Determine outliers using the interquartile range, describe and interpret the shape and features of a distribution (symmetry, skewness, modality, centre, spread and outliers) and compare data displays using these features

A full-mark response computes the IQR, $26 - 18 = 8$, then the upper fence, $Q_3 + 1.5 \times \text{IQR} = 26 + 1.5 \times 8 = 26 + 12 = 38$.

It then states the comparison explicitly: $44 > 38$, therefore $44$ is an outlier.

Markers award one mark for the IQR, one for correctly applying the $1.5 \times \text{IQR}$ rule to get the fence $38$, and one for the comparison and conclusion. A bare "yes" with no fence shown scores poorly, even if the answer is correct.

Part (a): the distribution is positively skewed (skewed to the right) - the long tail points to the right.

Part (b): the mean is larger than the median, because the small number of very high earners in the right tail pull the mean up, while the median (the middle position) is barely affected.

Part (c): the median better represents a typical worker, because it is resistant to the few extreme high incomes that distort the mean.

Markers reward the correct shape name, the correct mean-versus-median direction WITH the tail reasoning, and a justified choice of the median for a typical value. Naming the skew the wrong way (a common slip) loses the part (a) and part (b) marks together.

Part (a): IQR $= 30 - 24 = 6$; upper fence $= 30 + 1.5 \times 6 = 30 + 9 = 39$; since $41 > 39$, the value $41$ is an outlier.

Part (b): shape is positively skewed (the upper tail is stretched by the high value); centre is a median of $27$ degrees (preferred over the mean because of the outlier); spread is an IQR of $6$ degrees for the middle half, with a full range of $41 - 19 = 22$ degrees inflated by the outlier.

Markers award the outlier test (fence plus comparison), then one mark each for a correctly justified shape, centre and spread. Quoting the median rather than the mean for the centre of a skewed set is part of what is rewarded here.

Part (a) - interquartile range. The IQR is the upper quartile minus the lower quartile:

Part (b) - the two fences. The lower fence sits $1.5 \times \text{IQR}$ below $Q_1$ and the upper fence sits $1.5 \times \text{IQR}$ above $Q_3$. Here $1.5 \times 4 = 6$, so

Any value below $7$ or above $23$ would be flagged as an outlier. (Check: the fences sit one full $1.5 \times \text{IQR}$ step outside each quartile, so they should straddle the quartiles symmetrically, and $7$ and $23$ are each $6$ away from $13$ and $17$.)

Part (a) - mean versus median. In a positively skewed set the few large values in the tail pull the mean upward, while the median (a position, not a total) barely moves. So the mean is greater than the median.

Part (b) - the tail. Positive skew means the data is stretched out towards the high (positive) end, so the long tail points to the right. The bulk of the data is bunched on the left with a few large values trailing off to the right. (Memory hook: the skew is named for the direction the tail points, so "positive / right skew" has its tail on the right.)

Count the peaks. Two clear, separate peaks means the distribution is bimodal.

Interpret the peaks. Two peaks usually signals two groups combined into one data set. For heights, a plausible explanation is that the data mixes two subgroups, for example male and female students, each clustering around its own typical height. (A single-peak set is unimodal; no clear peak is sometimes called uniform.)

Find the IQR. Subtract the quartiles:

Find the upper fence. A high value is tested against the upper fence, $Q_3 + 1.5 \times \text{IQR}$. Here $1.5 \times 6 = 9$, so

Compare and conclude. The value $40$ is greater than the upper fence $26$, so

means $40$ is an outlier by the $1.5 \times \text{IQR}$ rule. (For completeness the lower fence is $11 - 9 = 2$, and the smallest value $8$ is above $2$, so there is no low outlier. State the fence you cross, then the comparison: that is the line markers reward.)

Find the IQR. Subtract the quartiles:

Find both fences. With $1.5 \times 6 = 9$:

Compare each extreme value. The smallest value is $3$ and the largest is $46$:

so both fall outside their fences. The outliers are $3$ minutes and $46$ minutes. (Check: every other value lies between $13$ and $37$, so exactly two points are flagged, one at each end. Always test both fences, not just the obvious big value.)

Part (a) - the IQR. Subtract the quartiles:

Part (b) - test the value $72$. A high value is tested against the upper fence. With $1.5 \times 10 = 15$:

Since

the value $72$ is an outlier.

Part (c) - describe the distribution. Work through the four features in order:

• Shape: ignoring the outlier the values rise fairly evenly, but the long upper tail (one value far above the rest) makes the data positively skewed (skewed to the right).
• Centre: the median is $31$ minutes; the median is the better measure of centre here because the outlier would inflate the mean.
• Spread: the IQR is $10$ minutes (the middle half of commuters are within a $10$ minute band); the full range is $72 - 22 = 50$ minutes, stretched by the outlier.
• Outliers: there is one outlier at $72$ minutes, well above the upper fence of $51$; this is a genuine value (a very long commute, perhaps a transport delay) rather than an error, so it should be kept but noted.

So the travel times are positively skewed with a median of $31$ minutes, an IQR of $10$ minutes, and one high outlier at $72$ minutes. (Check: the median $31$ sits inside the quartiles $26$ and $36$ as it must, and only the single value $72$ crosses a fence.)

Part (a) - test Class A for outliers. First the IQR:

With $1.5 \times 6 = 9$, the fences are

The smallest value is $4$ and the largest is $20$, and

so both extremes lie inside the fences: Class A has no outliers.

Part (b) - shape of each class. For Class A the values are spread fairly evenly and the mean and median are close, so the shape is roughly symmetric. For Class B the marks bunch up near the top (the maximum is $20$) with a tail of lower marks trailing down to $12$, so Class B is negatively skewed (skewed to the left).

Part (c) - compare centre and spread. Class B has the higher centre (median $17$ versus $10$), so Class B performed better overall. Class B is also more consistent: its IQR is $Q_3 - Q_1 = 18 - 15 = 3$, smaller than Class A's IQR of $6$, so Class B's middle marks are more tightly clustered. (Check: comparing like with like, both statements use the median for centre and the IQR for spread, which is the safe pairing when a set may be skewed.)