Functional Skills Level 2 — Data and Statistics

Estimated Mean from Grouped Data

← Previous Back to Course Next Lesson →

Learning Objectives

Real-World Applications

Subscribe to keep learning

You’ve seen the first part of this lesson for free. Log in and subscribe to unlock the rest, plus every other lesson, worksheet and checkpoint paper.

See Plans →
Why Grouped Data Gives an Estimate

When data is grouped into class intervals (e.g. $10 \leq x < 20$), we do not know the individual values within each group. The best single value to represent all the data in that class is the midpoint — the value halfway between the lower and upper boundary.

Midpoint = (lower boundary + upper boundary) ÷ 2
Grouped Frequency Table — Finding Midpoints Class Interval Frequency (f) Midpoint (m) 10 ≤ x < 20 4 15 20 ≤ x < 30 9 20 ≤ x < 30 9 25 30 ≤ x < 40 11 35 40 ≤ x < 50 6 45 Midpoint = (lower + upper) ÷ 2, e.g. (10 + 20) ÷ 2 = 15 Why We Use the Midpoint Class: 20 ≤ x < 30 — we know 9 values are in here... ? ? ? ? ? Best estimate: midpoint = 25
Key point: We cannot find the exact mean from grouped data because we do not know the individual values. We use the midpoint as our best single representative for each class.
✔ Quick Check emgd1 — Midpoints and why we estimate
Question 1 of 10
The Estimated Mean Formula

Once we have the midpoint for each class, we multiply each midpoint by its frequency to get $fm$, then use the formula:

$$\text{Estimated mean} = \frac{\Sigma fm}{\Sigma f}$$

Where $m$ = midpoint and $f$ = frequency. The four steps are: find each midpoint, multiply midpoint × frequency, sum all $fm$ values, then divide by the total frequency.

Worked Example: Heights of 30 Students Class (cm) Freq (f) Midpoint (m) f × m 140 ≤ h < 150 5 145 725 150 ≤ h < 160 12 155 1860 160 ≤ h < 170 9 165 1485 170 ≤ h < 180 4 175 700 TOTALS Σf = 30 Σfm = 4770 Applying the Formula Σfm = 4770 Σf = 30 ÷ Estimated mean 4770 ÷ 30 = 159 cm

Full Worked Example — Heights of 30 students

Step 1 — Find midpoints
140–150: m=145, 150–160: m=155, 160–170: m=165, 170–180: m=175
Step 2 — Calculate fm
5×145=725, 12×155=1860, 9×165=1485, 4×175=700
Step 3 — Sum fm
Σfm = 725 + 1860 + 1485 + 700 = 4770
Step 4 — Divide by Σf
Estimated mean = 4770 ÷ 30 = 159 cm
✔ Quick Check emgd2 — Applying the estimated mean formula
Question 1 of 10
Interpreting the Estimated Mean

The result is an estimate — always state this in your answer. Compare the estimated mean to the modal class (the class with the highest frequency) to check your answer is reasonable. If they are in very different parts of the data, re-check your working.

Cinema Customer Ages — Bar Chart 0–9 10–19 20–29 30–39 40+ 8 22 30 ← modal 18 7 Comparing Measures for Cinema Ages Estimated Mean ≈ 23.8 years old Modal Class 20–29 f = 30 Median Class 20–29 30th value falls here

Cinema ages example — Worked calculation

Ages (years): 0–9 (f=8, m=4.5), 10–19 (f=22, m=14.5), 20–29 (f=30, m=24.5), 30–39 (f=18, m=34.5), 40–49 (f=7, m=44.5)

fm values
36 + 319 + 735 + 621 + 311.5 = 2022.5
Σf
8 + 22 + 30 + 18 + 7 = 85
Estimated mean
2022.5 ÷ 85 ≈ 23.8 years (1 d.p.)
Interpretation
The estimated mean age is approximately 23.8 years. This falls within the modal class (20–29), which suggests a reasonable answer.
Always write "estimated mean" — never just "mean" when working from grouped data. Calling it the exact mean would be inaccurate.
✔ Quick Check emgd3 — Interpretation and full problems
Question 1 of 10

Practice: Estimated Mean from Grouped Data

Use each table to calculate the estimated mean, then click Check. Give answers to 1 decimal place.

1. Homework time (mins), 30 pupils

Time (mins)fmf×m
0 < t ≤ 104520
10 < t ≤ 20815120
20 < t ≤ 301125275
30 < t ≤ 40735245
Total30660

2. Age (years) of 20 people surveyed

Agefmf×m
0 < a ≤ 103515
10 < a ≤ 2061590
20 < a ≤ 301125275
Total20380

3. Distance run (km) by 25 athletes

Distancefmf×m
0 < d ≤ 552.512.5
5 < d ≤ 10127.590
10 < d ≤ 15812.5100
Total25202.5

4. Weight of parcels (kg), 15 parcels

Weightfmf×m
0 < w ≤ 2414
2 < w ≤ 47321
4 < w ≤ 64520
Total1545

5. Texts sent per day, 40 pupils

Textsfmf×m
0 < t ≤ 201010100
20 < t ≤ 401830540
40 < t ≤ 601250600
Total401240

6. Waiting time (mins) at a bus stop, 20 people

Timefmf×m
0 < t ≤ 542.510
5 < t ≤ 10107.575
10 < t ≤ 15612.575
Total20160

7. Age of trees in a park (years), 25 trees

Agefmf×m
0 < a ≤ 2051050
20 < a ≤ 401030300
40 < a ≤ 601050500
Total25850

8. Customers per hour, 30 hours recorded

Customersfmf×m
0 < c ≤ 106530
10 < c ≤ 201515225
20 < c ≤ 30925225
Total30480

9. Length of phone calls (mins), 20 calls

Lengthfmf×m
0 < l ≤ 5102.525
5 < l ≤ 1067.545
10 < l ≤ 15412.550
Total20120

10. Weekly grocery spend (£), 25 households

Spendfmf×m
0 < s ≤ 40520100
40 < s ≤ 801260720
80 < s ≤ 1208100800
Total251620

⚠ Things to Watch Out For

  • Using class boundaries instead of midpoints: The estimated mean uses the MIDPOINT of each class, not the boundary values. For the class 10–20, use 15, not 10 or 20.
  • Dividing by number of rows instead of total frequency: Divide Σfm by Σf (sum of all frequencies), not by the number of class intervals.
  • Calling it the exact mean: Always say "estimated mean" — you cannot calculate the exact mean from grouped data because individual values are unknown.
  • Open-ended classes: A class like "40 and over" has no upper boundary. You must make a reasonable assumption for the midpoint (e.g. treat it as 40–50, midpoint 45), and state your assumption.
  • Midpoint calculation error: For 20 ≤ x < 30: midpoint = (20 + 30) ÷ 2 = 25. Students sometimes write 20 + 10 = 30, or forget to divide by 2.
What would you like to do next?
Need 1:1 support? Book a Functional Skills Maths tutor — 1-to-1 online lessons from £45/hour.
Privacy Policy · Terms & Conditions