The lengths of 250 leaves of a certain type of plant are measured, correct to the nearest centimetre. The results are summarised in the table below. Length (cm) 5–9 10–14 15–19 20–24 25–29 30–39 Frequency 18 28 60 72 48 24

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

1. [1] At least 4 of 46, 106, 178, 226, 250 cumulative frequencies correct. May be implied by accurate plotting if scale suitable. May be by data table.
2. [1] Linearly scaled axes correctly labelled cumulative frequency (cf) (from 0 to 250), length (oe) and cm (from 5 to 39.5) – or a suitable title, with at least 3 values identified on each. Axes can be the other way round. Axes must be more than 50% of grid.
3. [1] At least 4 points correctly plotted at class upper end points (9.5, 14.5, 19.5, 24.5, 29.5, 39.5) on scaled axes.
4. [1] All points plotted correct, curve drawn (within tolerance) and joined to (4.5,0) and not going above 250 vertically within range. A0 if straight line segments used.
5. Length (cm) <9.5, <14.5, <19.5, <24.5, <29.5, <39.5. CF 18, 46, 106, 178, 226, 250
6. [1] Clear indication of use of graph at cf 155 is required.
7. [1] Must be an increasing cf graph. Expect an answer in the range 22.5 < l < 23.5 from correct graph.
8. 250 × 0.62 = 155. Line drawn from 155 on cf axis to meet graph at l = 23.
9. [1] At least 5 correct midpoints seen, may be unsimplified, may be in calculation, may be by data table.
10. [1] Correct unsimplified mean formula using their 6 midpoints (not upper bound, lower bound, upper limits, lower limits, cw, fd, f or cf and must be within class) condone 1 error. If midpoints correct accept 5190/250 or (126+336+1020+1584+1296+828)/250
11. [1] Accept 20.76, 19/25, 38/50 or 76/100 not improper fraction If M1 withheld, SC1 for 20.76 oe WWW.
12. Midpoints 7, 12, 17, 22, 27, 34.5. Mean = (7×18 + 12×28 + 17×60 + 22×72 + 27×48 + 34.5×24) / 250 = (126 + 336 + 1020 + 1584 + 1296 + 828) / 250 = 5190 / 250 = 20.76