01 · Explicit interval
Find the midpoint of 12 ≤ t < 18 minutes.
Hint
Average the endpoints.
Worked solution
(12 + 18)/2 = 15 minutes.
Understand · explore · practise
Estimates from grouped data
Estimate a mean from class midpoints, locate a median class and recognise information lost by grouping. Compare possible original datasets with the same frequencies.
Before you startFrequency-table averages and class boundaries.
01 / Grouping hides the individual values
An interval is not a single measurement.
Grouped mean estimate = Σfm / Σf, where m is each class midpoint.
Replacing all values in a class by its midpoint is an approximation.
The selector shows three possible lists that produce the same grouped table. Their exact means differ, but their midpoint estimates agree. Grouped data alone cannot recover the individual measurements.
Same groups, different exact meansExplore
| Class | Frequency | Midpoint |
|---|---|---|
| 0 ≤ x < 10 | 2 | 5 |
| 10 ≤ x < 20 | 3 | 15 |
| 20 ≤ x < 30 | 1 | 25 |
Possible original values: 5, 5, 15, 15, 15, 25.
Exact mean for this possible list = 80/6 ≈ 13.333333.
The midpoint estimate from the grouped table stays 80/6 ≈ 13.333333 for every option. The table alone cannot tell which original list occurred.
02 / Find a representative class value
Use the actual interval boundaries.
For a bounded interval with endpoints L and U, midpoint m = (L + U)/2. The midpoint is halfway across the interval. It is not necessarily the average of the observations in that interval.
02 · Rounded classes
Lengths recorded to the nearest centimetre are grouped as 10–14 and 15–19 inclusive. Give the boundaries and midpoints.
Hint
Move half a centimetre outside each extreme recorded value.
Worked solution
First: 9.5 ≤ length < 14.5, midpoint 12. Second: 14.5 ≤ length < 19.5, midpoint 17.
03 / Weight the midpoints by frequency
A wide class is not automatically a heavy weight.
Classes 0–10, 10–20, 20–30 have frequencies 2, 3, 1; each includes its lower boundary only.Worked example
Midpoints: 5, 15, 25
Use interval centres.
Σfm = 2×5 + 3×15 + 1×25 = 80
Frequencies supply the weights.
Σf = 6; estimated mean = 80/6 ≈ 13.33
Say estimated because the exact values are hidden.
03 · Estimate a mean
The two rounded length classes in question 2 have frequencies 3 and 5. Estimate the mean length.
Hint
Use midpoints 12 and 17.
Worked solution
(3×12 + 5×17)/8 = 121/8 = 15.125 cm, an estimate.
04 · Unequal widths
Intervals 0 ≤ x < 10 and 10 ≤ x < 30 have frequencies 4 and 6. Estimate the mean.
Hint
Midpoints are 5 and 20; use frequencies 4 and 6.
Worked solution
(4×5 + 6×20)/10 = 14. The second class is wider, but the weights remain the observation counts.
04 / Why the answer is an estimate
Many original lists fit one table.
Watch: one grouped table can hide different means
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Midpoint replacement works exactly if the mean within every occupied class equals its midpoint. A uniform distribution within each class is one useful modelling assumption supporting the estimate, but the actual observations need not satisfy it.
05 · Two possible lists
A class 10 ≤ x < 20 contains two observations. Give two possible lists with different exact means but the same midpoint estimate.
Hint
Try both values near the bottom, then both near the top.
Worked solution
Possible lists are 10, 10 (exact mean 10) and 19, 19 (exact mean 19). Both give midpoint estimate 15 from the grouped table.
06 · Precision is not knowledge
Does writing an estimated mean to ten decimal places make it an exact mean?
Hint
The information lost by grouping remains lost.
Worked solution
No. More decimal places refine the arithmetic on midpoint substitutes, not knowledge of the original values.
05 / Locate the class containing the centre
Use cumulative frequencies and check middle ranks.
In the model, cumulative frequencies are 2, 5, 6. The third and fourth observations both fall in 10 ≤ x < 20, so the median lies in that interval. Estimating its position inside the class requires an additional assumption; the next interpolation lesson covers that.
If the two middle observations fall in different classes, their average need not lie in either class. State what is known instead of blindly reporting one row.
07 · Median class
Classes 0 ≤ x < 5, 5 ≤ x < 10, 10 ≤ x < 15 have frequencies 3, 8, 4. Locate the median class.
Hint
There are 15 values; find rank 8.
Worked solution
Cumulative counts are 3, 11, 15. Rank 8 is in 5 ≤ x < 10.
08 · Middle ranks split
Four observations occupy 0 ≤ x < 10 and four occupy 20 ≤ x < 30. Why can you not say the median is an observed value in the second class?
Hint
The fourth and fifth observations lie in different classes.
Worked solution
The median averages those two middle observations. That average is not necessarily an observed value and may lie in the gap. For middle values 8 and 22, the median is 15.
06 / Be explicit about modal class
Frequency and concentration are different ideas.
When asked for the modal class by frequency, identify the class containing the most observations. With unequal class widths, the class with the greatest frequency need not be the class with the greatest frequency density (frequency ÷ width). Histogram comparisons use density; do not confuse those two criteria.
09 · Highest frequency
In the model, which class has the highest frequency?
Hint
Compare 2, 3 and 1.
Worked solution
10 ≤ x < 20 contains three observations, the most of the three classes.
10 · Frequency versus density
Class A has width 5 and frequency 10; class B has width 20 and frequency 20. Compare their frequencies and frequency densities.
Hint
Density is count per unit width.
Worked solution
B has the greater frequency (20 versus 10). A has the greater density: 10/5 = 2 versus 20/20 = 1.
07 / Check measurement conventions and open classes
Not every label has the same boundaries.
Age in completed years is rounded down, not to the nearest year. A reported age range 20–29 corresponds to exact ages 20 ≤ age < 30, with midpoint 25. Open-ended classes have no defined finite midpoint unless extra information supplies a boundary.
11 · An open-ended class
Why can a class “50 minutes or more” prevent a standard midpoint estimate of the overall mean?
Hint
Its upper endpoint is missing.
Worked solution
There is no defined midpoint for that class. You need more information or an explicitly justified assumption; inventing a midpoint silently is not valid.
12 · Age versus nearest year
Compare the midpoint for ages 20–29 in completed years with a measurement recorded to the nearest year grouped as 20–29.
Hint
The recording conventions imply different boundaries.
Worked solution
Completed ages: [20,30), midpoint 25. Nearest-year measurements: [19.5,29.5), midpoint 24.5 under the usual half-up convention.
08 / Name the assumptions
Estimate, locate and qualify.
- Determine the true class boundaries.
- Use finite midpoints and frequency weights for an estimated mean.
- Use cumulative counts to locate middle ranks.
- Distinguish highest frequency from highest density.
- State what grouping prevents you from knowing exactly.
Section 1 of 8 · Grouping hides the individual values