.
- Using class boundaries instead of midpoints for a grouped mean, or presenting the estimate as an exact answer.
- Joining a frequency polygon to the class ends instead of the midpoints, or drawing a line of best fit that must pass through the first and last points.
Exam tip: When asked to compare two sets of data, give one comment on an average and one on the spread, each with numbers, and write the comment in the context of the question, for example "the buses are typically later but more consistent".
Check yourself
- Find the mean, median, mode and range of 6, 9, 9, 12, 14.
Answer
Mean \(=50\div5=10\). Median \(=9\) (the middle value). Mode \(=9\). Range \(=14-6=8\). - In a survey of 40 people, 14 travel by bus. What angle represents bus in a pie chart?
Answer
\(\tfrac{14}{40}\times360^\circ=126^\circ\). - The mean of 8 numbers is 12. A ninth number is added and the mean becomes 13. What is the ninth number?
Answer
The total of the 9 numbers is \(9\times13=117\). The total of the first 8 is \(8\times12=96\). The ninth number is \(117-96=21\). - A stem and leaf diagram has: 1 | 2 5 8; 2 | 0 3 3 7; 3 | 1 4 (key: 2 | 3 means 23). Find the median.
Answer
The values are 12, 15, 18, 20, 23, 23, 27, 31, 34. There are 9 values, so the median is the 5th, which is 23.