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AP Statistics  /  Unit 1: Exploring One-Variable Data and Collecting Data  /  Topic 1.9
NUM8ERS study notes · Topic 1.9

Comparisons of the Distributionsfor One Quantitative Variable

Compare two groups with clear graphs and evidence. Learn to explain differences in typical values and variability, then use z-scores to understand where an individual value stands within its own group.

2026–27 curriculum4 worked examples8 practice questionsGraphs + comparisons + z-scores

By the end of this lesson, you should be able to:

  • Compare distributions of the same quantitative variable using appropriate graphs and summaries.
  • Write a direct comparison of center, variability, shape and unusual features in context.
  • Support or challenge a claim using evidence from more than one distribution.
  • Calculate and interpret z-scores, including negative values.
  • Distinguish a higher raw value from a higher relative position.

Before you start: Know how to describe a distribution, find numerical summaries and read a boxplot. Review Topic 1.8 if boxplots are still unfamiliar.

First time learning this? Begin with the two-school example, then work through the comparison checklist and z-score number line.

Here to revise? Use the quick checklist, then try the practice questions before opening the solutions.

01 · Start here

The concept in 60 seconds

A comparison connects two or more groups using the same quantitative variable. Instead of writing separate descriptions and leaving the reader to work out the difference, say how the groups are similar or different and give evidence.

Compare whole distributions

Ask: which group has a higher typical value? Which has more variability? How do their shapes and unusual features compare?

Example: “Birch High has a higher median travel time than Cedar High: 2727 versus 19 min19 \,\mathrm{min}.”

Compare individual positions

Ask: how far is this value from its own group’s mean, measured in standard deviations? A z-score answers that question.

Example: a score 22 standard deviations above its test mean has a higher standardized position than one only 11 standard deviation above its test mean.

Keep the two tasks distinct: a boxplot comparison describes groups. A z-score describes an individual value relative to a reference distribution.

Higher is not always preferable. A higher test score may be desirable; a shorter race time may be desirable. Explain what “better” means in the context before making a judgment.

02 · One variable, two groups

Compare school travel times

2 fictional schools2\text{ fictional schools} record students’ one-way home-to-school travel time on the Tuesday being studied, rounded to the nearest minute. We use the same 2020 Cedar High observations from earlier lessons and a new sample of 1212 Birch High observations.

See the recorded values

Cedar High, n=20n = 20: 88, 1010, 1212, 1414, 1414, 1515, 1616, 1818, 1818, 1818, 2020, 2222, 2424, 2525, 2626, 2828, 3030, 3232, 3535, 4242.

Birch High, n=12n = 12: 1818, 2020, 2222, 2424, 2424, 2626, 2828, 3030, 3030, 3232, 3434, 3636.

2 school samples2\text{ school samples} on the same minute scale

11 dot represents 11 student · stacks show repeated times

Aligned dotplots of Cedar and Birch school travel timesCedar values [8, 10, 12, 14, 14, 15, 16, 18, 18, 18, 20, 22, 24, 25, 26, 28, 30, 32, 35, 42]. Birch values [18, 20, 22, 24, 24, 26, 28, 30, 30, 32, 34, 36]. Both axes run from 5 to 45 minutes; each dot is one student. Cedar has median 19, Birch median 27. The samples overlap. 5 10 15 20 25 30 35 40 45 Recorded one-way travel time (minutes) Cedar High · n = 20 5 10 15 20 25 30 35 40 45 Recorded one-way travel time (minutes) Birch High · n = 12

On a small screen, scroll the plot sideways to read the full scale.

The dotplots retain every observation. Compare their positions and patterns on the shared numerical scale rather than the total number of dots. Invented teaching data.

The Cedar observations extend farther toward the low end and have a longer upper tail. Birch’s observations are arranged symmetrically around 27 min27 \,\mathrm{min}. The groups overlap: a higher typical time at Birch does not mean every Birch student travels longer than every Cedar student.

Compare the medians and the middle halves

Modified boxplots · actual non-outlier whisker endpoints · 1.5×IQR⁡1.5\times\operatorname{IQR} rule

Compare the medians and the middle halvesShared numerical scale. Cedar: n=20; five-number summary 8, 14.5, 19.0, 27.0, 42; IQR 12.5; fences [-4.25, 45.75]; modified whiskers [8, 42]; flagged observations []. Birch: n=12; five-number summary 18, 23.0, 27.0, 31.0, 36; IQR 8.0; fences [11.0, 43.0]; modified whiskers [18, 36]; flagged observations []. 5 10 15 20 25 30 35 40 45 Recorded one-way travel time (minutes) Cedar n = 20 Birch n = 12 Median 19 Median 27

On a small screen, scroll the plot sideways to read the full scale.

