Guide · For SS 1 – SS 3 students · 7 min read
Statistics for SS 2: from frequency tables to standard deviation
Frequency tables, mean, median and mode, grouped data, the ogive, quartiles and standard deviation, all worked through with the same two sets of class results so you can see how each idea builds on the last.
By the MathPlug team, Reflex System · Published
Statistics is one of the friendliest topics in senior secondary maths. The methods follow the same pattern every time, the arithmetic is mostly adding and multiplying, and once you can build a good table, most of the marks are already yours.
In this guide we use two small sets of class results and carry them through every section: first a frequency table of quiz scores, then a grouped table of test marks. You will see the mean, median, mode, ogive, quartiles and standard deviation come out of the same data, one column at a time.
Start with a frequency table
Ada's class of 20 students wrote a quiz marked out of 10. Instead of listing all 20 scores, we count how many students got each score. That count is the frequency, written f. The score itself is x.
| Score (x) | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|
| Frequency (f) | 2 | 3 | 3 | 2 | 6 | 3 | 1 |
Always check the total first: 2 + 3 + 3 + 2 + 6 + 3 + 1 = 20. If Σf (the sum of the frequencies) does not match the number of students, something has been miscounted, and every later answer will be wrong.
💡 Tip: When you make the table yourself from a list of raw data, use tally marks and cross each number off the list as you count it. It stops you counting a value twice.
Mean, median and mode from the table
The mean is the total of all the scores divided by the number of students. In a frequency table, each score x appears f times, so the total is found by multiplying: x̄ = ΣfxΣf.
Find the mean, median and mode of the quiz scores.
- Multiply each score by its frequency: fx = 4, 9, 12, 10, 36, 21, 8.
- Σfx = 4 + 9 + 12 + 10 + 36 + 21 + 8 = 100
- Mean = ΣfxΣf = 10020 = 5
- Mode: the score with the highest frequency is 6 (6 students), so the mode is 6.
- Median: with 20 scores, the middle is between the 10th and 11th. Running totals of f: 2, 5, 8, 10, 16, 19, 20.
- The 10th score is 5 (the running total reaches 10 at x = 5). The 11th score is 6 (scores 11 to 16 are all 6).
- Median = 5 + 62 = 5.5
Notice the three answers are different. That is normal. The mean uses every value, the median is the middle value, and the mode is simply the most common. Questions often ask for all three, so set out your table with an fx column from the start.
Grouped data and class midpoints
When marks are spread over a wide range, we group them into classes. Tunde's teacher recorded test marks out of 50 for 40 students:
| Marks | Frequency (f) | Midpoint (x) | fx |
|---|---|---|---|
| 1 – 10 | 4 | 5.5 | 22 |
| 11 – 20 | 8 | 15.5 | 124 |
| 21 – 30 | 14 | 25.5 | 357 |
| 31 – 40 | 10 | 35.5 | 355 |
| 41 – 50 | 4 | 45.5 | 182 |
| Total | 40 | 1040 |
We no longer know each exact mark, so we use the class midpoint to stand for every mark in that class. The midpoint of 11 – 20 is 11 + 202 = 15.5. Then the method is exactly the same as before:
Find the mean mark of the grouped data.
- Σf = 4 + 8 + 14 + 10 + 4 = 40
- Σfx = 22 + 124 + 357 + 355 + 182 = 1040
- Mean = 104040 = 26 marks
- The modal class (the class with the highest frequency) is 21 – 30.
Because midpoints are used, the mean of grouped data is an estimate. That is fine; it is what the question expects.
Cumulative frequency and the ogive
A cumulative frequency is a running total. It tells you how many students scored up to the end of each class. To plot it, we use the upper class boundary, which is halfway between one class and the next: 10.5, 20.5, 30.5 and so on.
| Marks | Upper class boundary | f | Cumulative frequency |
|---|---|---|---|
| 1 – 10 | 10.5 | 4 | 4 |
| 11 – 20 | 20.5 | 8 | 12 |
| 21 – 30 | 30.5 | 14 | 26 |
| 31 – 40 | 40.5 | 10 | 36 |
| 41 – 50 | 50.5 | 4 | 40 |
Plot each cumulative frequency against its upper class boundary, start the curve at (0.5, 0), and join the points with a smooth curve. This S-shaped graph is the ogive. Label both axes and use a sensible scale; examiners give marks for a neat, correct graph.
