GCSE Maths: Averages from grouped frequency tables (Foundation)
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Averages from grouped frequency tables
When data covers a wide range (heights, times, masses), it is put into classes (groups) in a grouped frequency table. For example, 12 students with heights between 150 cm and 160 cm are all in one class. The table tells us how many values are in each class, but we lose the exact values.
Because we no longer know the exact values, we cannot work out the exact mean, median or range. We can find: the modal class (the class with the highest frequency), the class that contains the median, and an estimate of the mean. To estimate the mean, we pretend that every value in a class is at the midpoint of the class. The midpoint is halfway between the two ends: add them and divide by 2. Then use the same method as for a frequency table: multiply the frequency by the midpoint, add up, and divide by the total frequency.
Median class. Add up the frequencies to find n, work out the position of the middle value (n + 1) ÷ 2, then add a running total of the frequencies. The median class is the class where the running total first reaches that position.
Key ideas
- The modal class is the class with the highest frequency (when the classes have equal width).
- Midpoint of a class = (lower end + upper end) ÷ 2.
- Estimated mean = total of (frequency × midpoint) ÷ total frequency.
- The median class contains the value at position (n + 1) ÷ 2: use a running total of frequencies.
- Grouped data gives only estimates, because the exact values are not known.
| Height | Frequency |
|---|---|
| 140 < h ≤ 150 cm | 4 |
| 150 < h ≤ 160 cm | 10 |
| 160 < h ≤ 170 cm | 14 |
| 170 < h ≤ 180 cm | 8 |
| 180 < h ≤ 190 cm | 4 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(140 + 150) ÷ 2 = 145
(150 + 160) ÷ 2 = 155
(160 + 170) ÷ 2 = 165
(170 + 180) ÷ 2 = 175
(180 + 190) ÷ 2 = 185 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Height Frequency (f) Midpoint f × midpoint 140 < h ≤ 150 cm 4 145 580 150 < h ≤ 160 cm 10 155 1550 160 < h ≤ 170 cm 14 165 2310 170 < h ≤ 180 cm 8 175 1400 180 < h ≤ 190 cm 4 185 740 Total 40 6580 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 40
Total of f × midpoint = 6580 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.6580 ÷ 40 = 164.5 cm
Check. The estimate should be inside the range of the classes, from 140 to 190, and 164.5 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
(a) Modal class.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The modal class is the class with the highest frequency.
- Look down the frequency column for the biggest number. The class that has the most values is the modal class.Frequencies: 4, 10, 14, 8, 4. The highest is 14.
- Read the class that goes with it. The answer is the class, not the frequency.Modal class = 160 < h ≤ 170 cm
Check. No other class has as many as 14 values. ✓
Watch out. Give the class (the group), not the frequency.
(b) Median class.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The median class is the class that contains the middle value.
- Find the total frequency. This is the number of values.4 + 10 + 14 + 8 + 4 = 40
- Find the position of the middle value. Middle position = (n + 1) ÷ 2.(40 + 1) ÷ 2 = 20.5
- Add a running total column. The running total shows how many values are in each class and all the classes before it.
Class Frequency Running total 140 < h ≤ 150 cm 4 4 150 < h ≤ 160 cm 10 14 160 < h ≤ 170 cm 14 28 170 < h ≤ 180 cm 8 36 180 < h ≤ 190 cm 4 40 - Find the class where the running total first reaches that position. Positions 15 to 28 are in this class, and 20.5 is one of them.Median class = 160 < h ≤ 170 cm
Check. Before this class there are 14 values, and 20.5 is more than that, but not more than 28. ✓
Watch out. The answer is the class, not a number. We cannot find the exact median from grouped data.
| Texts | Frequency |
|---|---|
| 0–4 | 3 |
| 5–9 | 8 |
| 10–14 | 12 |
| 15–19 | 5 |
| 20–24 | 2 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(0 + 4) ÷ 2 = 2
(5 + 9) ÷ 2 = 7
(10 + 14) ÷ 2 = 12
(15 + 19) ÷ 2 = 17
(20 + 24) ÷ 2 = 22 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Texts Frequency (f) Midpoint f × midpoint 0–4 3 2 6 5–9 8 7 56 10–14 12 12 144 15–19 5 17 85 20–24 2 22 44 Total 30 335 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 30
Total of f × midpoint = 335 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.335 ÷ 30 = 11.1667… = 11.2 to 1 d.p.
