Boxplots and distributions

Min $= 5$ , $Q_1 = 10$ (median of $5, 8, 12, 15$ : average of $8$ and $12$ ), median $= 18$ (5th value), $Q_3 = 23.5$ (median of $20, 22, 25, 30$ : average of $22$ and $25$ ), max $= 30$ .
IQR $= Q_3 - Q_1 = 23.5 - 10 = 13.5$ .
IQR $= 30 - 10 = 20$ . Lower fence $= 10 - 1.5 \times 20 = 10 - 30 = -20$ . Upper fence $= 30 + 1.5 \times 20 = 30 + 30 = 60$ .
Boxplot with whisker at $2$ , box from $8$ to $20$ , median line at $14$ , whisker to $28$ .
Positively skewed (the data is more spread out above the median than below).
$\dfrac{40}{100} = 0.4$ or $40\%$ .
$Q_3 = Q_1 + \text{IQR} = 25 + 12 = 37$ .
False. The median is only centred if the distribution is symmetric. In a skewed distribution, the median is closer to one quartile.

Five-number summary: min $= 152$ , $Q_1 = 160$ (median of positions 1—7), median $= 167$ (8th value), $Q_3 = 178$ (median of positions 9—15), max $= 198$ . IQR $= 178 - 160 = 18$ . Upper fence $= 178 + 1.5 \times 18 = 178 + 27 = 205$ . Lower fence $= 160 - 27 = 133$ . Both $195$ and $198$ are below $205$ , so there are no outliers.
Class X has a higher median ( $35$ vs $30$ ) and a smaller IQR ( $40 - 28 = 12$ vs $42 - 25 = 17$ ). Class X performed better overall and more consistently. Class Y has a higher maximum ( $50$ ) but also a lower minimum ( $20$ vs $15$ — actually Class Y min is higher). Both classes have similar ranges.
$P(\text{bus} \mid \text{Year 9}) = \dfrac{30}{60} = 0.5$ . $P(\text{bus} \mid \text{Year 10}) = \dfrac{15}{60} = 0.25$ . Year 9 students are twice as likely to catch the bus as Year 10 students.
House prices are often positively skewed: most houses cluster around a typical value, but a few very expensive properties pull the mean upward. The median is resistant to extreme values and better represents the “typical” house price. A few multi-million-dollar sales can raise the mean significantly without affecting most buyers’ experience.
Ordered: $3, 5, 7, 8, 10, 12, 14, 15, 50$ . $Q_1 = \dfrac{5 + 7}{2} = 6$ . $Q_3 = \dfrac{14 + 15}{2} = 14.5$ . IQR $= 14.5 - 6 = 8.5$ . Upper fence $= 14.5 + 1.5 \times 8.5 = 14.5 + 12.75 = 27.25$ . Since $50 > 27.25$ , the value $50$ is an outlier.

Correlation does not prove causation because a third variable could explain both. For example, students from families with higher socioeconomic status may be more likely to eat breakfast and have access to tutoring, quiet study spaces, and parental support. The breakfast itself may not cause higher scores; the underlying variable (family resources) may drive both outcomes.
Factory A is more consistent (IQR $= 0.8$ mm vs $2.5$ mm). Factory B has a median closer to the target of $50.0$ mm. Trade-off: Factory A produces bolts of very uniform length but slightly above target; Factory B hits the target on average but with much greater variability. If precision matters (e.g. safety-critical components), Factory A is preferable despite the slight offset, which could be corrected by recalibrating.
Sources of bias: (i) Self-selection bias — only people who chose to respond are counted; those with strong opinions may be overrepresented. (ii) Platform bias — users of that particular social media platform may not be representative of the general population (e.g. younger demographic, specific political leanings). Both could overestimate or underestimate true support depending on the platform’s user base.
Range uses only the two most extreme values, so a single outlier can make the range very large. IQR uses the middle $50\%$ and is resistant to outliers. Example: $\{10, 12, 14, 15, 16, 18, 100\}$ . Range $= 100 - 10 = 90$ (misleadingly large). IQR $= 18 - 12 = 6$ (reflects the actual spread of most data).

Adding $60$ : the minimum stays at $15$ (or whatever it was), the maximum becomes $60$ . IQR $= 30 - 15 = 15$ . Upper fence $= 30 + 1.5 \times 15 = 52.5$ . Since $60 > 52.5$ , yes, $60$ is an outlier. The median may shift slightly upward (from the average of the 10th and 11th values to the 11th value of the new 21-value set). $Q_1$ and $Q_3$ may shift slightly but the effect is small.
Symmetric: both whiskers are approximately equal length, extending evenly from the box. Positively skewed: the right whisker is much longer than the left; data extends further above $Q_3$ than below $Q_1$ . Both have the same box size (IQR $= 10$ ) and median ( $50$ ), but the skewed version has the median closer to $Q_1$ .
The argument has some merit: in a normal distribution, about $0.7\%$ of values lie beyond $Q_1 - 1.5 \times \text{IQR}$ or $Q_3 + 1.5 \times \text{IQR}$ , so we might expect roughly $0.007 \times 500 \approx 3$ — $4$ outliers. However, if the data is truly free of measurement errors and follows a tight distribution, it is possible (though unlikely) to have no outliers. The researcher should report the distribution shape and explain why outliers are absent.
With association: Sport-yes/Sleep-yes $= 30$ , Sport-yes/Sleep-no $= 10$ , Sport-no/Sleep-yes $= 15$ , Sport-no/Sleep-no $= 25$ . $P(\text{sleep} \mid \text{sport}) = \dfrac{30}{40} = 0.75$ , $P(\text{sleep} \mid \text{no sport}) = \dfrac{15}{40} = 0.375$ . These differ, showing an association. No association: Sport-yes/Sleep-yes $= 22.5$ , Sport-yes/Sleep-no $= 17.5$ , Sport-no/Sleep-yes $= 22.5$ , Sport-no/Sleep-no $= 17.5$ (using whole numbers: $23, 17, 22, 18$ ). Now $P(\text{sleep} \mid \text{sport}) \approx P(\text{sleep} \mid \text{no sport}) \approx 0.5625$ , so the variables are approximately independent.

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