Birch’s median is higher, 2727 versus 19 min19 \,\mathrm{min}. Cedar’s box is longer along the minute axis: IQR⁡\operatorname{IQR} 12.512.5 versus 8 min8 \,\mathrm{min}. Neither sample has a flagged outlier. Sample sizes are explicitly added labels, rather than information encoded by a standard box. Invented teaching data.
Calculated summaries of these fictional samples; all times and spreads are in minutes.
SummaryCedar HighBirch High
Sample size20201212
Median19192727
Q1Q_{1} to Q3Q_{3}14.514.5 to 27272323 to 3131
IQR⁡\operatorname{IQR}12.512.588
Minimum to maximum88 to 42421818 to 3636
Range34341818
Mean21.3521.3527.0027.00
Sample standard deviation, ss8.878.875.625.62

Predict: which school has longer typical travel times, and which has more variability in its middle half?

Check your prediction

Birch has longer typical travel times, using its median of 27 min27 \,\mathrm{min} versus Cedar’s 1919. Cedar has more variability in the middle half, because its IQR⁡\operatorname{IQR} is 12.5 min12.5 \,\mathrm{min} versus Birch’s 88. These are two different comparisons: a higher center does not automatically mean greater spread.

Quartiles here use the median of each half of the ordered data. For odd sample sizes, exclude the overall median from the halves. Modified boxplots use the 1.5×IQR⁡1.5\times\operatorname{IQR} outlier rule. Neither school sample has a flagged observation under that rule.

03 · Read the evidence

Choose a graph and compare the important features

SOCS—shape, outliers, center and spread—is a useful reminder. Also look for clusters or gaps when the graph displays them. Let the question and available evidence determine what your answer needs; the mnemonic is not an instruction to invent missing information.

Center: which is typically higher?Name the median or mean and compare values with units. “Birch’s median is 8 min8 \,\mathrm{min} higher” is more informative than “Birch is higher.”
Spread: which varies more?Name the measure: IQR⁡\operatorname{IQR}, range or standard deviation. Different measures can lead to different comparisons.
Shape: how do patterns differ?Compare symmetry, skewness or peaks when visible. Identify the direction of a tail rather than the location of the tallest bar.
Unusual features: what stands out?Compare outliers, clusters and gaps that the display supports. If using a numerical outlier rule, name it.

What can each graph tell you?

Choose a display that retains the information needed for the question.
DisplayUseful forLimitation to remember
Aligned dotplotsSeeing individual values, repeated observations, clusters, gaps and overall patterns in small samples.Large samples can become crowded. Use a common numerical scale.
HistogramsComparing overall shape and variability, especially with larger samples.Bins hide exact values. Match bin boundaries and consider relative frequencies when sample sizes differ.
Back-to-back stem-and-leaf plotComparing two small datasets while retaining every recorded value.Read the two sides using the key; leaves on the left are usually listed in descending order.
Parallel boxplotsQuick comparisons of medians, IQRs, endpoints and flagged outliers. Asymmetry can suggest skewness.Standard boxes hide peaks, clusters, gaps, exact means and sample sizes. Use fuller graphs for detailed shape claims.

Read a back-to-back stem-and-leaf plot

Here is a separate teaching example of travel times from 2 small groups2\text{ small groups}. Stems are tens and leaves are ones. Every leaf represents 1 student\text{Every leaf represents }1\text{ student}.

Key: on the left, leaf 22 beside stem 11 means 12 min12 \,\mathrm{min} for Group A; on the right, stem 11 beside leaf 88 means 18 min18 \,\mathrm{min} for Group B.

Group A (n=10)∣Stem∣Group B (n=10)8  8  6  4  4  2∣1∣86  4  2  0∣2∣0  2  4  4  6  8∣3∣0  0  2\begin{array}{rcccl}\text{Group A }(n=10)&\mid&\text{Stem}&\mid&\text{Group B }(n=10)\\8\;8\;6\;4\;4\;2&\mid&1&\mid&8\\6\;4\;2\;0&\mid&2&\mid&0\;2\;4\;4\;6\;8\\&\mid&3&\mid&0\;0\;2\end{array}

Group A has 66 observations in the teens, while Group B has only 1\text{has only }1. Group B’s values generally sit higher on the common minute scale. Its 33 observations in the thirties are 3030, 3030 and 32 min32 \,\mathrm{min}.

