DOT PLOTS, HISTOGRAMS AND STEM-AND-LEAF PLOTS IN AP STATISTICS
How to Read, Interpret and Compare Distributions: The Pile–Slice–Say Method for US and UK Students
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THE PIZZA DELIVERY TEST: WHY AVERAGES CAN BE MISLEADING
Two pizza delivery apps make almost identical promises: your food should arrive in about half an hour.
You record 15 delivery times from each app.
App A — Delivery times in minutes
22, 24, 25, 26, 26, 27, 28, 28, 28, 29, 29, 30, 31, 33, 38
App B — Delivery times in minutes
15, 18, 20, 22, 25, 27, 28, 29, 31, 33, 35, 38, 41, 45, 50
At first glance, the results look remarkably similar.
| Statistic | App A | App B |
|---|---|---|
| Number of deliveries | 15 | 15 |
| Median | 28 min | 29 min |
| Mean | 28.27 min | 30.47 min |
| Minimum | 22 min | 15 min |
| Maximum | 38 min | 50 min |
| Range | 16 min | 35 min |
The median delivery times differ by just one minute.
But imagine ordering dinner.
With App A, most deliveries arrive within a relatively narrow window. With App B, you might receive your food in 15 minutes or wait 50 minutes.
The averages alone do not reveal this difference.
That is why statisticians use graphical displays of quantitative data.
A dot plot reveals individual observations. A histogram shows how observations are distributed across intervals. A stem-and-leaf plot organizes the numbers while preserving their original values.
These graphs help answer questions that a single average cannot:
Where do most observations occur?
Is the distribution symmetric or skewed?
Are there unusually high or low values?
How much variation exists?
Which of two groups is more consistent?
These are important skills in AP Statistics, GCSE Mathematics, IB Mathematics and A-level Statistics.
This article develops one practical method for reading, constructing and comparing statistical graphs.
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WHAT YOU WILL LEARN
By the end, you should be able to:
Read and interpret a dot plot.
Describe the shape of a statistical distribution.
Identify symmetric, skewed, unimodal, bimodal and approximately uniform distributions.
Read a histogram and understand frequency and class intervals.
Explain why histogram bin width matters.
Construct a stem-and-leaf plot from raw data.
Make split-stem and back-to-back stem-and-leaf diagrams.
Calculate quartiles and the interquartile range.
Identify potential outliers using the 1.5 × IQR rule.
Compare two distributions using shape, outliers, center and spread.
Understand frequency density in unequal-width histograms.
Write better statistical explanations for AP Statistics free-response questions.
PART 1: THE PILE–SLICE–SAY METHOD
Instead of memorizing disconnected definitions, use three questions whenever you encounter a graph of quantitative data.
1. PILE: WHERE DOES THE DATA COLLECT?
Look for concentrations of observations.
Are most values grouped around 20? Are there several peaks? Is one side of the graph much longer than the other?
The concentration of values reveals the general shape of the distribution.
2. SLICE: HOW HAS THE DATA BEEN ORGANIZED?
Different graphs organize observations differently.
Dot plot: individual observations remain visible.
Stem-and-leaf plot: observations are separated into stems and leaves.
Histogram: observations are grouped into numerical intervals called bins or classes.
Understanding this organization tells you what information the graph preserves and what information it hides.
3. SAY: HOW DO YOU DESCRIBE WHAT YOU SEE?
Use the familiar AP Statistics checklist:
SOCS = Shape, Outliers, Center, Spread
Shape: Is the distribution symmetric, skewed or multimodal?
Outliers: Are any observations unusually distant from the rest?
Center: Where is the typical value?
Spread: How widely are the observations distributed?
Always describe the actual variable.
For example, write:
"The distribution of pizza delivery times is skewed to the right."
Do not simply write:
"The data is skewed."
Adding context makes your statistical explanation clearer.
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PART 2: DOT PLOTS — HOW TO READ AND INTERPRET INDIVIDUAL VALUES
What is a dot plot in statistics?
A dot plot is a graphical representation in which each observation is represented by a dot above its corresponding value on a number line.
If a value occurs four times, four dots are stacked above that value.
Unlike a histogram, a dot plot does not group observations into intervals.
This makes it particularly useful for small or moderately sized quantitative data sets.
When should you use a dot plot?
