Why Summary Measures Belong in Pairs
When comparing two distributions, it is not enough to report two centers or two measures of spread. The choice of summaries affects what the comparison says. As in Choosing Appropriate Summary Statistics, inspect the distributions first: shape and unusual values help determine which summaries describe them well.
A useful convention is to pair the mean with the standard deviation, and the median with the interquartile range (IQR). These pairs work together because each tells a compatible story about center and spread. The mean is an arithmetic balance point, and the standard deviation describes distances from that mean. When the number of observations is odd, the median is the middle observation; when it is even, the median is the average of the two middle observations. The IQR describes the width of the middle half.
This convention is not a rule that makes other calculations impossible. You could report a mean and an IQR for one group, for example. But comparing that group’s mean with another group’s median would compare different descriptions of center. It would be difficult to explain the result as a difference between the groups’ centers. Likewise, comparing a standard deviation with an IQR would compare different descriptions of spread.
The mean and standard deviation are both affected by extreme observations. They tend to be useful when distributions are roughly symmetric and do not have strong outliers. The median and IQR are resistant, so they are often more useful when a distribution is skewed or has outliers. As discussed in Resistant Versus Non-Resistant Statistics, the choice should fit the data—not just the calculation that is easiest to make.
What Each Pair Helps You Describe
With a mean-and-standard-deviation comparison, the difference between the means describes how far apart the groups’ arithmetic averages are, in the variable’s units. The difference between the standard deviations describes how far apart their spreads around those means are, also in the variable’s units. Standard deviation is not a guarantee that all observations lie within a particular distance of the mean.
With a median-and-IQR comparison, the difference between the medians describes how far apart the groups’ middle values are. The difference between the IQRs compares the widths of their middle halves. An IQR does not describe the full range of the observations; the minimum and maximum may be much farther away.
As in Comparing Centers Using Medians and Means and Comparing Spread Using IQR and Standard Deviation, name both groups and the feature being compared. A difference can be positive or negative depending on the subtraction order. State the order so a reader knows which group has the greater summary.
Worked Example: Comparing Symmetric Distributions With Means and Standard Deviations
A fictional school compares the number of minutes students at two after-school clubs spend reading during a session. The observations are:
Club Cedar: 8, 9, 10, 11, 12 minutes
Club Maple: 10, 11, 12, 13, 14 minutes
Question: Which summary pair is appropriate, and how do the centers and spreads compare?
Choose the pair. Both lists are symmetric around their middle values and have no observations far from the rest. A mean-and-standard-deviation pair is reasonable. We can also check that the medians and IQRs tell a compatible story.
Calculate the means. Cedar’s sum is \(8+9+10+11+12=50\), so its mean is \(50/5=10\) minutes. Maple’s sum is \(10+11+12+13+14=60\), so its mean is \(60/5=12\) minutes. The mean difference, Maple minus Cedar, is \(12-10=2\) minutes.
Calculate the sample standard deviations. For Cedar, the squared deviations from the mean are \(4,1,0,1,4\), which sum to 10. Thus
For Maple, the squared deviations from its mean are also \(4,1,0,1,4\), so its sample standard deviation is also \(\sqrt{10/4}\approx1.581\) minutes. The spread difference is \(1.581-1.581=0\) minutes.
Check the other pair. Each median is the middle observation: 10 minutes for Cedar and 12 minutes for Maple. Using the median-of-halves quartile method, Cedar has \(Q_1=8.5\) and \(Q_3=11.5\), so its IQR is \(11.5-8.5=3\) minutes. Maple has \(Q_1=10.5\) and \(Q_3=13.5\), so its IQR is also 3 minutes.
Conclude. Maple’s mean is 2 minutes higher than Cedar’s, while the groups have the same sample standard deviation, about 1.581 minutes. Their medians also differ by 2 minutes, and their IQRs are both 3 minutes. In context, Maple students in these observations spent more time reading on average and at the median, while the two groups had equal spread by either matching measure.