Check: what are the medians of these 2 groups\text{these }2\text{ groups}?

Group A’s middle 2 values\text{middle }2\text{ values} are 1818 and 1818, so its median is 18 min18 \,\mathrm{min}. Group B’s middle 2 are\text{middle }2\text{ are} 2424 and 2626, so its median is 25 min25 \,\mathrm{min}. Group B’s median is 7 min7 \,\mathrm{min} higher. These values belong to this stem-and-leaf example, rather than to the Cedar–Birch samples.

04 · Make a fair comparison

Compare numerical summaries and graph scales fairly

Use the same feature on both sides

Compare a median with a median, an IQR⁡\operatorname{IQR} with an IQR⁡\operatorname{IQR}, and a standard deviation with a standard deviation. A statement such as “Cedar’s range is greater than Birch’s IQR⁡\operatorname{IQR}” mixes two different summaries and does not directly compare their variability using one measure.

Median and IQR⁡\operatorname{IQR} are often informative when a distribution is skewed or contains outliers because they are resistant to extreme values. Mean and standard deviation describe the average and variation around it and are often useful for relatively symmetric distributions without strong outliers. If a question asks for a specific summary, use it and explain any relevant limitation.

The school table supports a comparison of means and standard deviations as well as medians and IQRs. These numbers were calculated from the observations; they were not read directly from the standard boxplots.

Common axes and bins keep the visual comparison honest

Graphs of the same variable should use compatible units and numerical scales. Histograms should use the same bin boundaries when comparing shapes. Check the labels before judging which distribution looks wider.

Different counts, matching proportions

Top: frequencies · bottom: percentages · matching 1010-minute bins

Frequency and relative-frequency histograms for unequal samplesGroup A sample [5, 12, 14, 16, 18, 22, 24, 26, 28, 35] has 10 observations, with counts 1,4,4,1 in bins [0,10),[10,20),[20,30),[30,40). Group B sample [5, 5, 12, 12, 14, 14, 16, 16, 18, 18, 22, 22, 24, 24, 26, 26, 28, 28, 35, 35] has 20 observations, with counts 2,8,8,2. Both relative frequency sequences are 10%,40%,40%,10%. Top plots share 0–10 count axis; bottom plots share 0–50 percent axis. 0 10 20 30 40 Recorded wait (minutes) 0 2 4 6 8 10 Frequency 1 4 4 1 Group A · n = 10 0 10 20 30 40 Recorded wait (minutes) 0 2 4 6 8 10 Frequency 2 8 8 2 Group B · n = 20 0 10 20 30 40 Recorded wait (minutes) 0% 10% 20% 30% 40% 50% Relative frequency 10% 40% 40% 10% Group A · n = 10 0 10 20 30 40 Recorded wait (minutes) 0% 10% 20% 30% 40% 50% Relative frequency 10% 40% 40% 10% Group B · n = 20

On a small screen, scroll the plot sideways to read the full scale.

Group B’s count bars are 2 times as tall2\text{ times as tall}. The matching percentage bars show the same proportional pattern in these bins; more observations do not by themselves mean greater spread. Invented teaching data.

In this separate fictional example, Group B has 2 times as many observations2\text{ times as many observations} as Group A. Every frequency bar is 2 times as tall2\text{ times as tall}, but the percentage in each matching bin is the same: 10%10\%, 40%40\%, 40%40\% and 10%10\%. Taller count bars alone do not show greater variability or a higher center.

Relative frequency=bin countgroup sample size\text{Relative frequency}=\frac{\text{bin count}}{\text{group sample size}}

Group A’s [10,20)[10,20) minute bin: 410=0.40=40%\frac{4}{10}=0.40=40\%.

Group B’s matching bin: 820=0.40=40%\frac{8}{20}=0.40=40\%.

The frequency panels share one count scale; the relative-frequency panels share one percentage scale. Matching bins use 0≤time<100\le\text{time}<10, 10≤time<2010\le\text{time}<20, 20≤time<3020\le\text{time}<30 and 30≤time<40 min30\le\text{time}<40\,\mathrm{min}.

Compare a claim with the feature it actually concerns

“More consistent” concerns variability, so use an appropriate spread measure. “Typically longer” concerns center. “Every observation is larger” concerns the actual ordering and overlap, so medians alone cannot establish it. “Caused by the school” needs information about how the study was designed; the graphs alone do not show causation.

05 · Put an individual value in context

Understand and calculate z-scores

A z-score tells you how many standard deviations a value lies above or below its distribution’s mean. Start with the difference from the mean, then divide by the standard deviation to put that difference on a common scale.

