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, each row gave an exact value. Here members ticked time bands instead. In 20 ≤ t < 40, 20 is included and 40 is excluded, so exactly 40 minutes goes in the next band. The five bands have 6, 21, 15, 12 and 6 members, sixty in all. Counts show how many visits fall in each band, not exact times or which band a named member chose.
To estimate a mean, choose each band's midpoint as its stand-in: add the boundaries and divide by 2. For 20 ≤ t < 40, (20 + 40) / 2 = 30. The five midpoints are 10, 30, 50, 70 and 90. A midpoint is equally far from both boundaries, but need not equal the real class mean. A value can be at most half a band's width from its midpoint. Lower boundaries give smaller estimates than midpoints; upper boundaries give larger ones.
Use the same method as before, 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.
Midpoints replace unknown values, so 47 is an estimate even with exact arithmetic. A class's actual total is its actual mean times its frequency. If that mean is below its midpoint, the class total is overestimated; other classes may offset this. The estimate is exact if all values equal their class midpoints, but errors can also cancel. Each value's error is at most half the band width. Narrower bands lower this bound, not necessarily the actual mean error.
Find the modal class beside the largest frequency: 20 ≤ t < 40 has 21 members. Exact modes are unknown. Running totals are 6, 27, 42, 54, 60. For odd count n, locate (n + 1) / 2. For even n, this is a half position between n / 2 and n / 2 + 1: find both observations and average their values. Here 30.5 lies between positions 30 and 31, both in 40 ≤ t < 60, so the median lies there. If the two middle observations are in different bands, their average's band may be unknown.
Use midpoints times frequencies, add, then divide by total frequency to estimate the mean. The modal class has the largest frequency. Running totals locate the middle observation or pair.
Practice questions and four readings were written for this dataset. The retained video, "Estimating mean and median in data displays", is published by Khan Academy. Its license is not specified.
The visits question gave exact numbers, so its frequency table gave an exact mean. The second question asked how long members stay on site and offered bands to tick. The sixty members' answers were counted 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. No exact times were recorded.
The 21 members in 20 ≤ t < 40 need one value to stand for them. Which value, and why?
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