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0580/42 Feb/Mar 2023, Question 2: Estimating the Mean for Grouped Data

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 15 marks

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The question, in short

Use a reaction-time frequency table for 100 students, a box-and-whisker plot of heights, and a grouped frequency table of 200 potato masses to find averages, probability, spread and a histogram.

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

Part (ai) · 1 mark

State the most common reaction time from the frequency table.

  1. 1Highest frequency is 32, for 7 seconds
  2. 2Mode = 7

Answer7

Part (aii) · 1 mark

Find the median reaction time for the 100 students.

  1. 1Median is between the 50th and 51st values
  2. 2Cumulative frequencies: 3, 35, 54, …
  3. 3The 50th and 51st values are both 8
  4. 4Median = 8

Answer8

Part (aiii) · 3 marks

Calculate the mean reaction time from the table.

  1. 1Σfx = 3×6 + 32×7 + 19×8 + 29×9 + 11×10 + 6×11
  2. 2= 18 + 224 + 152 + 261 + 110 + 66 = 831
  3. 3Mean = 831 ÷ 100 = 8.31

Answer8.31

Part (aiv) · 2 marks

For two students chosen at random, find the probability both have reaction times of at least 9 seconds.

  1. 1Students with 9 seconds or more = 29 + 11 + 6 = 46
  2. 2P(both) = 46/100 × 45/99
  3. 3= 2070/9900 = 23/110

Answer23/110

Part (bi) · 1 mark

Using the box-and-whisker plot, find the range of the heights.

  1. 1Range = highest height − lowest height (the ends of the whiskers)
  2. 2Range = 53 cm

Answer53

Part (bii) · 1 mark

Using the box-and-whisker plot, find the interquartile range of the heights.

  1. 1IQR = upper quartile − lower quartile (the ends of the box)
  2. 2IQR = 20 cm

Answer20

Part (ci) · 4 marks

Estimate the mean mass of the 200 potatoes from the grouped frequency table.

  1. 1Midpoints: 80, 155, 250
  2. 2Σfx = 60 × 80 + 99 × 155 + 41 × 250
  3. 3= 4800 + 15345 + 10250 = 30395
  4. 4Mean = 30395 ÷ 200 = 151.975 g

Answer151.975

Part (cii) · 2 marks

Complete the histogram for the potato masses, using the given bar for 50 < m ≤ 110.

  1. 1Frequency density = frequency ÷ class width
  2. 2110 < m ≤ 200: 99 ÷ 90 = 1.1
  3. 3200 < m ≤ 300: 41 ÷ 100 = 0.41
  4. 4Draw a bar from 110 to 200 of height 1.1 and a bar from 200 to 300 of height 0.41

AnswerCorrect histogram completed with widths 110 to 200 and 200 to 300 and heights 1.1 and 0.41

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