Population parameters: z=x−μσz=\frac{x-\mu}{\sigma}

Using sample summaries: z=x−xˉsz=\frac{x-\bar{x}}{s}

Read the symbols before substituting numbers.
SymbolMeaning
xxThe individual value being standardized.
μ\mu, pronounced “mu”The population mean.
σ\sigma, pronounced “sigma”The population standard deviation.
xˉ\bar{x} and ssThe sample mean and sample standard deviation, used when appropriate population parameters are unknown.

Use a positive standard deviation. If all observations are identical, the standard deviation is 0\text{standard deviation is }0 and the formula involves division by 0\text{division by }0, so a z-score is undefined.

The same value on raw and standardized scales

Reference population: μ=80 points\mu = 80 \,\mathrm{points} · σ=10 points\sigma = 10 \,\mathrm{points}

A 95-point score has z-score positive 1.5A raw-score number line runs 55 to 115 points. Scores 60,70,80,90,100,110 correspond to z values -2,-1,0,1,2,3. A highlighted value 95 corresponds to z=1.5 because (95-80)/10=1.5. The graph shows positions only, with no assumed distribution curve. 60 70 80 90 100 110 Raw score (points) z = -2 z = -1 z = 0 z = +1 z = +2 z = +3 95 points z = +1.5

On a small screen, scroll the plot sideways to read the full scale.

Each 1010-point step is 1 population standard deviation1\text{ population standard deviation}. The highlighted score is 15 points15 \,\mathrm{points} above the mean, or 1.51.5 standard deviations above it. No Normal shape is assumed. Invented teaching data.

Calculate first, then interpret

  1. Identify the reference group: make sure the mean and standard deviation belong to the distribution containing the value.
  2. Subtract the mean: here, 95−80=15 points95-80=15\,\mathrm{points} above the mean.
  3. Divide by the standard deviation: 1510=1.5\frac{15}{10}=1.5.
  4. Interpret: this 9595-point score is 1.51.5 standard deviations above the reference population mean.

Points divided by points cancel, so a z-score has no original measurement units. You report its meaning in standard deviations, rather than saying “z=1.5 pointsz = 1.5 \,\mathrm{points}.”

Positive zzThe value is above its reference mean.
Negative zzThe value is below its reference mean. A negative z-score can describe a positive measurement.
z=0z = 0The value equals the reference mean. It need not equal the median.
Larger ∣z∣\lvert z\rvertThe value is farther from its mean in standard-deviation units. Keep the sign when comparing which position is higher.

A z-score does not automatically give a percentile

You can standardize a value even when its distribution is not Normal. But z=1z = 1 does not automatically mean a particular percentage lies below it. A percentile or probability requires information about the distribution’s shape or its actual observations.

Can we work backward from a z-score?

Yes. Rearranging gives x=μ+zσx=\mu+z\sigma. If μ=80 points\mu = 80 \,\mathrm{points}, σ=10 points\sigma = 10 \,\mathrm{points} and z=−1.2z = -1.2, then x=80+(−1.2)(10)=68 pointsx=80+(-1.2)(10)=68\,\mathrm{points}. Check: 6868 is below the mean, so its negative z-score makes sense.

06 · Follow the reasoning

Worked examples

Example 1: compare two complete distributions

Task: compare the Cedar and Birch travel-time samples using the two graphs and numerical summaries above.

  1. Center: Birch’s median is 27 min27 \,\mathrm{min} versus Cedar’s 1919, so Birch has longer typical travel times by this measure.
  2. Variability: Cedar’s IQR⁡\operatorname{IQR} is 12.5 min12.5 \,\mathrm{min} versus Birch’s 88; Cedar’s range is also larger, 3434 versus 18 min18 \,\mathrm{min}.
  3. Shape: Cedar shows a longer upper tail, while Birch’s recorded times are symmetric around 27 min27 \,\mathrm{min}. Use the dotplots for the detailed pattern.
  4. Outliers: neither sample contains observations beyond its 1.5×IQR⁡1.5\times\operatorname{IQR} fences.
  5. Scope: describe these recorded samples. The displays alone do not establish a difference for all students or explain its cause.

Model response: “For these samples, Birch students have longer typical one-way travel times than Cedar students, with medians of 2727 and 19 min19 \,\mathrm{min}, respectively. Cedar’s times vary more in the middle half, with an IQR⁡\operatorname{IQR} of 12.5 min12.5 \,\mathrm{min} compared with 8 min8 \,\mathrm{min} at Birch. Cedar has a longer upper tail, while Birch’s sample is symmetric. Neither sample has a flagged outlier under the 1.5×IQR⁡1.5\times\operatorname{IQR} rule.”

