Use the class midpoint as the value: With grouped data the exact values are unknown, so each value is taken as the class midpoint x.
The estimated mean is then Σfx ÷ Σf with x = midpoint.
IB-style question — estimated mean
Waiting times (minutes) are grouped: 0–10 → 3, 10–20 → 20, 20–30 → 8, 30–40 → 9, 40–50 → 10 (50 people).
Estimate the mean waiting time.
Step by step
- Midpoints are 5, 15, 25, 35, 45. Form Σfx.
- Divide by Σf = 50.
Final answer
Estimated mean ≈ 25.6 minutes.
Midpoint = (lower + upper) ÷ 2: Find each midpoint first (e.g. (10 + 20)/2 = 15), then weight by the frequency.
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Tallest bar, and where the middle falls: The modal class has the greatest frequency.
The median class is the class containing the (n/2)-th value — find it from a running total (cumulative frequency).
IB-style question — modal & median class
Using the same data (3, 20, 8, 9, 10 for 0–10, …, 40–50; n = 50), state the modal class and the median class.
Step by step
- Modal class = greatest frequency.
- Running totals 3, 23, 31, 40, 50; the n/2 = 25th value sits in…
Final answer
Modal class 10 ≤ t < 20; median class 20 ≤ t < 30.
Modal ≠ median class in general: They are often different classes — find each one separately (frequency for the mode, running total for the median).
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Set the estimated-mean formula equal to the given mean: If a frequency is unknown, write Σfx ÷ Σf = (given mean) with x = midpoints, then solve the equation for the unknown frequency.
IB-style question — missing frequency
Grouped data has classes 0–10, 10–20, 20–30 with frequencies 4, k, 6 and an estimated mean of 16.
Find k.
Step by step
- Midpoints 5, 15, 25; write the mean equation.
- Simplify and solve.
Final answer
k = 10.
Cross-multiply, then collect k: Multiply both sides by (Σf), expand, and gather the k-terms — it becomes a simple linear equation.