The late times for 7 buses has been given, but we need all 8.
However, we can work towards a solution using symbols and logic. If we list the late times in order symbolically we have:
b₁, b₂, b₃, b₄, b₅, b₆, b₇, b₈, b₉, b₁₀.
If the median is 4.5, b₅+b₆=9 because the median is the average, in this case, of the 5th and 6th data elements.
Also, if the minimum late time is 1 minute, b₁=b₂=1, and since the mode is also 1, b₃=1 because 1 must be the most recurring data value (mode). We therefore assume that at least one of the remaining two buses must have been a minute late. The values 8, 8, 9, 9 are all greater than 4.5 so they must be included in b₆-b₁₀, the last 5 values.
We know because the data is ordered that b₅≥2, and that b₄=2. If b₅=2, b₆=7 because b₅+b₆=9. If b₅=3, b₆=6, if b₅=4, b₆=5. No other values for b₅ and b₆ are possible.
Here are possible datasets:
- 1, 1, 1, 2, 2, 7, 8, 8, 9, 9.
- 1, 1, 1, 2, 3, 6, 8, 8, 9, 9.
- 1, 1, 1, 2, 4, 5, 8, 8, 9, 9.
Each dataset has a mean of 4.8 minutes, a median of 4.5 minutes and a mode of 1 minute, and one of the two afternoon buses was a minute late. The other afternoon bus and the bus for which we are not given the late time could have been 2, 3, 4, 5, 6 or 7 minutes late. In other words, if we had the missing late data, we could determine the late time of the other bus in the pair that came later.
