Outliers, Box Plots, and Comparing Distributions

A data set has $Q_1 = 18$ and $Q_3 = 30$. Using the rule "an outlier is a value more than $1.5 \times IQR$ beyond the nearer quartile," is the value 47 an outlier?

• No — the upper fence is $30 + 1.5(12) = 48$, and $47 < 48$

  Correct. $IQR = 30 - 18 = 12$, and $1.5 \times 12 = 18$, so the upper fence is $30 + 18 = 48$. 47 sits just inside it. This is deliberately a close call: the fence and the value differ by only 1, exactly why the comparison has to be shown in full rather than judged by eye.

• BYes — 47 is greater than $Q_3$ (30)

  Every value above $Q_3$ is, by definition, in the upper quarter of the data — that alone doesn't make it unusual. An outlier has to clear the FENCE, a full $1.5 \times IQR$ beyond $Q_3$, not just clear $Q_3$ itself.

• CYes — using Q1's fence instead: $18 + 1.5(12) = 36$, and $47 > 36$

  This tests 47 against a fence built from $Q_1$, but 47 sits in the UPPER part of the data, so it has to be tested against the fence built from $Q_3$ — the nearer quartile — not $Q_1$.

• DNo — because 47 isn't the maximum value in the data

  Whether a value is the overall maximum is irrelevant to whether it's an outlier — an outlier is decided purely by its distance from the nearer quartile, measured against the fence, whatever else is in the data set.

Traps tested: Any value past q3 treated as outlier · Wrong quartile fence used · Outlier status confused with being the extreme value

For this data (Q1 = 40, Q3 = 52), a value is defined as an outlier if it lies more than 2 × IQR beyond the nearer quartile.

Is the value 73 an outlier, under the rule as stated in THIS question?

• No — the upper fence is $52 + 2(12) = 76$, and $73 < 76$

  Correct. $IQR = 52 - 40 = 12$, and the question's own rule uses a multiplier of 2, not 1.5: $52 + 2(12) = 76$, and 73 falls short of it.

• BYes — the upper fence is $52 + 1.5(12) = 70$, and $73 > 70$

  This applies the standard $1.5 \times IQR$ rule from memory instead of the $2 \times IQR$ rule this specific question states. It's a confirmed, real error: "a small number of candidates opted to apply the more commonly used outlier formula... rather than the one quoted in the question" (Jun 2022, Q1(b)) — and here, unlike in that example, using the wrong multiplier changes the actual conclusion, not just the working.

• CYes — any value more than $Q_3$ plus the IQR itself ($52 + 12 = 64$) counts as an outlier

  This is a third formula, matching neither the rule this question gives nor the more familiar $1.5 \times IQR$ default — it isn't derived from anything stated in the question.

• DCannot be determined without seeing the full data set

  The rule plus $Q_1$ and $Q_3$ is everything the calculation needs — the individual raw values play no part in deciding whether 73 clears the fence.

Traps tested: Remembered formula overrides the rule given in the question · Invented rule not matching the question · Overclaims uncertainty

A box plot has $Q_1 = 15$, median $= 19$, $Q_3 = 21$, with whiskers reaching 5 (minimum) and 24 (maximum, not an outlier). What does this suggest about the skew of the distribution?

• Negative (left) skew — both the box (median $-$ Q1 $= 4$, against Q3 $-$ median $= 2$) and the whiskers (Q1 $-$ min $= 10$, against max $-$ Q3 $= 3$) show more spread on the LOW side

  Correct. Both comparisons agree: the lower half of the box is wider, and the lower whisker is longer, so more of the spread sits below the median than above it — a longer left tail, i.e. negative skew.

• BPositive (right) skew

  The upper half of the box (median to $Q_3$) is the SMALLER of the two gaps here, and the upper whisker is the SHORTER of the two — both point toward more spread on the low side, not the high side. Positive skew would need the opposite pattern.

• CRoughly symmetric, since the median sits inside the box, not at either edge

  The median sitting somewhere inside the box is true of almost every box plot and says nothing about symmetry on its own — comparing the actual gaps (4 against 2, and 10 against 3) is what decides it, and neither pair is close to equal.

• DNothing can be concluded about skew from a box plot — only from the raw data

  Reading skew direction from a box plot's proportions is exactly the spec 2.4 skill being tested here — a legitimate, examinable comparison, not a shortcut that needs the raw data to back it up.

Traps tested: Skew direction reversed · Skew read from visual impression not figures · Box plot skew reading dismissed

A data set has $Q_1 = 10$, $Q_3 = 25$. Using $1.5 \times IQR$, the upper fence is 47.5. The highest value in the data that is NOT an outlier is 40; one value, 55, is an outlier. Where does the upper whisker end, and where is 55 shown?

• The whisker ends at 40; 55 is plotted separately, beyond the end of the whisker

  Correct. The whisker only ever reaches an actual value in the data that survives the outlier rule; 55 fails that rule, so it is shown as an isolated point instead.

• BThe whisker ends at 47.5 (the fence); 55 is plotted separately

  47.5 is a boundary used to TEST values, not a value that exists in the data — a whisker only ever ends at a real data point, and the confirmed record shows exactly this error happening in practice.

• CThe whisker ends at 55, and no point is plotted separately

  This folds the outlier back into the ordinary range, defeating the entire purpose of identifying it — an outlier is shown SEPARATELY, precisely because it isn't a natural continuation of the rest of the data.

• DThe whisker ends at $Q_3$ (25); there is no upper whisker at all once an outlier exists

  An outlier shortens the whisker to the nearest genuine non-outlier value — it doesn't remove the whisker. Here that value is 40, well above $Q_3$.

Traps tested: Whisker drawn to fence not to data · Outlier included in the whisker · Whisker omitted entirely once an outlier exists

Data set A has a higher median (42 minutes) than data set B (35 minutes). Which response to "compare the typical time recorded by A and B" would be awarded full marks?

• "Data set A had the higher median (42 minutes, against 35 minutes for B), so on average the times recorded in A were longer."

  Correct — it names the statistic being compared, gives both figures, and states the direction. This is the exact shape of answer a real mark scheme requires: "a reference to a named statistic and supporting figures" (Jun 2024, Q1(e)).

• B"Data set A has a higher average."

  No figures, and 'average' doesn't say which one. A confirmed real error record puts it plainly: "surprisingly too many students failed to give supporting figures" (Jun 2024, Q1(e)) — this is exactly that failure.

• C"The two data sets are broadly similar."

  This doesn't even attempt the comparison the question asked for (typical time), let alone give a direction or a figure — a technically-vague observation like this scores nothing.

• D"Data set A shows a positive correlation with data set B."

  Correlation measures the relationship between two paired VARIABLES measured on the same individuals — it doesn't apply to comparing two separate groups' typical values at all, so this answer is off-topic rather than merely imprecise.

Traps tested: Comparison missing supporting figures · Comparison too vague no direction · Irrelevant statistical concept invoked