A bicycle repair shop's two mechanics repair bikes in a week. Mechanic A repairs \(15\) bikes, mean time \(22\) minutes. Mechanic B repairs \(25\) bikes, me...
Question 1 Report
A bicycle repair shop's two mechanics repair bikes in a week. Mechanic A repairs \(15\) bikes, mean time \(22\) minutes. Mechanic B repairs \(25\) bikes, mean time \(30\) minutes. The box plots show the spread for each.
- Write a formula for the combined mean repair time, in terms of the numbers repaired and mean times. (2)
- Work out the combined mean repair time for all 40 bikes. (2)
- State, with a reason, which mechanic is more consistent. (2)
- The shop advertises a typical time under half an hour. Is this fair? Give a reason. (1)
Answer Details
Combining two means correctly requires weighting each mechanic's mean by how many bikes they repaired, not simply averaging the two means; the box plots then let interquartile ranges be compared directly to judge consistency, independently of the combined mean.
- \[ \text{combined mean} = \frac{n_A\bar{x}_A+n_B\bar{x}_B}{n_A+n_B} \] [2 marks]
- \[ = \frac{15(22)+25(30)}{40} = \frac{330+750}{40} = \frac{1080}{40} = 27 \text{ minutes} \] [2 marks]
- Mechanic A has the smaller interquartile range, \(8\) minutes compared with Mechanic B's \(15\) minutes, so Mechanic A has more consistent repair times. [2 marks]
- Yes, the claim is fair, since the combined mean of \(27\) minutes is less than the advertised half hour (\(30\) minutes). [1 mark]
Because Mechanic B repaired more bikes (\(25\) against \(15\)) and took longer on average, the combined mean of \(27\) minutes sits closer to Mechanic B's \(30\)-minute average than to Mechanic A's \(22\)-minute average; simply averaging \(22\) and \(30\) to get \(26\) would ignore this and understate the true combined mean slightly.
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