0580/42 Oct/Nov 2022, Question 3: Estimating the Mean for Grouped Data
Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 15 marks
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The question, in short
Using grouped distance data for 200 cars, estimate a mean, compare two versions of the final interval, work out histogram block heights, and find probabilities for randomly chosen cars.
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Part (ai) · 4 marks
Estimate the mean distance from Kai's grouped frequency table.
- 1Mid-points: 90, 125, 175, 250, 350
- 2Σfx = 7×90 + 33×125 + 76×175 + 52×250 + 32×350 = 42255
- 3Mean = 42255 ÷ 200 = 211.275 km
Answer211.275
Part (aii) · 2 marks
Using Ann's table with a different last interval, find how much her mean estimate differs from Kai's, showing working.
- 1Only the last mid-point changes: 350 → 330, a change of 20
- 2Difference = 32 × 20 ÷ 200 = 3.2 km
Answer3.2 or −3.2
Part (aiii) · 3 marks
Given the 200 < d ≤ 300 block is 2.6 cm high, find the heights of the blocks for 80 < d ≤ 100, 150 < d ≤ 200 and 300 < d ≤ 400.
- 1Frequency density 200 < d ≤ 300: 52 ÷ 100 = 0.52 → 2.6 cm, so height = 5 × frequency density
- 280 < d ≤ 100: 7 ÷ 20 = 0.35 → 1.75 cm
- 3150 < d ≤ 200: 76 ÷ 50 = 1.52 → 7.6 cm
- 4300 < d ≤ 400: 32 ÷ 100 = 0.32 → 1.6 cm
Answer80 < d ≤ 100: 1.75 cm 150 < d ≤ 200: 7.6 cm 300 < d ≤ 400: 1.6 cm
Part (b) · 1 mark
Find the probability that a single randomly chosen car travelled more than 300 km.
- 1P(more than 300 km) = 32/200 = 4/25
Answer4/25
Part (ci) · 2 marks
Find the probability that two randomly chosen cars both travelled 150 km or less.
- 1150 km or less: 7 + 33 = 40 cars
- 2P = 40/200 × 39/199 = 1560/39800 = 39/995
Answer39/995
Part (cii) · 3 marks
Find the probability that one chosen car went over 200 km while the other went 100 km or less.
- 1More than 200 km: 52 + 32 = 84 cars, 100 km or less: 7 cars
- 2P = 2 × 84/200 × 7/199 = 1176/39800
- 3= 147/4975
Answer147/4975
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Also searched as: 0580/42/O/N/22 · 0580_w22_qp_42 · 0580 w22 42 · question 3.
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