Example 2: a lower raw score can have a higher relative position

Task: Aria scores 92 points92 \,\mathrm{points} on Test A, where the reference population has μ=80\mu = 80 and σ=8 points\sigma = 8 \,\mathrm{points}. Ben scores 8888 on Test B, where μ=70\mu = 70 and σ=6 points\sigma = 6 \,\mathrm{points}. Who has the higher standardized position?

Aria: z=92−808=1.5z=\frac{92-80}{8}=1.5.

Ben: z=88−706=3z=\frac{88-70}{6}=3.

Compare positions using each test’s own reference group

Test A: μ=80\mu = 80, σ=8 points\sigma = 8 \,\mathrm{points} · Test B: μ=70\mu = 70, σ=6 points\sigma = 6 \,\mathrm{points}

Aria and Ben compared on a common z-score axisAria scores92 on Test A with population mean80 and standard deviation8, so z=1.5. Ben scores88 on Test B with mean70 and standard deviation6, so z=3. Aria has the higher raw score; Ben has the higher standardized position. The horizontal axes are z scores, not raw test points. −1 0 1 2 3 4 Standardized position, z Aria Test A Ben Test B 92 points · z = +1.5 88 points · z = +3

On a small screen, scroll the plot sideways to read the full scale.

Both points use the same standardized scale. Ben’s lower raw score lies farther above his test’s mean in standard-deviation units. Invented teaching data.

Answer: Ben has the higher standardized position because 3>1.53>1.5. His score is 33 standard deviations above Test B’s mean, whereas Aria’s is 1.51.5 standard deviations above Test A’s mean. Aria has the higher raw score, 9292 versus 8888.

This compares positions relative to different reference groups. It does not prove that one student has greater overall ability, and without distribution information it does not establish their percentile ranks.

Example 3: lower values can be desirable

Task: in 2 fictional race categories2\text{ fictional race categories}, Casey finishes in 59 s59 \,\mathrm{s} and Drew in 51 s51 \,\mathrm{s}. Casey’s reference population has μ=65 s\mu = 65 \,\mathrm{s} and σ=3 s\sigma = 3 \,\mathrm{s}; Drew’s has μ=55 s\mu = 55 \,\mathrm{s} and σ=4 s\sigma = 4 \,\mathrm{s}. Compare their positions.

Casey: z=59−653=−2z=\frac{59-65}{3}=-2.

Drew: z=51−554=−1z=\frac{51-55}{4}=-1.

Answer: Casey is 22 standard deviations below the category mean, while Drew is 11 below. Since shorter times are desirable, Casey’s result is farther below the category mean on the standardized scale. Drew still has the shorter actual time, 5151 versus 59 s59 \,\mathrm{s}.

Do not turn −2-2 into +2+2 when ranking positions: −2-2 is the lower standardized position. Its absolute value, 22, tells us its distance from the mean.

Example 4: recover the original value

Task: a package’s mass has z=−1.5z = -1.5 relative to a population with mean 100 g100 \,\mathrm{g} and standard deviation 8 g8 \,\mathrm{g}. Find the package’s mass.

x=μ+zσx=\mu+z\sigma

x=100+(−1.5)(8)=100−12=88 gx=100+(-1.5)(8)=100-12=88\,\mathrm{g}.

Check: 88−1008=−1.5\frac{88-100}{8}=-1.5. The mass is positive even though the z-score is negative; the minus sign tells us the mass is below the mean.

07 · Make your reasoning clear

Write a comparison and justify a claim

Use three ingredients: name both groups, state a comparative relationship, and support it with the relevant feature or calculation. Add the variable and units so the meaning is clear.

“Group ___ has a [higher/lower/similar] ___ than Group ___, with ___ versus ___ [units].”

“The claim is [supported/not supported] for these data because ___.”

Improve a vague statement

Vague: “Birch is higher and Cedar is bigger.”

Clear: “Birch’s median one-way travel time is 8 min8 \,\mathrm{min} higher than Cedar’s, while Cedar has the larger IQR⁡\operatorname{IQR}, 12.512.5 versus 8 min8 \,\mathrm{min}.”

“Bigger” could refer to sample size, center or variability. Naming the measure removes the ambiguity.

Evaluate the claim the question actually makes

Claim: “Birch students usually travel longer, but their travel times are more consistent.”