Dot plots work especially well when:
The data set is relatively small.
Individual observations matter.
Values repeat frequently.
You want to identify clusters and gaps.
You want to compare two small groups visually.
Examples include:
Number of siblings per student.
Goals scored in a series of matches.
Number of mistakes made on a test.
Delivery times for a small number of orders.
Daily temperatures over a short period.
Example: App A as a dot plot
Recall the delivery times:
22, 24, 25, 26, 26, 27, 28, 28, 28, 29, 29, 30, 31, 33, 38
The frequency table gives the exact number of dots above each value.
| Delivery time | Number of dots |
|---|---|
| 22 | 1 |
| 23 | 0 |
| 24 | 1 |
| 25 | 1 |
| 26 | 2 |
| 27 | 1 |
| 28 | 3 |
| 29 | 2 |
| 30 | 1 |
| 31 | 1 |
| 32 | 0 |
| 33 | 1 |
| 34–37 | 0 |
| 38 | 1 |
A dot plot constructed from this table would have its tallest stack at 28 minutes.
How to interpret a dot plot step by step
Step 1: Read the horizontal axis.
Identify the variable and its units. Here, the variable is delivery time, measured in minutes.
Step 2: Find the tallest stack.
The tallest stack is at 28 minutes, making 28 the mode.
Step 3: Identify clusters.
Most deliveries are concentrated between 25 and 31 minutes.
Step 4: Look for gaps.
There are no observations at 34, 35, 36 or 37 minutes.
Step 5: Look for isolated observations.
The observation at 38 minutes is separated from the main cluster.
Step 6: Count the dots.
There are 15 observations, so n = 15.
Describing the shape of App A
A suitable statistical description is:
"The distribution of delivery times is unimodal, with its highest concentration around 28 minutes. It has a slight right-hand tail, with a relatively isolated observation at 38 minutes."
Notice that this description identifies the variable, its units, the peak and the direction of the tail.
Understanding the main distribution shapes
SYMMETRIC DISTRIBUTION
A symmetric distribution has approximately similar shapes on either side of its center.
The left and right sides need not be perfectly identical.
RIGHT-SKEWED DISTRIBUTION
A right-skewed distribution has a longer tail extending towards larger values.
Most observations are concentrated towards the lower end.
Examples include some income distributions and waiting-time distributions.
LEFT-SKEWED DISTRIBUTION
A left-skewed distribution has a longer tail extending towards smaller values.
Most observations are concentrated towards the higher end.
An example might be a test on which most students score highly but a few obtain very low marks.
UNIMODAL DISTRIBUTION
A distribution with one main peak.
BIMODAL DISTRIBUTION
A distribution with two distinct peaks.
This can suggest that two different populations or groups have been combined.
For example, the commuting times of cyclists and car drivers might produce two concentrations.
APPROXIMATELY UNIFORM DISTRIBUTION
A distribution in which observations occur at roughly similar frequencies across the displayed values or intervals.
A finite sample from a uniform process will not normally be perfectly flat.
The most important rule about skewness
Skewness is named after the direction of the tail, not the location of the tallest peak.
Tail towards larger values = right skew.
Tail towards smaller values = left skew.
This distinction is particularly important in statistics examinations.
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PART 3: HISTOGRAMS — HOW TO DESCRIBE SHAPE, CENTER AND SPREAD
What is a histogram?
A histogram displays quantitative data by grouping observations into numerical intervals called bins or classes.
The horizontal axis represents the numerical intervals.
The vertical axis represents frequency, relative frequency or frequency density, depending on the histogram.
The bars normally touch because the intervals represent consecutive sections of a numerical scale.
Histogram versus bar chart
This is a common examination question.
| Feature | Histogram | Bar chart |
|---|---|---|
| Data | Quantitative | Categorical |
| Horizontal axis | Numerical intervals | Categories |
| Bars | Normally touching | Usually separated |
| Main purpose | Show a distribution | Compare categories |
| Individual values | Grouped into bins | Category totals |
For example:
Delivery times grouped into 10-minute intervals: histogram.
Number of people choosing different pizza toppings: bar chart.
Do not confuse the two.
Worked histogram example: Travel times
Suppose 40 people report their journey times to school or work.
