The seventh payout equals the original average, so the instinct is that "we added a value right at the center, so nothing about the spread or shape changes." That instinct is wrong on two of the four statements, and seeing exactly where it breaks is the whole problem. Reason about each statistic by its own rule rather than treating them as one bundle that moves or freezes together.
Start with the mean (statement I). The original six payouts sum to six times the average, and the new payout equals that average, so the new total is seven times the average over seven people. The mean is unchanged, so I is forced true. This is the bait: because the added value sits at the mean, it is tempting to assume the rest of the distribution is equally undisturbed.
The range (statement III) is the "which statistic cannot move" case. The average of a set whose values are not all equal always lies strictly between the smallest and largest values, so the added value is an interior point, never a new minimum or maximum. The largest payout and the smallest payout are both still present, so the range is untouched. III is forced true.
The standard deviation (statement IV) is the counterintuitive one. The new payout sits exactly at the mean, so its distance from the mean is zero, and since the mean did not move, every original payout keeps its same distance from the mean. The total of the squared distances is therefore unchanged, but it is now spread over seven payouts instead of six, so the average squared distance shrinks and the standard deviation strictly decreases. Because the original payouts were not all equal, that total is genuinely positive, so the spread really does drop rather than stay flat. Adding a data point lowered the spread, which is the opposite of the usual "more data spreads things out" reflex. IV is forced true.
The median (statement II) is the only free one, because the median is decided by rank position, not by magnitude. Try a symmetric set such as 10, 20, 30, 40, 50, 60, whose average is 35; insert 35 and the sorted middle value is still 35, so the median holds. Now try a right-skewed set such as 10, 20, 30, 40, 50, 90, whose average is 40; insert 40 and the new middle value jumps to 40. The median stayed in one case and rose in the other, so it is not forced. II is not a must.
So I, III, and IV are forced and II is free, which is choice (C). The most common wrong answer is (B), I, II, and III only: it comes from the single mental model that adding a value at the center freezes everything, so it wrongly keeps the median locked and wrongly assumes the spread is unchanged. Choice (E) banks all three genuinely forced statements (the mean, the range, and the smaller spread) correctly, but commits the first of those two errors: it wrongly keeps the median locked, assuming that adding a value equal to the mean must fix the median, when the median is rank-dependent and genuinely free to move. Choice (A) is the trap for the strong solver who does everything right except the standard deviation, correctly forcing the mean and the range, correctly freeing the median, but assuming "same center, so same spread" and dropping IV. Choice (D) credits only the statistics a solver imagines moved, the median and the standard deviation, and overlooks that the two most obvious statistics, the mean and the range, are the ones that are actually locked.
The reusable method: when one value is added to a data set, never reason about the statistics as a single block. Evaluate the mean by the shift formula, the range by whether the new value is a new extreme, the standard deviation by what happens to the squared distances over the new count, and the median by rank position. A value added exactly at the mean leaves the mean and range fixed, strictly shrinks the standard deviation, and leaves the median free.