Reading the Recycling Data
A Grade 6 class records how many plastic bottles each of four classrooms collects in a week. Room A reports 48 bottles. Room B reports 36. Room C reports 60. Room D reports 12, but the student recording it says, "I only counted Monday and Tuesday, then multiplied by 2." The class average is announced as 39 bottles. A visitor claims, "Room C collected more than twice what Room B collected, so Room C is the top collector." The teacher wants the class to decide whether the average of 39 is a fair summary and whether the visitor's claim is logically sound.
1. Evaluate whether the announced average of 39 is a fair summary of the data. Identify any problem with how Room D's number was produced and explain how it affects the average.
Read the worked answer
The arithmetic mean is (48 + 36 + 60 + 12) ÷ 4 = 156 ÷ 4 = 39, so the calculation is correct.
But fairness is a separate question from arithmetic. Room D's 12 was not a full-week count - it was two days doubled, which assumes both remaining days matched the first two. That is an estimate, not a measurement, so the 39 is built on one estimated value among three measured ones.
Also, the mean is pulled by the spread: the values range from 12 to 60, and no room actually collected near 39 (the closest, 36, is 3 away; 48 is 9 away). The mean is a legitimate summary of the total, but it does not describe a typical room well here. A fair answer should state the mean and note that Room D's figure is an estimate that could shift the average up or down once the full week is counted.
2. Assess the visitor's claim that Room C is the top collector because 60 is more than twice 36. Is the reasoning valid? Explain.
Read the worked answer
The arithmetic part is true: 2 × 36 = 72, and 60 is not more than 72 - so the visitor's premise is actually false. 60 is less than twice 36.
Even setting that error aside, the conclusion "Room C is the top collector" happens to be true (60 is the largest value), but the reasoning given does not establish it. Being more than twice Room B's count would not by itself make C the top collector unless we also compared C to every other room. A valid argument must compare C to all rooms: 60 > 48, 60 > 36, and 60 > 12, so C is the top collector.
The lesson: a conclusion can be correct while the reasoning offered for it is flawed, and a numerical comparison must be checked before it is used as evidence.