The grouped frequency table is:
| Travel time (minutes) | Frequency |
|---|---|
| 0 to under 10 | 3 |
| 10 to under 20 | 9 |
| 20 to under 30 | 12 |
| 30 to under 40 | 8 |
| 40 to under 50 | 5 |
| 50 to under 60 | 2 |
| 60 to under 70 | 1 |
| Total | 40 |
The histogram would have seven adjoining bars.
The tallest bar represents the 20 to under 30 minute interval, with a frequency of 12.
PILE: Where are most observations?
The largest concentration is between 20 and 30 minutes.
The frequencies decrease towards the higher intervals.
There are only three observations in the intervals from 50 to under 70 minutes.
This suggests a distribution with one main peak and a longer tail towards larger travel times.
SLICE: What do the intervals tell us?
Each class has a width of 10 minutes.
The histogram shows how many observations fall inside each interval.
However, it does not reveal the exact travel time of any individual person.
For example, the bar from 20 to under 30 minutes tells us that 12 people fall within that interval. It does not tell us whether those people travelled for 21, 24 or 29 minutes.
That is an important limitation of grouped data.
SAY: Describe the distribution
Shape: The distribution is unimodal and appears right-skewed, with its highest frequency in the 20 to under 30 minute interval.
Center: The median lies in the 20 to under 30 minute interval.
Spread: The observations extend across class intervals from 0 to under 70 minutes. The actual minimum and maximum cannot be identified from the grouped table alone.
Unusual observations: The upper intervals contain relatively few observations, but the grouped table does not provide enough information to establish whether any individual observation is an outlier.
How do you estimate the median from a histogram?
This is a frequently misunderstood question.
You cannot normally read the exact median from a histogram.
You can identify the interval containing the median by using cumulative frequencies.
For the travel-time example:
| Interval | Frequency | Cumulative frequency |
|---|---|---|
| 0–under 10 | 3 | 3 |
| 10–under 20 | 9 | 12 |
| 20–under 30 | 12 | 24 |
| 30–under 40 | 8 | 32 |
| 40–under 50 | 5 | 37 |
| 50–under 60 | 2 | 39 |
| 60–under 70 | 1 | 40 |
There are 40 observations.
The median is the average of the 20th and 21st observations.
The cumulative frequency reaches 12 before the 20–30 interval and 24 after it.
Therefore, both middle observations lie in the 20 to under 30 minute interval.
The exact median is unknown.
Estimating the median using interpolation
If we assume that observations are approximately evenly distributed within the median class, we can estimate the median.
Estimated median:
20 + [(20 − 12) ÷ 12] × 10
= 20 + (8 ÷ 12) × 10
≈ 26.7 minutes.
This is an estimate, not an exact value. Its accuracy depends on the assumption about how observations are distributed inside the class.
Estimating the mean from grouped data
The mean can be estimated using class midpoints.
Formula:
Estimated mean = Σ(f × midpoint) ÷ Σf
For our example:
| Class | Midpoint | Frequency | f × midpoint |
|---|---|---|---|
| 0–10 | 5 | 3 | 15 |
| 10–20 | 15 | 9 | 135 |
| 20–30 | 25 | 12 | 300 |
| 30–40 | 35 | 8 | 280 |
| 40–50 | 45 | 5 | 225 |
| 50–60 | 55 | 2 | 110 |
| 60–70 | 65 | 1 | 65 |
| Total | 40 | 1,130 |
Estimated mean:
1,130 ÷ 40 = 28.25 minutes.
This is also an estimate because every observation within a class has been represented by its midpoint.
Mean versus median in skewed distributions
The mean is sensitive to extreme observations.
The median depends on the middle position and is much less affected by extreme values.
As a useful rule of thumb:
Right-skewed distribution: mean is often greater than median.
Left-skewed distribution: mean is often less than median.
Approximately symmetric distribution: mean and median are often reasonably close.
These are tendencies, not universal laws.
Which summary should you use?
| Distribution | Preferred summaries |
|---|---|
| Approximately symmetric, without influential outliers | Mean and standard deviation |
| Skewed | Median and IQR |
| Contains influential outliers | Median and IQR |
| Strongly asymmetric | Median and IQR |
Why histogram bin width changes the picture
One of the most important ideas in graphical statistics is that the appearance of a histogram depends on how the intervals are selected.
Use App A's 15 delivery times again.