Check. The estimate should be inside the range of the classes, from 0 to 24, and 11.2 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Think about what information is missing when data is grouped.
- Recall what grouped data tells us. We know how many heights are in each class, but not where in the class each height is.For example, the 14 students in 160 < h ≤ 170 could be anywhere from just above 160 to 170.
- Recall what we did to find the mean. We used the midpoint of each class, which is a guess.The midpoint is an assumption.
- Conclude. The answer depends on the assumption, so it is an estimate, not the exact mean.Alex is wrong.
Check. If all 14 students in a class were at the bottom of the class, the true mean would be lower. ✓
Watch out. Grouped data gives an estimate. Say “estimate” in your answer.
Watch out
For grouped data multiply the frequency by the midpoint of the class, and divide by the total frequency. Do not use the class limits, and do not divide by the number of classes. The answer is only an estimate.Exercise 24E Averages from grouped frequency tables
Calculator allowed| Class | Frequency |
|---|---|
| 0 < x ≤ 10 | 5 |
| 10 < x ≤ 20 | 9 |
| 20 < x ≤ 30 | 4 |
| 30 < x ≤ 40 | 2 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The modal class is the class with the highest frequency.
- Look down the frequency column for the biggest number. The class that has the most values is the modal class.Frequencies: 5, 9, 4, 2. The highest is 9.
- Read the class that goes with it. The answer is the class, not the frequency.Modal class = 10 < x ≤ 20
Check. No other class has as many as 9 values. ✓
Watch out. Give the class (the group), not the frequency.
| Class | Frequency |
|---|---|
| 10 < x ≤ 20 | 6 |
| 20 < x ≤ 30 | 12 |
| 30 < x ≤ 40 | 8 |
| 40 < x ≤ 50 | 4 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The modal class is the class with the highest frequency.
- Look down the frequency column for the biggest number. The class that has the most values is the modal class.Frequencies: 6, 12, 8, 4. The highest is 12.
- Read the class that goes with it. The answer is the class, not the frequency.Modal class = 20 < x ≤ 30
Check. No other class has as many as 12 values. ✓
Watch out. Give the class (the group), not the frequency.
| Class | Frequency |
|---|---|
| 100 < x ≤ 120 | 8 |
| 120 < x ≤ 140 | 10 |
| 140 < x ≤ 160 | 15 |
| 160 < x ≤ 180 | 7 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The modal class is the class with the highest frequency.
- Look down the frequency column for the biggest number. The class that has the most values is the modal class.Frequencies: 8, 10, 15, 7. The highest is 15.
- Read the class that goes with it. The answer is the class, not the frequency.Modal class = 140 < x ≤ 160
Check. No other class has as many as 15 values. ✓
Watch out. Give the class (the group), not the frequency.
| Class | Frequency |
|---|---|
| 0 < x ≤ 10 | 5 |
| 10 < x ≤ 20 | 9 |
| 20 < x ≤ 30 | 4 |
| 30 < x ≤ 40 | 2 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The median class is the class that contains the middle value.
- Find the total frequency. This is the number of values.5 + 9 + 4 + 2 = 20
- Find the position of the middle value. Middle position = (n + 1) ÷ 2.(20 + 1) ÷ 2 = 10.5
- Add a running total column. The running total shows how many values are in each class and all the classes before it.
