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4MA1/2HR Jan 2022, Question 14: Tree Diagrams for Dependent Events

Pearson Edexcel International GCSE Mathematics A 4MA1 · Paper 2HR (Higher) · Calculator · 8 marks

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The question, in short

From a cumulative frequency graph of 60 algebra marks out of 110, estimate median, count at 58 or less, and interquartile range; compare spread with geometry IQR 9; find probability two random students both scored above 64.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 1 mark

Use the cumulative frequency graph to estimate the median algebra mark.

  1. 1The median is the 30th value, because half of 60 is 30
  2. 2Read across from 30 on the cumulative frequency axis to the curve, then down
  3. 3The median algebra mark is about 48

Answer48

Part (b) · 1 mark

Use the graph to estimate how many students scored 58 marks or fewer.

  1. 1Read up from 58 on the mark axis to the curve, then across
  2. 2The cumulative frequency is about 46, so 46 students scored 58 or less

Answer46

Part (c) · 2 marks

Use the graph to estimate the interquartile range of the algebra marks.

  1. 1The lower quartile is the 15th value, a quarter of 60: reading across gives 40
  2. 2The upper quartile is the 45th value, three quarters of 60: reading across gives 56
  3. 3Interquartile range = 56 − 40 = 16

Answer16

Part (d) · 1 mark

Decide whether Luis is right that algebra marks are more spread out than geometry marks (IQR 9), giving a reason.

  1. 1The algebra interquartile range is 16
  2. 2The geometry interquartile range is 9
  3. 316 is bigger than 9, so the algebra marks are more spread out

AnswerYes, because the algebra IQR is larger than the geometry IQR

Part (e) · 3 marks

Using the graph, estimate the probability that two students chosen at random both scored above 64.

  1. 1Read up from 64 on the mark axis: the cumulative frequency is 50
  2. 2So 60 − 50 = 10 students scored more than 64
  3. 3P(first student is one of them) = 10/60
  4. 4After that one is taken there are 9 left out of 59, so P(second) = 9/59
  5. 5(10/60) × (9/59) = 90/3540 = 3/118

Answer3/118

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