Real questions, real marking

Albert's corrected score of 37 now falls above the original upper quartile of 47, so it no longer sits in the bottom quarter — the lower quartile itself shifts up to the new value 35

States the new lower quartile is 35, not the original 33 — the single most commonly missed mark on this part, per the examiner report.

New IQR = Q3 − new Q1 = 47 − 35 = 12 (down from the original 14), so the outlier fences must be recalculated using this new IQR

Recalculates the IQR using the NEW lower quartile — a smaller IQR than before — and attempts to use it to re-derive the outlier fences, rather than reusing the original fences.

Lower fence = 35 − 1.5 × 12 = 35 − 18 = 17 (upper fence = 47 + 18 = 65 is unchanged, since Q3 never moved)

At least the correct lower outlier boundary of 17 is seen — this is the number the whole recalculation was building toward.

Draw a box with exactly two whiskers, median unchanged at 42 (Beth's and Albert's corrections don't touch the median)

Draws the box-and-whisker structure itself — two whiskers, median still at 42 — independent of which quartile values are plotted; this mark is available even to a candidate who never spots the Q1 shift.

Box drawn with its lower edge at the recalculated 35 (not the original 33) and its upper edge unchanged at 47

The box's lower quartile edge must specifically be at the candidate's own recalculated value AND that value must actually differ from 33 — a box still drawn at 33 cannot earn this follow-through mark, even if every other line is correct.

Plot exactly two outliers, at 11 and 15 — both below the new lower fence of 17, with no overlap onto the whisker itself

Exactly two outliers plotted, at 11 and 15, marked as points distinct from the whisker rather than absorbed into it.

Lower whisker drawn to 18 — the lowest data value that is NOT an outlier under the new 17 fence — and upper whisker extended to Beth's corrected 60

Lower whisker ends at the lowest non-outlier value (18, since the fence itself is 17) and the upper whisker is redrawn out to the new maximum of 60.

E(X) = 1/2 > 0 describes the spinner's own long-run average score, not Amar's chance of winning THIS game — the quantity that actually decides whether he's more likely to win or lose is P(Amar wins), found in part (d) to be 7/16, which is less than 0.5

Identifies that the deciding factor is P(win) compared with 0.5 — not the sign of E(X) — and states that P(win) < 0.5 (follow through on their own part (d) value, provided it is genuinely below 0.5).

Since P(Amar wins) = 7/16 < 0.5, he is more likely to lose than win, so Amar should not play the game — disagree with him

States the conclusion that Amar should not play (or explicitly 'disagree') — dependent on the M1 reasoning about P(win) actually being present, not asserted on its own.