Class Frequency Running total 0 < x ≤ 10 5 5 10 < x ≤ 20 9 14 20 < x ≤ 30 4 18 30 < x ≤ 40 2 20 - Find the class where the running total first reaches that position. Positions 6 to 14 are in this class, and 10.5 is one of them.Median class = 10 < x ≤ 20
Check. Before this class there are 5 values, and 10.5 is more than that, but not more than 14. ✓
Watch out. The answer is the class, not a number. We cannot find the exact median from grouped data.
| Class | Frequency |
|---|---|
| 10 < x ≤ 20 | 6 |
| 20 < x ≤ 30 | 12 |
| 30 < x ≤ 40 | 8 |
| 40 < x ≤ 50 | 4 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The median class is the class that contains the middle value.
- Find the total frequency. This is the number of values.6 + 12 + 8 + 4 = 30
- Find the position of the middle value. Middle position = (n + 1) ÷ 2.(30 + 1) ÷ 2 = 15.5
- Add a running total column. The running total shows how many values are in each class and all the classes before it.
Class Frequency Running total 10 < x ≤ 20 6 6 20 < x ≤ 30 12 18 30 < x ≤ 40 8 26 40 < x ≤ 50 4 30 - Find the class where the running total first reaches that position. Positions 7 to 18 are in this class, and 15.5 is one of them.Median class = 20 < x ≤ 30
Check. Before this class there are 6 values, and 15.5 is more than that, but not more than 18. ✓
Watch out. The answer is the class, not a number. We cannot find the exact median from grouped data.
| Class | Frequency |
|---|---|
| 100 < x ≤ 120 | 8 |
| 120 < x ≤ 140 | 10 |
| 140 < x ≤ 160 | 15 |
| 160 < x ≤ 180 | 7 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The median class is the class that contains the middle value.
- Find the total frequency. This is the number of values.8 + 10 + 15 + 7 = 40
- Find the position of the middle value. Middle position = (n + 1) ÷ 2.(40 + 1) ÷ 2 = 20.5
- Add a running total column. The running total shows how many values are in each class and all the classes before it.
Class Frequency Running total 100 < x ≤ 120 8 8 120 < x ≤ 140 10 18 140 < x ≤ 160 15 33 160 < x ≤ 180 7 40 - Find the class where the running total first reaches that position. Positions 19 to 33 are in this class, and 20.5 is one of them.Median class = 140 < x ≤ 160
Check. Before this class there are 18 values, and 20.5 is more than that, but not more than 33. ✓
Watch out. The answer is the class, not a number. We cannot find the exact median from grouped data.
| Class | Frequency |
|---|---|
| 0 < x ≤ 10 | 5 |
| 10 < x ≤ 20 | 9 |
| 20 < x ≤ 30 | 4 |
| 30 < x ≤ 40 | 2 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(0 + 10) ÷ 2 = 5
(10 + 20) ÷ 2 = 15
(20 + 30) ÷ 2 = 25
(30 + 40) ÷ 2 = 35 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Class Frequency (f) Midpoint f × midpoint 0 < x ≤ 10 5 5 25 10 < x ≤ 20 9 15 135 20 < x ≤ 30 4 25 100 30 < x ≤ 40 2 35 70 Total 20 330 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 20
Total of f × midpoint = 330 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.330 ÷ 20 = 16.5
Check. The estimate should be inside the range of the classes, from 0 to 40, and 16.5 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
| Class | Frequency |
|---|---|
| 10 < x ≤ 20 | 6 |
| 20 < x ≤ 30 | 12 |
| 30 < x ≤ 40 | 8 |
| 40 < x ≤ 50 | 4 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(10 + 20) ÷ 2 = 15
(20 + 30) ÷ 2 = 25
(30 + 40) ÷ 2 = 35
(40 + 50) ÷ 2 = 45 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Class Frequency (f) Midpoint f × midpoint 10 < x ≤ 20 6 15 90 20 < x ≤ 30 12 25 300 30 < x ≤ 40 8 35 280 40 < x ≤ 50 4 45 180 Total 30 850 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 30
Total of f × midpoint = 850 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.850 ÷ 30 = 28.3333… = 28.3 to 1 d.p.