Why Skewness and Outliers Can Change the Comparison
A high outlier can pull a mean upward and increase a standard deviation, even when most observations remain near their original values. The median and IQR depend more on the ordered middle observations, so a single extreme value often has little effect on them. This difference matters when describing groups: the two pairs may give noticeably different impressions of center and spread.
The goal is not to hide one set of summaries. Instead, choose the pair that best represents the distributions, and explain why it fits. If the data are skewed or contain an outlier, the median and IQR are usually the more suitable pair for comparing typical location and middle-half spread. If the question specifically asks about means, report them, but acknowledge that an extreme value may influence the comparison.
Worked Example: An Outlier Changes the Mean Comparison
A fictional community garden compares the number of tomatoes harvested from 10 plants in each of two plots:
Plot A: 2, 3, 4, 4, 5, 5, 6, 7, 8, 30 tomatoes
Plot B: 2, 3, 4, 5, 5, 6, 7, 8, 9, 20 tomatoes
Question: What does each summary pair say about the groups, and which pair is more appropriate?
Inspect the data. Both plots have a high value well above most of their observations. The distributions have long right tails, so the median and IQR are more resistant summaries of their centers and middle-half spreads.
Calculate the means and sample standard deviations. Plot A has sum 74 and sum of squared observations 1144. Its mean is \(74/10=7.4\) tomatoes. Using the computational form of the sample variance,
The arithmetic checks because \(74^2/10=547.6\) and \(1144-547.6=596.4\). Plot B has sum 69 and sum of squared observations 709. Its mean is \(69/10=6.9\) tomatoes, and
Here, \(69^2/10=476.1\), and \(709-476.1=232.9\), confirming the substitution. By the mean, Plot A is higher by \(7.4-6.9=0.5\) tomatoes. Its standard deviation is higher by about \(8.140-5.087=3.053\) tomatoes.
Calculate the medians and IQRs. Plot A’s median is the average of its fifth and sixth values: \((5+5)/2=5\) tomatoes. Its lower half is 2, 3, 4, 4, 5, with median \(Q_1=4\); its upper half is 5, 6, 7, 8, 30, with median \(Q_3=7\). Thus, Plot A’s IQR is \(7-4=3\) tomatoes.
Plot B’s median is \((5+6)/2=5.5\) tomatoes. Its lower half has median \(Q_1=4\), and its upper half has median \(Q_3=8\), so its IQR is \(8-4=4\) tomatoes. By these resistant summaries, Plot B’s median is \(0.5\) tomatoes higher, and its middle-half spread is 1 tomato wider.
Conclude. The mean suggests Plot A has the higher average harvest, but the medians suggest Plot B has the higher center. The high values influence the means, particularly Plot A’s value of 30. Because both distributions are right-skewed and contain high values, the median-and-IQR comparison better describes their typical centers and middle-half spreads: Plot B’s median is 0.5 tomato higher, and its IQR is 1 tomato wider. These statements describe the observed plants, not a cause for the difference.
Using the Same Pair for Both Groups
Choosing one summary pair for each group separately can lead to an awkward comparison. Imagine that one group looks roughly symmetric, but the other has a strong right tail. The mean and standard deviation might suit the first group, while the median and IQR might suit the second. Yet a direct comparison of those different summaries would not give a clean center-to-center or spread-to-spread comparison.
In that situation, state what the shapes suggest and consider whether the resistant pair—median and IQR—can describe both groups usefully. Medians and IQRs can be calculated for any quantitative distribution, even if the distribution is symmetric. The point is to use summaries that remain informative for both groups, not to force the same shape on them.
If the task requires means and standard deviations, calculate and compare those matching measures, but note when skewness or outliers may influence them. Do not replace one group’s mean with its median just because the two values are far apart. The difference may reveal something important about the group’s shape, but it is not a valid comparison of the same measure of center.
