Unit 1 · 1.1

Frequency distributions

Classes, width rounded up, boundaries ∓ 0.5, relative & cumulative frequency

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Rule cards

n = total number of data values = sum of all the frequencies.

Quantitative = numbers you measure or count; qualitative = categories (grades, colors).
Frequency f = how many times a value occurs.
'Most data values' = biggest f; 'fewest' = smallest f.

Traps

Gave the data value when the question asked how many.
'how many days had 3 calls?' → answered 3
✓ the answer is the frequency f in that row
Gave a percent when the question asked for a count.
f ÷ n × 100 when only f was asked
✓ percent of a class = f ÷ n × 100 only when a percent is asked

Width W = range ÷ number of classes, ALWAYS rounded UP to the next whole number; first class starts at the lowest value.

(1) The number of classes is given.
(2) R = max − min.
(3) W = R ÷ classes → ALWAYS round UP to the next whole number.
(4) First class A to B with A = lowest value, B = A + W; next class starts at B.
(5) Boundaries: subtract 0.5 from each end so no value lands in two classes (7–12 → 6.5–11.5).
Check: same width, no overlap, frequencies add up to n.

Traps

Rounded the class width down (or to the nearest).
49 ÷ 6 = 8.17 → 8
✓ always UP: 8.17 → 9
Started the first class somewhere other than the lowest value.
start at 460 because it looks round
✓ the first class starts at the lowest data value, 465
Used the class limits where boundaries were needed.
465–474 used as boundaries
✓ subtract 0.5 from each end: 464.5–473.5
Left out a class with frequency 0.
skip a class nobody falls in
✓ every class is listed; f = 0 is allowed

Cumulative frequency is a running total: the first equals the first f, the last equals n.

Traps

Listed the plain frequencies as cumulative.
f = 9, 7, 7 → cumulative 9, 7, 7
✓ running total 9, 16, 23, … ending at n = 40
Mixed limits and boundaries when finding the midpoint.
(464.5 + 474) ÷ 2
✓ midpoint = (lower + upper) ÷ 2: 464.5–473.5 → 469