These questions cover processing a dataset, choosing a graph, anomalies, conclusions and evaluation. All data is invented. Write your answer first, then open the worked solution.
Use the fieldwork sample and graph planner and the mistake log and retest queue alongside the set. The module overview is fieldwork analysis and evaluation.
Questions
Q1. Seven litter counts are 14, 9, 17, 12, 13, 9 and 20. Find the mean, median, mode and range.
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Total: 14 + 9 + 17 + 12 + 13 + 9 + 20 = 94. Mean: 94 ÷ 7 = 13.43…, so 13.4.
Sorted: 9, 9, 12, 13, 14, 17, 20. The median is the 4th value, 13. The mode is 9 and the range is 20 − 9 = 11.
Q2. Six sand-dune heights are 22, 18, 30, 26, 24 and 20 cm. Find the median and the mean.
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Sorted: 18, 20, 22, 24, 26, 30. With six values the median is the mean of the 3rd and 4th: (22 + 24) ÷ 2 = 23 cm. The total is 140, so the mean is 140 ÷ 6 = 23.33…, or 23.3 cm.
Q3. Of 70 people interviewed, 42 say they travel to the centre by bus. What percentage is that, and what pie chart angle represents it?
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42 ÷ 70 × 100 = 60%. The angle is 60 × 3.6 = 216°. Check: 216 ÷ 360 = 0.6.
Q4. Name the most suitable display for each: (a) distance from a river mouth against pebble size at ten sites; (b) the number of days with each wind direction; (c) the share of three land uses.
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(a) A scatter graph with a line of best fit, because there are two measurements per site. (b) A wind rose, since directions form a circle. (c) A pie chart or divided bar, since the data is parts of a whole.
Q5. Pedestrian counts run from 35 to 84. Suggest a vertical scale for graph paper that has 12 cm available, and say how tall the highest point would be.
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Use 0 to 90 with 1 cm for each 10 pedestrians, which needs 9 cm and fits in 12 cm. The point at 84 is 84 ÷ 10 = 8.4 cm tall.
Q6. Site counts are 40, 38, 35, 36, 70 and 33. Identify the anomaly and find the mean with and without it.
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The value 70 is far above the others, which lie between 33 and 40.
With it: 40 + 38 + 35 + 36 + 70 + 33 = 252, and 252 ÷ 6 = 42. Without it: 252 − 70 = 182, and 182 ÷ 5 = 36.4.
Q7. The aim is “to investigate whether river depth increases downstream”. Depth is 12 cm at site 1 and 30 cm at site 5. Calculate the change and the percentage increase.
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Change: 30 − 12 = 18 cm. Percentage: 18 ÷ 12 × 100 = 150%.
Q8. A student writes: “The results prove that deeper rivers are faster everywhere.” The aim was about depth downstream. Give two faults.
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First, it answers a different question, because the aim was about depth, not speed. Second, it overclaims with “prove” and “everywhere”. A better version says the results support or do not support the aim for the sites measured.
Q9. A line of best fit passes through (0, 90) and (2, 30), with distance in km and pedestrian count. Find its gradient and say what it means.
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Gradient = (30 − 90) ÷ (2 − 0) = −60 ÷ 2 = −30. It means the line predicts a fall of 30 pedestrians for each extra kilometre. This is only a description of the fitted line for these sites.
Q10. A student writes: “My results were poor because it started raining and my group argued.” Rewrite it as a specific improvement based only on a limit in the method, which was one count of five minutes at each of six sites.
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For example: “Each site was counted once for five minutes, so one unusual count could affect the result. Counting three times at each site and using the mean would make the evidence more reliable.” The rewritten answer uses the recorded method and does not rely on events that are not in the data.
If you got these wrong
| Where the error happened | Lesson to revisit |
|---|---|
| Mean, median, range or percentage went wrong | Process a small original dataset |
| Graph, scale or display choice was wrong | Select a graph that answers the question |
| Anomaly was ignored or deleted | Identify anomalous evidence transparently |
| Conclusion drifted from the aim | Link a conclusion to the investigation aim |
| Evaluation was vague or invented | Propose a specific improvement |
Log each miss in the mistake log and retest queue. When you want a teacher to watch your reasoning on unseen datasets, see our online one-to-one Geography tuition.