Check. The estimate should be inside the range of the classes, from 10 to 50, and 28.3 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
| Class | Frequency |
|---|---|
| 100 < x ≤ 120 | 8 |
| 120 < x ≤ 140 | 10 |
| 140 < x ≤ 160 | 15 |
| 160 < x ≤ 180 | 7 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(100 + 120) ÷ 2 = 110
(120 + 140) ÷ 2 = 130
(140 + 160) ÷ 2 = 150
(160 + 180) ÷ 2 = 170 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Class Frequency (f) Midpoint f × midpoint 100 < x ≤ 120 8 110 880 120 < x ≤ 140 10 130 1300 140 < x ≤ 160 15 150 2250 160 < x ≤ 180 7 170 1190 Total 40 5620 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 40
Total of f × midpoint = 5620 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.5620 ÷ 40 = 140.5
Check. The estimate should be inside the range of the classes, from 100 to 180, and 140.5 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
| Waiting time | Frequency |
|---|---|
| 0 < t ≤ 10 min | 3 |
| 10 < t ≤ 20 min | 8 |
| 20 < t ≤ 30 min | 14 |
| 30 < t ≤ 40 min | 10 |
| 40 < t ≤ 50 min | 5 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The modal class is the class with the highest frequency.
- Look down the frequency column for the biggest number. The class that has the most values is the modal class.Frequencies: 3, 8, 14, 10, 5. The highest is 14.
- Read the class that goes with it. The answer is the class, not the frequency.Modal class = 20 < t ≤ 30 min
Check. No other class has as many as 14 values. ✓
Watch out. Give the class (the group), not the frequency.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The median class is the class that contains the middle value.
- Find the total frequency. This is the number of values.3 + 8 + 14 + 10 + 5 = 40
- Find the position of the middle value. Middle position = (n + 1) ÷ 2.(40 + 1) ÷ 2 = 20.5
- Add a running total column. The running total shows how many values are in each class and all the classes before it.
Class Frequency Running total 0 < t ≤ 10 min 3 3 10 < t ≤ 20 min 8 11 20 < t ≤ 30 min 14 25 30 < t ≤ 40 min 10 35 40 < t ≤ 50 min 5 40 - Find the class where the running total first reaches that position. Positions 12 to 25 are in this class, and 20.5 is one of them.Median class = 20 < t ≤ 30 min
Check. Before this class there are 11 values, and 20.5 is more than that, but not more than 25. ✓
Watch out. The answer is the class, not a number. We cannot find the exact median from grouped data.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(0 + 10) ÷ 2 = 5
(10 + 20) ÷ 2 = 15
(20 + 30) ÷ 2 = 25
(30 + 40) ÷ 2 = 35
(40 + 50) ÷ 2 = 45 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Waiting time Frequency (f) Midpoint f × midpoint 0 < t ≤ 10 min 3 5 15 10 < t ≤ 20 min 8 15 120 20 < t ≤ 30 min 14 25 350 30 < t ≤ 40 min 10 35 350 40 < t ≤ 50 min 5 45 225 Total 40 1060 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 40
Total of f × midpoint = 1060 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.1060 ÷ 40 = 26.5 min
Check. The estimate should be inside the range of the classes, from 0 to 50, and 26.5 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
| Goals | Frequency |
|---|---|
| 0–4 | 6 |
| 5–9 | 9 |
| 10–14 | 7 |
| 15–19 | 3 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(0 + 4) ÷ 2 = 2
(5 + 9) ÷ 2 = 7
(10 + 14) ÷ 2 = 12
(15 + 19) ÷ 2 = 17 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Goals Frequency (f) Midpoint f × midpoint 0–4 6 2 12 5–9 9 7 63 10–14 7 12 84 15–19 3 17 51 Total 25 210 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 25
Total of f × midpoint = 210 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.210 ÷ 25 = 8.4
Check. The estimate should be inside the range of the classes, from 0 to 19, and 8.4 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
| Length | Frequency |
|---|---|
| 0 < t ≤ 10 min | 6 |
| 10 < t ≤ 30 min | 10 |
| 30 < t ≤ 60 min | 4 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(0 + 10) ÷ 2 = 5
(10 + 30) ÷ 2 = 20
(30 + 60) ÷ 2 = 45 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Length Frequency (f) Midpoint f × midpoint 0 < t ≤ 10 min 6 5 30 10 < t ≤ 30 min 10 20 200 30 < t ≤ 60 min 4 45 180 Total 20 410 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 20
Total of f × midpoint = 410 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.410 ÷ 20 = 20.5 min
Check. The estimate should be inside the range of the classes, from 0 to 60, and 20.5 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.