The samples support “longer” using the medians: 2727 versus 19 min19 \,\mathrm{min}. They support “more consistent” in the middle half using Birch’s smaller IQR⁡\operatorname{IQR}: 88 versus 12.5 min12.5 \,\mathrm{min}. The range comparison also points to less overall spread at Birch in these samples.

What changes if the claim says “every Birch student travels longer”?

That stronger claim is not supported. Birch has an 1818-minute observation, while Cedar has a 4242-minute observation. The groups overlap. A comparison of medians describes typical values rather than proving a statement about every observation.

Describe what you observed: “These samples differ” is supported by the displayed data. Generalizing to populations requires information about data collection. Calling a difference “statistically significant” requires an appropriate inferential analysis.

There may be many possible explanations for a difference in school travel times. Suggesting a cause without study-design evidence would go beyond these graphs. For now, connect the claim to the measurements you can actually compare.

08 · Find and fix mistakes

Common mistakes and how to fix them

Check the meaning of your answer as well as its arithmetic.
MistakeWhy it failsBetter approach
Describe each group without comparing them.The reader has to infer the relationship.Use “higher than,” “less variable than” or “similar to,” with both group names.
Treat taller frequency bars as greater spread.Bar height may reflect a larger sample rather than a wider distribution.Check sample sizes, numerical scales, bin boundaries and relative frequencies.
Read a mean or 22 peaks from standard boxplots.Those details are not normally displayed.Use supplied summaries for means, and dotplots or histograms for detailed shape features.
Call one group “more spread out” without naming a measure.Range, IQR⁡\operatorname{IQR} and standard deviation measure different aspects of variation.Identify the measure and compare its values for both groups.
Use z=xσz=\frac{x}{\sigma} or z=μ−xσz=\frac{\mu-x}{\sigma}.The first omits the mean; the second reverses the intended sign.Use value−meanstandard deviation\frac{\text{value}-\text{mean}}{\text{standard deviation}}, then check above or below the mean.
Say z=−2z = -2 means a negative measurement.The sign describes a difference from the mean, rather than the original value.Say “22 standard deviations below the reference mean.”
Convert every z-score to a percentile using a Normal table.The distribution may not be Normal.Interpret the standardized distance. Use a percentile model only when justified.
Claim a sample difference proves causation or a population difference.Graphical differences alone do not establish those conclusions.Keep the claim within the data and study design provided.

Find the error: “A value with z=−2.4z = -2.4 has a higher standardized position than a value with z=1.1z = 1.1 because 2.42.4 is larger.”

Reveal the correction

The first value is farther from its own mean, because ∣−2.4∣>∣1.1∣\lvert-2.4\rvert>\lvert1.1\rvert. But its standardized position is lower, because −2.4<1.1-2.4<1.1. Signed zz compares position; absolute zz compares distance.

09 · Try first, reveal second

Practice questions with hints and solutions

Write a sentence with each numerical answer. You may use a calculator for arithmetic; you still need to choose the correct comparison and explain its meaning.

1. Compare center and variability

For 22 samples of daily reading time, Group A has median 24 min24 \,\mathrm{min} and IQR⁡\operatorname{IQR} 12 min12 \,\mathrm{min}. Group B has median 30 min30 \,\mathrm{min} and IQR⁡\operatorname{IQR} 8 min8 \,\mathrm{min}. Which group has the higher typical time, and which varies more in the middle half?

Hint

Use the median for the first question and the IQR⁡\operatorname{IQR} for the second.

Solution and explanation

Group B has the higher median, by 30−24=6 min30-24=6\,\mathrm{min}. Group A has more variability in the middle half, with IQR⁡\operatorname{IQR} 1212 versus 8 min8 \,\mathrm{min}. A higher center does not imply greater variability.

2. When two spread measures disagree

Route A’s recorded waits have IQR⁡\operatorname{IQR} 4 min4 \,\mathrm{min} and range 30 min30 \,\mathrm{min}, including a flagged high outlier. Route B’s have IQR⁡\operatorname{IQR} 8 min8 \,\mathrm{min} and range 16 min16 \,\mathrm{min}, with no flagged outliers. A student says, “Route A is more variable in every way.” Assess the statement.

Hint

Compare IQR⁡\operatorname{IQR} with IQR⁡\operatorname{IQR} and range with range. What can the outlier do to the range?

Solution and explanation

The statement is too broad. Route A has the larger range, 3030 versus 16 min16 \,\mathrm{min}, but Route B has the larger IQR⁡\operatorname{IQR}, 88 versus 4 min4 \,\mathrm{min}. Route B varies more in the middle half, while Route A spans a larger full interval. Route A’s high outlier helps explain why its range can be large despite a compact middle half.