Use the ogive to estimate the median, the quartiles and the interquartile range.
- With 40 students: median at 12 × 40 = 20th, lower quartile Q₁ at 14 × 40 = 10th, upper quartile Q₃ at 34 × 40 = 30th.
- From 20 on the cumulative frequency axis, go across to the curve and down: the 20th student lies between 20.5 (cf 12) and 30.5 (cf 26). Reading by proportion: 20.5 + 814 × 10 ≈ 20.5 + 5.7 = 26.2.
- Q₁: the 10th student lies between 10.5 (cf 4) and 20.5 (cf 12): 10.5 + 68 × 10 = 10.5 + 7.5 = 18.0.
- Q₃: the 30th student lies between 30.5 (cf 26) and 40.5 (cf 36): 30.5 + 410 × 10 = 30.5 + 4 = 34.5.
- Interquartile range = Q₃ − Q₁ = 34.5 − 18.0 = 16.5 marks
💡 Tip: Values read from a hand-drawn curve will be slightly different, perhaps 26 instead of 26.2. That is accepted. What matters is drawing your reading lines clearly on the graph so the examiner can see where your answer came from.
Measuring spread: range and interquartile range
An average tells you the centre. A measure of spread tells you how far the data stretches. Two classes can have the same mean and very different spreads.
- Range = highest value − lowest value. For the quiz: 8 − 2 = 6. It is quick, but one unusual score can make it misleading.
- Interquartile range = Q₃ − Q₁. It measures the spread of the middle half of the data, so it ignores extreme values. For the test marks it was 16.5.
- Semi-interquartile range is half of the interquartile range: 16.5 ÷ 2 = 8.25. Some questions ask for this instead, so read carefully.
Standard deviation with a full table
Standard deviation measures how far, on average, the values sit from the mean. A small standard deviation means the scores are bunched together; a large one means they are spread out. The method is a table with a few extra columns. Let d = x − x̄ (the deviation from the mean). Then SD = √(Σfd2 ÷ Σf).
| x | f | d = x − 5 | d2 | fd2 |
|---|---|---|---|---|
| 2 | 2 | −3 | 9 | 18 |
| 3 | 3 | −2 | 4 | 12 |
| 4 | 3 | −1 | 1 | 3 |
| 5 | 2 | 0 | 0 | 0 |
| 6 | 6 | 1 | 1 | 6 |
| 7 | 3 | 2 | 4 | 12 |
| 8 | 1 | 3 | 9 | 9 |
| Total | 20 | 60 |
Calculate the standard deviation of the quiz scores.
- Σfd2 = 18 + 12 + 3 + 0 + 6 + 12 + 9 = 60
- Variance = Σfd2Σf = 6020 = 3
- Standard deviation = √3 ≈ 1.73 (to 2 decimal places)
- Find the mean first, and keep it exact.
- Square every deviation, so negatives become positive.
- Multiply by the frequency before adding. Forgetting f is the most common slip.
- Divide by Σf, then take the square root last.
For grouped data, use the class midpoints as x and follow exactly the same steps. Our standard deviation formula page and the mean deviation and standard deviation lesson have more practice.
Questions students ask
- How do you find the mean from a frequency table?
- Multiply each value by its frequency to get fx, add the fx column to get Σfx, and divide by the total frequency Σf. For grouped data, use the class midpoints as the values.
- How do you find the median from an ogive?
- Find half the total frequency, go across from that point on the cumulative frequency axis to the curve, then straight down to the horizontal axis. The value you read is the median. Do the same at a quarter and three quarters of the total for the quartiles.
- What is the difference between variance and standard deviation?
- Variance is the average of the squared deviations from the mean, Σfd2 ÷ Σf. Standard deviation is the square root of the variance, so it is in the same units as the data.
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