Method. To find the mean, add up all the values and divide by how many values there are.
- Add up all the values. The mean uses the total of all the data.12 + 18 + 21 + 25 + 27 + 31 + 34 + 36 + 41 + 45 = 290
- Count how many values there are. The mean shares the total equally between all the values, so we need to know how many there are.There are 10 values.
- Divide the total by the number of values. mean = total ÷ number of values.290 ÷ 10 = 29
Check. Multiply back: 29 × 10 = 290, which is the total. ✓
Watch out. The mean is not the middle value. Add every value, and divide by how many values there are (not by the largest value).
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(10 + 20) ÷ 2 = 15
(20 + 30) ÷ 2 = 25
(30 + 40) ÷ 2 = 35
(40 + 50) ÷ 2 = 45 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Age Frequency (f) Midpoint f × midpoint 10 ≤ a < 20 years 2 15 30 20 ≤ a < 30 years 3 25 75 30 ≤ a < 40 years 3 35 105 40 ≤ a < 50 years 2 45 90 Total 10 300 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 10
Total of f × midpoint = 300 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.300 ÷ 10 = 30
Check. The estimate should be inside the range of the classes, from 10 to 50, and 30 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Put the two answers side by side.
- Write both answers. Exact mean = 29
Estimate = 30 - Find the difference. The estimate uses midpoints rather than the real ages.1 apart
- Comment. The estimate is close, but not equal, because grouping loses the exact values.The estimate is close to the true mean but not the same.
Check. The estimate is within a few years of the true mean. ✓
Watch out. The estimate can be higher or lower than the true mean.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Think about what a grouped table tells us and what it hides.
- What does the table show? It shows only how many values are in each class.For example, 14 values are between 160 and 170.
- What is missing? We cannot tell where in the class each value lies, so we cannot say which value is exactly in the middle or exactly the biggest or smallest.The exact values are lost.
Check. We can only say which class the median is in. ✓
Watch out. Use “estimate” for the mean, and “class” for the median.
| Mass | Frequency |
|---|---|
| 0 < m ≤ 20 g | 8 |
| 20 < m ≤ 40 g | 18 |
| 40 < m ≤ 60 g | 14 |
| 60 < m ≤ 80 g | 10 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The median class is the class that contains the middle value.
- Find the total frequency. This is the number of values.8 + 18 + 14 + 10 = 50
- Find the position of the middle value. Middle position = (n + 1) ÷ 2.(50 + 1) ÷ 2 = 25.5
- Add a running total column. The running total shows how many values are in each class and all the classes before it.
Class Frequency Running total 0 < m ≤ 20 g 8 8 20 < m ≤ 40 g 18 26 40 < m ≤ 60 g 14 40 60 < m ≤ 80 g 10 50 - Find the class where the running total first reaches that position. Positions 9 to 26 are in this class, and 25.5 is one of them.Median class = 20 < m ≤ 40 g
Check. Before this class there are 8 values, and 25.5 is more than that, but not more than 26. ✓
Watch out. The answer is the class, not a number. We cannot find the exact median from grouped data.
Method. “More than 40 g” means the classes 40 < m ≤ 60 and 60 < m ≤ 80.
- Pick the classes above 40 g. 40 < m ≤ 60: 14
60 < m ≤ 80: 10 - Add. 14 + 10 = 24
Check. Parcels up to 40 g: 8 + 18 = 26, and 26 + 24 = 50. ✓
Watch out. Read the class boundaries carefully.