Bin width of 10 minutes
20 to under 30: 11 observations
30 to under 40: 4 observations
This creates a simple two-bar summary.
Bin width of 5 minutes
20 to under 25: 2
25 to under 30: 9
30 to under 35: 3
35 to under 40: 1
The main concentration becomes much clearer.
Bin width of 2 minutes
22 to under 24: 1
24 to under 26: 2
26 to under 28: 3
28 to under 30: 5
30 to under 32: 2
32 to under 34: 1
34 to under 36: 0
36 to under 38: 0
38 to under 40: 1
The distribution now reveals more detail, including the isolated observation at 38.
All three histograms represent exactly the same 15 observations.
The difference is the choice of intervals.
Very wide bins can conceal important patterns. Very narrow bins can make random fluctuations look like meaningful peaks.
Choose intervals that reveal the overall structure without creating a misleading impression.
Histogram class boundaries
A common convention is to include the left boundary and exclude the right boundary.
Therefore:
20 belongs to 20 to under 30.
29 belongs to 20 to under 30.
30 belongs to 30 to under 40.
Always follow the class-interval notation given in the question.
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PART 4: STEM-AND-LEAF PLOTS — BUILDING A GRAPH WITHOUT LOSING THE DATA
What is a stem-and-leaf plot?
A stem-and-leaf plot separates each observation into two parts:
Stem: the leading digit or digits.
Leaf: the final digit.
For example:
27 becomes 2 | 7.
Here, 2 is the stem and 7 is the leaf.
A stem-and-leaf plot organizes numerical observations while retaining the original values.
It is particularly useful for small and moderately sized data sets.
In the UK, the same representation is commonly called a stem-and-leaf diagram.
How to construct a stem-and-leaf plot
Use App B's delivery times:
15, 18, 20, 22, 25, 27, 28, 29, 31, 33, 35, 38, 41, 45, 50
Step 1: Arrange the data in ascending order.
The values are already sorted.
Step 2: Identify the stems.
For two-digit observations, use the tens digit.
Step 3: List every required stem.
Include empty stems between the smallest and largest observations.
Step 4: Write the leaves in ascending order.
The units digits become the leaves.
Step 5: Add a key.
A key is essential.
Example:
2 | 7 = 27 minutes.
App B stem-and-leaf plot
Stem | Leaves
1 | 5 8
2 | 0 2 5 7 8 9
3 | 1 3 5 8
4 | 1 5
5 | 0Key: 2 | 7 represents 27 minutes.
Count the leaves:
2 + 6 + 4 + 2 + 1 = 15.
The total matches the number of observations.
How to interpret the stem-and-leaf diagram
Shape: The observations are concentrated in the 20s and 30s, with fewer observations at the higher values.
Center: The eighth observation is 29, so the median is 29 minutes.
Spread: The minimum is 15 and the maximum is 50.
Range = 50 − 15 = 35 minutes.
Unusual observations: The observation at 50 is relatively high, but whether it qualifies as an outlier requires a separate calculation.
Split stem-and-leaf plots
Sometimes a single stem contains too many leaves.
For example, a stem containing the leaves:
0 1 2 3 4 5 6 7 8 9
does not show much detail.
A split-stem plot solves this problem.
Each stem is divided into two rows:
First row: leaves 0–4.
Second row: leaves 5–9.
App B becomes:
Stem | Leaves
1 |
1 | 5 8
2 | 0 2
2 | 5 7 8 9
3 | 1 3
3 | 5 8
4 | 1
4 | 5
5 | 0
5 |Key: 2 | 7 = 27 minutes.
The empty rows are retained to make the structure clear.
The key should explain that each stem has been divided into two rows.
Back-to-back stem-and-leaf plots
A back-to-back stem-and-leaf diagram is especially useful when comparing two small data sets.
The two groups share the same stems.
One group's leaves appear on the left, while the other group's leaves appear on the right.
Using App A and App B:
App A leaves | Stem | App B leaves
| 1 | 5 8
9 9 8 8 8 7 6 6 5 4 2 | 2 | 0 2 5 7 8 9
8 3 1 0 | 3 | 1 3 5 8
| 4 | 1 5
| 5 | 0Key:
Right side: 2 | 7 = 27 minutes for App B.
Left side: 9 | 2 = 29 minutes for App A.
For the left-hand side, leaves are written in descending order from left to right so that the values increase as you move towards the stem.