3. Compare unequal sample sizes

Matching histogram bins contain 1515 of Group A’s 3030 observations and 4040 of Group B’s 8080 observations. Does Group B have a greater proportion in that bin?

Hint

Divide each bin count by its own group’s total.

Solution and explanation

No. Both proportions are 50%50\%. Group A: 1530=0.50\frac{15}{30}=0.50. Group B: 4080=0.50\frac{40}{80}=0.50. Group B has more observations in the bin because its sample is larger; its proportion is the same.

4. Calculate and interpret a negative z-score

A 7272-point score belongs to a reference population with μ=80 points\mu = 80 \,\mathrm{points} and σ=5 points\sigma = 5 \,\mathrm{points}. Find and interpret its z-score.

Hint

The score is below the mean, so expect a negative answer.

Solution and explanation

z=72−805=−85=−1.6z=\frac{72-80}{5}=\frac{-8}{5}=-1.6. The score is 1.61.6 standard deviations below the reference population mean. The score itself is positive: 72 points72 \,\mathrm{points}.

5. Higher position or farther away?

Observation P has z=−2.2z = -2.2 and observation Q has z=1.7z = 1.7 in their respective reference distributions. Which has the higher standardized position? Which is farther from its own mean?

Hint

Use the signed scores for position, then their absolute values for distance.

Solution and explanation

Q has the higher standardized position, because 1.7>−2.21.7>-2.2. P is farther from its reference mean, because 2.2>1.72.2>1.7 in standard-deviation units. These answers address different questions.

6. Compare scores from different tests

Lina scores 84 points84 \,\mathrm{points} on a test with population mean 7272 and standard deviation 6 points6 \,\mathrm{points}. Omar scores 9090 on another test with population mean 8080 and standard deviation 8 points8 \,\mathrm{points}. Who has the higher raw score? Who has the higher standardized position?

Hint

Calculate a separate z-score using each test’s own mean and standard deviation.

Solution and explanation

Omar has the higher raw score, 9090 versus 8484. Lina’s z=84−726=2z=\frac{84-72}{6}=2; Omar’s z=90−808=1.25z=\frac{90-80}{8}=1.25. Lina has the higher standardized position: 22 standard deviations above her test’s mean compared with Omar’s 1.251.25. These scores alone do not identify either student’s exact percentile.

7. Read the two sides of a stem-and-leaf plot

Use the back-to-back travel-time plot in Section 3. How many students in each group have times of at least 20 min20 \,\mathrm{min}? What proportion is that within each group?

Hint

Count leaves on stems 22 and 33 on each side. Both samples have 1010 observations.

Solution and explanation

Group A has 44 students on stem 22 and 00 on stem 33: 410=40%\frac{4}{10}=40\%. Group B has 66 on stem 22 and 33 on stem 33: 910=90%\frac{9}{10}=90\%. Group B has the greater proportion with recorded travel times of at least 20 min20 \,\mathrm{min}.

8. What does a z-score actually establish?

A shop’s delivery-time distribution is strongly right-skewed. A delivery has z=0z = 0. A student says, “This delivery is at the median, so exactly 50%50\% of deliveries take less time.” Is that conclusion justified?

Hint

Which center appears in the z-score formula?

Solution and explanation

No. z=0z = 0 means the delivery time equals the reference mean. It does not establish that the time equals the median or that exactly half of the deliveries are shorter. In a right-skewed distribution, the mean is typically above the median. The actual observations or an appropriate distribution model are needed to determine a percentile.

10 · Keep the essentials

Quick revision

Whole-group comparisonSame variable, clear group names, compatible scales and units. Compare important features directly.
Match the summariesMedian with median; IQR⁡\operatorname{IQR} with IQR⁡\operatorname{IQR}; mean with mean; standard deviation with standard deviation.
Use graph evidence carefullyBoxplots show summaries; dotplots and histograms reveal more about shape. Different sample sizes can change count heights.
Population z-scorez=x−μσz=\frac{x-\mu}{\sigma}, with σ>0\sigma>0. Use the correct reference population.
Read the signPositive: above the mean. Negative: below it. Zero: equal to the mean.
Separate position and distanceHigher signed zz means a higher standardized position. Larger ∣z∣\lvert z\rvert means farther from the reference mean.

Quick questions students often ask

Must I mention every SOCS feature in every answer?

Answer the question asked. For a full distribution comparison, address the important features the evidence supports. For a question only about center and variability, focus on those. Do not invent peaks or outliers to fill a checklist.