Method. Find the number in the first class, write it as a fraction of 50, then as a percentage.
- Find the number of parcels that weigh up to 20 g. This is the first class.8
- Write as a fraction of 50. 850
- Convert to a percentage. Multiply by 2, so that the denominator is 100.850 = 16100 = 16%
Check. 16% of 50 = 0.16 × 50 = 8. ✓
Watch out. The whole is the total of all the frequencies (50).
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(0 + 20) ÷ 2 = 10
(20 + 40) ÷ 2 = 30
(40 + 60) ÷ 2 = 50 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Class Frequency (f) Midpoint f × midpoint 0 < x ≤ 20 8 10 80 20 < x ≤ 40 14 30 420 40 < x ≤ 60 18 50 900 Total 40 1400 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 40
Total of f × midpoint = 1400 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.1400 ÷ 40 = 35.0
Check. The estimate should be inside the range of the classes, from 0 to 60, and 35.0 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
| Time | Frequency |
|---|---|
| 40 < t ≤ 50 s | 3 |
| 50 < t ≤ 60 s | 9 |
| 60 < t ≤ 70 s | 12 |
| 70 < t ≤ 80 s | 6 |
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The modal class is the class with the highest frequency.
- Look down the frequency column for the biggest number. The class that has the most values is the modal class.Frequencies: 3, 9, 12, 6. The highest is 12.
- Read the class that goes with it. The answer is the class, not the frequency.Modal class = 60 < t ≤ 70 s
Check. No other class has as many as 12 values. ✓
Watch out. Give the class (the group), not the frequency.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The median class is the class that contains the middle value.
- Find the total frequency. This is the number of values.3 + 9 + 12 + 6 = 30
- Find the position of the middle value. Middle position = (n + 1) ÷ 2.(30 + 1) ÷ 2 = 15.5
- Add a running total column. The running total shows how many values are in each class and all the classes before it.
Class Frequency Running total 40 < t ≤ 50 s 3 3 50 < t ≤ 60 s 9 12 60 < t ≤ 70 s 12 24 70 < t ≤ 80 s 6 30 - Find the class where the running total first reaches that position. Positions 13 to 24 are in this class, and 15.5 is one of them.Median class = 60 < t ≤ 70 s
Check. Before this class there are 12 values, and 15.5 is more than that, but not more than 24. ✓
Watch out. The answer is the class, not a number. We cannot find the exact median from grouped data.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.
- Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.(40 + 50) ÷ 2 = 45
(50 + 60) ÷ 2 = 55
(60 + 70) ÷ 2 = 65
(70 + 80) ÷ 2 = 75 - Multiply each frequency by its midpoint. This estimates the total of the values in that class.
Time Frequency (f) Midpoint f × midpoint 40 < t ≤ 50 s 3 45 135 50 < t ≤ 60 s 9 55 495 60 < t ≤ 70 s 12 65 780 70 < t ≤ 80 s 6 75 450 Total 30 1860 - Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.Total frequency = 30
Total of f × midpoint = 1860 - Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.1860 ÷ 30 = 62 s
Check. The estimate should be inside the range of the classes, from 40 to 80, and 62 is. It is only an estimate because the real values are not known. ✓
Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. The total of all the values is the mean multiplied by the number of values.
- Recall the estimated mean. From the last question.Estimated mean = 62 s
- Multiply by the number of runners. total = mean × number of values.62 × 30 = 1860 seconds
Check. Dividing back: 1860 ÷ 30 = 62. ✓
Watch out. This total is also only an estimate, because the mean was an estimate.
Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.
Method. Think about what information is still missing when data is grouped.
- Recall the method. We assume every value is at the midpoint of its class.The midpoint is an assumption.
- Think about narrow classes. With narrow classes the midpoint is closer to the real values, so the estimate is better.The error gets smaller.
- Conclude. The exact values are still unknown, so the answer is still an estimate.Lee is wrong.
Check. Only the original list of data gives the exact mean. ✓
Watch out. Say “estimate” for the mean of grouped data.