What does the back-to-back plot reveal?
App A has most observations concentrated in the 20s.
App B has observations extending from the teens through the 50s.
The difference in spread is immediately visible.
This is an example of how a graphical comparison can communicate information more effectively than a list of averages.
Stem-and-leaf plots with decimals
The same method works with decimal values.
Suppose the observations are:
3.4, 3.7, 4.1, 4.6, 4.8
The diagram is:
Stem | Leaves
3 | 4 7
4 | 1 6 8Key: 3 | 4 = 3.4.
For three-digit values:
125, 131, 148
you could use:
Stem | Leaves
12 | 5
13 | 1
14 | 8Key: 12 | 5 = 125.
Common stem-and-leaf mistakes
Writing leaves in the wrong order.
Forgetting the key.
Omitting empty stems.
Putting more than one digit in a leaf when the chosen key requires single-digit leaves.
Using inconsistent stems.
Forgetting to check that the number of leaves equals the sample size.
Which graph should you choose?
| Situation | Suitable graph |
|---|---|
| Small data set with repeated values | Dot plot |
| Small data set where exact values matter | Stem-and-leaf plot |
| Comparing two small groups | Back-to-back stem-and-leaf plot |
| Large quantitative data set | Histogram |
| Data with many distinct numerical values | Histogram |
These are useful guidelines, not rigid limits. The best graph depends on the size, type and purpose of the data.
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PART 5: OUTLIERS AND THE 1.5 × IQR RULE
An outlier is an observation that is unusually far from the rest of a distribution.
A point that looks unusual on a dot plot is not automatically an outlier under a formal statistical rule.
AP Statistics commonly uses the 1.5 × IQR rule.
What is the interquartile range?
The interquartile range measures the spread of the middle 50% of observations.
IQR = Q₃ − Q₁
Where:
Q₁ is the first quartile.
Q₃ is the third quartile.
The 1.5 × IQR outlier rule
First calculate the interquartile range.
Then calculate two boundaries, called fences.
Lower fence:
Q₁ − 1.5 × IQR
Upper fence:
Q₃ + 1.5 × IQR
An observation below the lower fence or above the upper fence is classified as a potential outlier under this rule.
Worked example: Is App A's 38-minute delivery an outlier?
Sorted observations:
22, 24, 25, 26, 26, 27, 28, 28, 28, 29, 29, 30, 31, 33, 38
There are 15 observations.
The median is the eighth observation:
Median = 28.
Exclude the median when dividing the remaining observations into two equal halves.
Lower half:
22, 24, 25, 26, 26, 27, 28
Therefore:
Q₁ = 26.
Upper half:
28, 29, 29, 30, 31, 33, 38
Therefore:
Q₃ = 30.
Calculate the IQR:
IQR = 30 − 26 = 4.
Calculate the upper fence:
30 + 1.5 × 4 = 36.
Calculate the lower fence:
26 − 1.5 × 4 = 20.
The fences are 20 and 36.
The observation 38 is above 36.
Therefore, 38 minutes is an outlier under the 1.5 × IQR rule.
The value 22 is not an outlier because it lies above the lower fence.
Worked example: Does App B have any outliers?
Sorted observations:
15, 18, 20, 22, 25, 27, 28, 29, 31, 33, 35, 38, 41, 45, 50
Median = 29.
Lower half:
15, 18, 20, 22, 25, 27, 28
Q₁ = 22.
Upper half:
31, 33, 35, 38, 41, 45, 50
Q₃ = 38.
IQR = 38 − 22 = 16.
Upper fence:
38 + 1.5 × 16 = 62.
Lower fence:
22 − 1.5 × 16 = −2.
Every observation lies between −2 and 62.
Therefore, App B has no outliers under this rule.
An important statistical lesson
App B contains more extreme-looking delivery times than App A.
Yet App B has no outliers under the 1.5 × IQR rule.
Why?
Because its overall spread is much larger.
The outlier rule measures unusual observations relative to the distribution itself. It does not apply one fixed maximum or minimum to every data set.
Should outliers be removed?
Never delete an outlier automatically.
First investigate its origin.
Possible explanations include:
A recording or measurement error.
An unusual but genuine observation.
An exceptional event.
A different population or process.
An outlier can have a substantial effect on the mean and standard deviation.
