The seventh payout equals the original average, so the instinct is that a value was added right at the center and therefore nothing about the spread or shape can change. That instinct is wrong on two of the four statements, and seeing exactly where it breaks is the whole problem. The reliable approach is to judge each statistic by its own rule rather than treating the four as one bundle that must move or freeze together.
Start with the mean (statement I). The original six payouts sum to 6 × their average, and the new payout equals that average, so the new total is 7 × the average shared among seven people. The mean is unchanged, so I must be 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 case where a statistic cannot move. 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 a new maximum. The largest payout and the smallest payout are both still present, so the range is untouched, and III must be 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 the same distance from the mean as before. The total of the squared distances is therefore unchanged, but it is now averaged 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 reflex that more data spreads things out. So IV must be true.
The median (statement II) is the only free one, because the median is decided by rank position, not by magnitude. Take a symmetric set such as 10, 20, 30, 40, 50, 60, whose average is 35. Insert 35, and the middle value of the sorted seven is still 35, so the median holds. Now take a set with one unusually large value, 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, and II is not a must.
That leaves I, III, and IV forced and II free, which is choice (C). To confirm nothing higher slips in, note that II genuinely fails in the second example above, so any answer that keeps the median must be rejected.
The reusable method is to resist reasoning about the statistics as a single block whenever one value is added to a data set. Evaluate the mean by the totals, 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.