Worked Example: Comparing Commute Times With a Resistant Pair
A fictional student group records one-way commute times, in minutes, for eight students using each of two routes:
Route A: 12, 14, 15, 16, 18, 20, 22, 43 minutes
Route B: 10, 12, 14, 15, 16, 18, 20, 25 minutes
Question: Compare center and spread, and explain why the choice of pair matters.
Choose and calculate the median-and-IQR pair. Route A has a high value of 43 minutes compared with most of its other times, so its distribution has a long right tail. The median and IQR give a resistant comparison. Route A’s median is \((16+18)/2=17\) minutes. Its lower half, 12, 14, 15, 16, has median \(Q_1=(14+15)/2=14.5\); its upper half has median \(Q_3=(20+22)/2=21\). Therefore, its IQR is \(21-14.5=6.5\) minutes.
Route B’s median is \((15+16)/2=15.5\) minutes. Its lower half has median \(Q_1=(12+14)/2=13\), and its upper half has median \(Q_3=(18+20)/2=19\). Therefore, its IQR is \(19-13=6\) minutes. Route A’s median is \(17-15.5=1.5\) minutes higher, and its IQR is \(6.5-6=0.5\) minute wider.
See what the mean-and-standard-deviation pair would say. Route A’s sum is 160 minutes and its sum of squared times is 3878. Its mean and sample standard deviation are
For Route B, the sum is 130 minutes and the sum of squared times is 2270. Thus,
The mean difference is \(20-16.25=3.75\) minutes, and the standard-deviation difference is about \(9.841-4.743=5.098\) minutes. The mean and standard deviation are pulled upward for Route A by the 43-minute commute. The calculations are consistent: \(160^2/8=3200\), leaving \(678\) for the variance numerator; \(130^2/8=2112.5\), leaving \(157.5\), and \(157.5/7=22.5\).
Conclude. The mean-and-standard-deviation pair describes a larger difference between the routes than the median-and-IQR pair does. Because Route A has a long right tail, the resistant pair is more appropriate for comparing typical commute time and the spread of the middle half. In these observations, Route A’s median is 1.5 minutes higher and its IQR is 0.5 minute wider than Route B’s. The mean-and-standard-deviation results can also be reported if relevant, with the influence of the long commute made clear.
Common Mistakes and AP Exam Tips
- Comparing different measures of center. A mean from one group and a median from another do not form a direct comparison of centers. Compare mean with mean or median with median.
- Comparing different measures of spread. An IQR and a standard deviation describe spread differently. Compare IQR with IQR or standard deviation with standard deviation.
- Choosing summaries before inspecting the distributions. Check shape and unusual values first. Skewness or outliers often favor medians and IQRs; roughly symmetric distributions without strong outliers often support means and standard deviations.
- Claiming that resistant summaries are always best. They are useful when extreme values or skewness are present, but the summary should still answer the question being asked. Explain your choice in relation to the distributions.
- Reporting a difference without its direction or units. State the subtraction order and variable units. “Route A’s median commute is 1.5 minutes higher” is clearer than “the difference is 1.5.”
- Assuming a difference in spread proves a difference in every part of the distributions. A larger IQR means a wider middle half, not necessarily a larger full range. A larger standard deviation does not mean every observation is farther from the mean.
A full-credit comparison names the groups, identifies the matching summaries, gives the difference with units and direction, and connects the choice of summaries to the distributions’ shapes or unusual values. Keep the conclusion focused on what those summaries describe.
Check Your Understanding
For each question, name the summary pair or comparison that fits the situation.
- Two roughly symmetric distributions have no strong outliers. Which center-and-spread pair is reasonable to compare?
- Two right-skewed distributions contain high outliers. Which pair is likely to give a more resistant comparison, and why?
- One group’s mean is 24 minutes and another group’s median is 21 minutes. Why is \(24-21\) not a direct comparison of the groups’ centers?
- Group A has an IQR of 8 units and Group B has an IQR of 5 units. What does the difference say about their middle halves, and what does it not establish about their full ranges?
- A group’s mean is much higher than its median. What feature of its distribution might help explain this, and which center-and-spread pair might be more appropriate for describing a typical value?