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0580/41 May/Jun 2022, Question 1: Estimating the Mean for Grouped Data

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 15 marks

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The question, in short

Use 15 midday temperatures for Suntown to complete a stem-and-leaf diagram, find median and upper quartile, calculate a pie chart sector angle, read a box-and-whisker plot, and work with grouped frequency tables.

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

Part (ai) · 2 marks

Complete the stem-and-leaf diagram for the 15 temperatures, using the given key where 1 | 7 means 17 °C.

  1. 1In order: 17, 17, 18, 18, 19, 19, 21, 21, 21, 21, 22, 23, 23, 24, 25
  2. 21 | 7 7 8 8 9 9
  3. 32 | 1 1 1 1 2 3 3 4 5

Answer1 | 7 7 8 8 9 9 2 | 1 1 1 1 2 3 3 4 5

Part (aii) · 1 mark

Find the median of the 15 temperatures.

  1. 115 values, so the median is the 8th value
  2. 2Median = 21 °C

Answer21

Part (aiii) · 1 mark

Find the upper quartile of the temperatures.

  1. 1Upper quartile = 3/4 × (15 + 1) = 12th value
  2. 2Upper quartile = 23 °C

Answer23

Part (aiv) · 2 marks

Calculate the pie chart sector angle representing the days with temperature 18 °C.

  1. 118 °C occurs on 2 days out of 15
  2. 2Angle = 2/15 × 360 = 48°

Answer48

Part (bi) · 1 mark

From the box-and-whisker plot of apple masses in grams, find the median.

  1. 1The median is the line inside the box
  2. 2Median = 120 g

Answer120

Part (bii) · 1 mark

From the box-and-whisker plot, find the range of the masses.

  1. 1Lowest = 40 g, highest = 170 g
  2. 2Range = 170 − 40 = 130 g

Answer130

Part (biii) · 1 mark

From the box-and-whisker plot, find the interquartile range.

  1. 1Lower quartile = 80 g, upper quartile = 140 g
  2. 2Interquartile range = 140 − 80 = 60 g

Answer60

Part (ci) · 4 marks

Using the grouped frequency table for 200 students' homework times, calculate an estimate of the mean time.

  1. 1Mid-values: 50, 70, 85, 95, 125
  2. 2Σfx = 6×50 + 10×70 + 70×85 + 84×95 + 30×125
  3. 3= 300 + 700 + 5950 + 7980 + 3750 = 18 680
  4. 4Mean = 18 680 ÷ 200 = 93.4 min

Answer93.4

Part (cii) · 2 marks

Given the height 17.2 cm for the 40 < t ≤ 90 bar, calculate the height of the 90 < t ≤ 150 bar on the histogram.

  1. 1Frequency density for 40 < t ≤ 90: 86 ÷ 50 = 1.72, shown as 17.2 cm (10 cm per unit)
  2. 2Frequency density for 90 < t ≤ 150: 114 ÷ 60 = 1.9
  3. 3Height = 1.9 × 10 = 19 cm

Answer19

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Also searched as: 0580/41/M/J/22 · 0580_s22_qp_41 · 0580 s22 41 · question 1.
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