A driver records the distance travelled in each of 150 journeys. These distances, correct to the nearest km, are summarised in the following table. Distance (km) | 0–4 | 5–10 | 11–20 | 21–30 | 31–40 | 41–60 Frequency | 12 | 16 | 32 | 66 | 20 | 4
- (a)Draw a cumulative frequency graph to illustrate the data.[4]
- (b)For 30% of these journeys the distance travelled is d km or more. Use your graph to estimate the value of d.[2]
- (c)Calculate an estimate of the mean distance travelled for the 150 journeys.[3]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2021). The mark or mark type is shown in brackets.
- [1] B1 Correct cumulative frequencies seen (may be by table or plotted accurately on graph), condone 12 not stated.
- [1] B1 Axes labelled 'distance (or d) [in] km' from 0 to 60 and 'cumulative frequency' (or cf) from 0 to 150.
- [1] M1 At least 5 points plotted at upper end points for d (allow upper boundary ±0·5) with a linear scale for distance, condone 0 - 4 interval inaccurate, no scale break on axis. Not bar graph/histogram unless clear indication of upper end point only of each bar.
- [1] A1 All plotted correctly at correct upper end points (4.5 etc.) with both scales linear (0 ≤ d ≤ 60, 0 ≤ cf < 150), curve drawn accurately joined to (0,0), cf line>150, no daylight if >150.
- [1] M1 105 seen or implied by indication on grid.
- [1] A1 Strict FT their increasing cumulative frequency graph, use of graph must be seen. If no clear evidence of use of graph: SC B1 FT correct value from their increasing cumulative frequency graph.
- [1] B1 At least 5 correct midpoints seen.
- [1] M1 Using 6 midpoint attempts (e.g. 2.25 ±0.5), condone one error not omission, multiplied by frequency, accept unevaluated, denominator either correct or their Σ frequencies.
- [1] A1 Evaluated, WWW, accept 21.5[866...].
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