Estimate a mean from grouped data using class midpoints, say why the answer is only an estimate, and identify the modal class and the class holding the median.
Last stop the visits table gave exact answers, because every row held one value. The second question is harder. It asked how long members stay on site, and nobody wrote 34 minutes. They ticked a band. In 20 ≤ t < 40 the ≤ end takes 20 itself and the < end stops just short of 40, so a visit of exactly 40 minutes goes in the next band. The five bands came back with 6, 21, 15, 12 and 6 members, sixty in all, and not one exact time among them.
You cannot add up times you do not have, so pick one value to stand for each band. The fairest single choice is the middle of the band, called the midpoint: lower boundary plus upper boundary, divided by 2. For 20 ≤ t < 40 that is (20 + 40) / 2 = 30. Do the same down the column and the five midpoints are 10, 30, 50, 70 and 90. Each one is now the value the whole band gets treated as having, exactly the way a value stood for its row last stop.
From here the machine is the one you already have, with the midpoint m in place of the value x. Multiply each midpoint by its frequency: 6 × 10 = 60, 21 × 30 = 630, 15 × 50 = 750, 12 × 70 = 840 and 6 × 90 = 540. Add them: 60 + 630 + 750 + 840 + 540 = 2820 minutes. Divide by how many members: 2820 / 60 = 47. So members stay about 47 minutes on average.
That 47 is an estimate, and no amount of care with the arithmetic makes it exact. The 21 members in the 20 ≤ t < 40 band are all being treated as 30-minute visitors. If most of them actually left after 22 minutes, the real mean is lower than 47. If most stayed until 38, it is higher. The estimate is only as good as the assumption that values sit evenly inside each band. Narrower bands would tighten the estimate, and only the actual times would make it exact.
Two more summaries, and both name a band rather than a number. The modal class is the band with the largest frequency, which is 20 ≤ t < 40 with 21 members. It is not 21, and there is no single mode, because no time was ever written down. For the median, run a total down the frequencies: 6, 27, 42, 54, 60. Position (60 + 1) / 2 = 30.5 sits past 27 and inside the block ending at 42. So the median lies in 40 ≤ t < 60. You get the band, not the minute.
When data arrives in bands you invent a value for each one: the midpoint, halfway between the boundaries. Multiply each midpoint by its frequency, add those, and divide by the total frequency for an estimated mean. It is an estimate because every member of a band is treated as sitting at its middle. The modal class is the band with the largest frequency, and a running total says which band holds the median.
The visits question gave exact numbers and the table handed over its mean without a fight. The second question is not so generous. It asked how long members stay on site, and it offered bands to tick. Sixty members came back as 6, 21, 15, 12 and 6 across the bands 0 to 20, 20 to 40, 40 to 60, 60 to 80 and 80 to 100 minutes. Not one exact time anywhere.
The 21 members in 20 ≤ t < 40 need one value to stand for them. Which value, and why?
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