Grouped Data and the Estimated Mean

1 · Why group data?

Here are the times, in minutes, that 40 pupils spent on their phone before going to sleep. Time is continuous data (it is measured, so it could be 27 minutes or 27.4 minutes), and nearly every value is different. A table with one row per value would be as long as the list.

Each group is a class interval (or just class). Its width is upper end − lower end. Choosing the width is a trade-off: narrow classes keep more detail but give a long, bumpy table; wide classes give a short table but hide the shape. Between 4 and 8 classes is usually sensible.

2 · Class intervals: ≤ and <

Every value must belong to exactly one class, so the class boundaries are written with inequalities. The class 20 ≤ t < 30 reads "t is at least 20 and less than 30": 20 is in it, 30 is not (30 starts the next class). Some questions write classes the other way round, 20 < t ≤ 30: now 20 is not in it but 30 is. Always read the signs; the value on the ≤ side is the one that's included.

With discrete (counted) data, classes are written like 0–4, 5–9, 10–14. There are no gaps to worry about, because a value like 4.5 can't happen.

3 · The modal class and the class containing the median

Once data is grouped, the exact values are hidden. We can't give the mode or the median as a number, but we can say which class they are in.

• The modal class is the class with the highest frequency. Answer with the class, e.g. "20 ≤ t < 30", never with the frequency 12.
• For the class containing the median, find the median position (n + 1) ÷ 2, then add up the frequencies as you go down the table (the running total, also called cumulative frequency) until you pass that position.

Try it

4 · Estimating the mean with midpoints

For an ungrouped table the mean is total of f × x ÷ total of f. In a grouped table there is no single x for each row. We know 12 pupils spent between 20 and 30 minutes, but not exactly how long. The fair guess is that, on average, they sat in the middle of the class. So we pretend every value is at the class midpoint:

midpoint = (lower end + upper end) ÷ 2

and then do exactly what you did for an ungrouped table, with the midpoint in place of x:

estimated mean = total of f × midpoint ÷ total of f

5 · Why it's only an estimate

The real 40 times are shown below as teal dots, sorted into their classes. The estimated mean uses only the table, so it can't see where the dots really sit inside each class. Slide every value to the bottom or top of its class, as if the data were different: the table doesn't change, so the estimate doesn't move, but the true mean does.

With the real data, the estimate (28.25 minutes) is very close to the true mean (1132 ÷ 40 = 28.3 minutes), because within each class some values sit below the midpoint and some above, and they roughly cancel out. But the table alone can't guarantee that. All it can promise is that the true mean lies somewhere between the "all at the bottom" mean and the "all at the top" mean. So in an exam, "explain why your answer is an estimate" earns its mark with: we don't know the exact values; we assumed every value in a class was at its midpoint.

6 · Estimating the range

The range is largest value − smallest value, but a grouped table hides both. The smallest value could be anywhere in the first class that has any data, and the largest anywhere in the last class with data. The usual estimate is the biggest the range could be:

estimated range = upper end of the last class − lower end of the first class

For the phone data that is 60 − 0 = 60 minutes. The true range is 58 − 3 = 55 minutes. Trap: ignore classes with frequency 0 at either end: if nobody is in them, the data doesn't reach them.

7 · Frequency polygons

A frequency polygon is a quick picture of a grouped table. For each class, plot a point at (midpoint, frequency), then join the points in order with straight lines. You don't join the last point back to the first, and you don't draw bars. Why the midpoint? It's the same assumption as the estimated mean: the point stands for the whole class, so it goes in the middle of it. If you draw the bar chart too, the polygon passes through the middle of the top of each bar.

Tap the grid to plot the six points for the phone-time table (tap a point again to remove it).

Frequency polygons are best for comparing two distributions: draw two on the same axes and they don't hide each other the way two bar charts would. That's on the comparing distributions page.

8 · Practice: estimate the mean and the range

Every question is new: continuous classes written both ways, discrete groups and unequal widths. If you get it wrong, the feedback tries to spot which mistake you made.

Worked examples

Quick check

What you've learned

• Continuous data is grouped into class intervals. In a ≤ x < b, a is included and b is not; in a < x ≤ b it's the other way round. Discrete groups look like 0–4, 5–9.
• The modal class has the highest frequency. The class containing the median is found with position (n + 1) ÷ 2 and running totals. Answer with a class, not a number of people.
• Midpoint = (lower end + upper end) ÷ 2, also for discrete groups (1–10 → 5.5) and unequal widths.
• Estimated mean = total of f × midpoint ÷ total of f. Never divide by the number of classes.
• It's an estimate because the exact values are hidden; we assume each value is at its class midpoint. The true mean lies between the mean using lower ends and the mean using upper ends.
• Estimated range = upper end of the last class − lower end of the first class, ignoring empty end classes.
• A frequency polygon plots (midpoint, frequency) and joins the points with straight lines.