Do overlapping boxplots mean the groups have the same center?

No. Boxes or whiskers can overlap while median lines differ. Describe the actual medians and IQRs. Overlap by itself also does not settle a question about statistical significance.

Can I calculate zz for a distribution that is not Normal?

Yes, provided the needed mean and positive standard deviation are defined. The score still measures standardized distance. Normal probability or percentile calculations require a justified Normal model.

Does z=2z = 2 always mean an outlier?

No universal outlier rule follows from that number alone. It means the value is 22 standard deviations above the mean. If the question asks for the 1.5×IQR⁡1.5\times\operatorname{IQR} rule, use quartiles and fences. The word “unusual” should be tied to a stated rule or distribution model.

11 · Put the ideas together

Final understanding check

2 fictional stores2\text{ fictional stores} record customer waits, in minutes. Each sample has 1010 observations:

Store A: 66, 88, 99, 1010, 1212, 1212, 1414, 1515, 1616, 3030
Store B: 88, 99, 1010, 1111, 1212, 1212, 1313, 1414, 1515, 1616

  1. Calculate the medians and IQRs, then assess: “Store A has longer typical waits and greater variability in the middle half.”
  2. Apply the 1.5×IQR⁡1.5\times\operatorname{IQR} rule. Explain how a modified boxplot would show each sample.
  3. For a separate reference population with mean wait 12 min12 \,\mathrm{min} and standard deviation 4 min4 \,\mathrm{min}, interpret a wait of 8 min8 \,\mathrm{min} using a z-score. Does it automatically identify a percentile?
Reveal the full solution and completed boxplots

1. Center and variability: both sample medians are 12 min12 \,\mathrm{min}. Store A has Q1=9Q_{1} = 9, Q3=15Q_{3} = 15 and IQR⁡=6 min\operatorname{IQR} = 6 \,\mathrm{min}. Store B has Q1=10Q_{1} = 10, Q3=14Q_{3} = 14 and IQR⁡=4 min\operatorname{IQR} = 4 \,\mathrm{min}. The claim’s “longer typical waits” part is not supported using medians; its “greater variability in the middle half” part is supported by Store A’s larger IQR⁡\operatorname{IQR}.

2. Outliers and boxplots: Store A’s fences are 9−1.5(6)=09-1.5(6)=0 and 15+1.5(6)=2415+1.5(6)=24. Its 3030-minute wait is flagged. Its modified whiskers end at 66 and 1616, and the 3030-minute wait appears as a separate point. Store B’s fences are 10−1.5(4)=410-1.5(4)=4 and 14+1.5(4)=2014+1.5(4)=20, so no observations are flagged; whiskers end at 88 and 1616.

Same median, different variability

Modified boxplots · actual non-outlier whisker endpoints · 1.5×IQR⁡1.5\times\operatorname{IQR} rule

Same median, different variabilityShared numerical scale. Store A: n=10; five-number summary 6, 9, 12.0, 15, 30; IQR 6; fences [0.0, 24.0]; modified whiskers [6, 16]; flagged observations [30]. Store B: n=10; five-number summary 8, 10, 12.0, 14, 16; IQR 4; fences [4.0, 20.0]; modified whiskers [8, 16]; flagged observations []. 5 10 15 20 25 30 35 Recorded customer wait (minutes) Store A n = 10 Store B n = 10 Median 12 Outlier 30 Median 12

On a small screen, scroll the plot sideways to read the full scale.

Both medians are 12 min12 \,\mathrm{min}. Store A has IQR⁡\operatorname{IQR} 6 min6 \,\mathrm{min} and a separately plotted 3030-minute outlier; Store B has IQR⁡\operatorname{IQR} 4 min4 \,\mathrm{min} and no flagged observations. Invented teaching data.

The full ranges are 30−6=24 min30-6=24\,\mathrm{min} for Store A and 16−8=8 min16-8=8\,\mathrm{min} for Store B. Store A’s outlier stays in the data and in its full range even though it is plotted beyond the whisker.

3. Standardized position: z=8−124=−1z=\frac{8-12}{4}=-1. This wait is 11 standard deviation below the separate reference population’s mean. Its percentile cannot be determined from zz alone without more distribution information.

Check your understanding: did you name the measure of center, compare IQRs directly, preserve the outlier and interpret the sign of zz?

12 · Keep building

Continue learning

You can now compare group patterns and individual relative positions. Next, return to the investigative question and plan what data are needed to answer it.

Previous: Topic 1.8 — Boxplots · Review z-scores · Back to the lesson overview