The median and IQR are generally less sensitive to extreme observations.
An outlier should be reported and interpreted, not silently discarded.
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PART 6: HOW TO COMPARE TWO DISTRIBUTIONS IN AP STATISTICS
Describing two graphs separately is not the same as comparing them.
A strong comparison identifies similarities and differences using the same statistical measures and the same context.
The same-scale rule
Before comparing two histograms or dot plots, check that the horizontal axes use the same scale.
For histograms, also check the class intervals and bin widths.
If one graph uses intervals of 5 minutes and another uses intervals of 10 minutes, the apparent shapes may not be directly comparable.
When sample sizes differ, consider relative frequencies rather than comparing raw counts.
A group of 300 observations will naturally produce larger frequency counts than a group of 30 in many situations.
Weak comparison versus strong comparison
Weak:
"App A has a median of 28 and App B has a median of 29."
This merely reports two numbers.
Stronger:
"App B's median delivery time is slightly higher than App A's, at 29 minutes compared with 28 minutes. However, App B has considerably greater variability, with an IQR of 16 minutes compared with just 4 minutes for App A."
This statement makes an actual comparison.
Comparing App A and App B using SOCS
SHAPE
App A has one main concentration around the upper 20s, with a high observation at 38 minutes.
App B has one main concentration in the 20s and a longer extension towards larger delivery times.
OUTLIERS
App A has one outlier at 38 minutes under the 1.5 × IQR rule.
App B has no outliers under the same rule.
CENTER
The medians are similar:
App A = 28 minutes.
App B = 29 minutes.
The means are:
App A = 28.27 minutes.
App B = 30.47 minutes.
The larger mean for App B is consistent with its greater concentration of high delivery times.
SPREAD
App A:
Range = 38 − 22 = 16 minutes.
IQR = 30 − 26 = 4 minutes.
Sample standard deviation ≈ 3.86 minutes.
App B:
Range = 50 − 15 = 35 minutes.
IQR = 38 − 22 = 16 minutes.
Sample standard deviation ≈ 10.05 minutes.
App B is substantially more variable by all three measures.
A complete comparison in context
"Both pizza delivery apps have similar median delivery times, with App A at 28 minutes and App B at 29 minutes. However, App B has considerably greater variability, with an IQR of 16 minutes compared with 4 minutes for App A. App A also has one high outlier at 38 minutes, while App B has none under the 1.5 × IQR rule. Therefore, App A provides more consistent delivery times in this sample."
This is a more meaningful conclusion than simply saying that one app is faster.
The data support a conclusion about consistency in the observed sample. They do not, by themselves, establish that App A will always be more reliable in the future.
Useful comparison vocabulary
Use precise language:
Higher than
Lower than
Approximately equal to
More variable than
Less spread out than
More consistent than
More concentrated than
More strongly skewed than
Whereas
In contrast
Unlike
Both distributions
Avoid writing two separate descriptions without connecting them.
Choosing the correct measures
| Situation | Compare |
|---|---|
| Roughly symmetric distributions without influential outliers | Means and standard deviations |
| Skewed distributions | Medians and IQRs |
| Distributions with outliers | Medians and IQRs |
| Any two distributions | Shape and unusual features as well |
Remember that identical medians do not mean two distributions are identical.
Two groups can have the same center but completely different shapes and spreads.
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PART 7: US AND UK STATISTICS — DIFFERENT TERMINOLOGY, RELATED METHODS
Students studying AP Statistics, GCSE Mathematics, IB Mathematics and A-level Statistics encounter many of the same graphical ideas.
Some terminology differs.
| US terminology | UK terminology |
|---|---|
| Stem-and-leaf plot | Stem-and-leaf diagram |
| Dot plot | Dot plot |
| Histogram | Histogram |
| Right-skewed | Positively skewed |
| Left-skewed | Negatively skewed |
| Interquartile range | Interquartile range |
| Frequency | Frequency |
| Class interval | Class interval |
The biggest additional point to understand is frequency density.
What is frequency density?
When histogram classes have unequal widths, the height of a bar cannot simply represent frequency.
Instead, the vertical axis represents frequency density.
Formula:
Frequency density = Frequency ÷ Class width
The area of each bar represents its frequency.
Worked frequency density example
Consider two intervals.
| Class interval | Frequency | Class width |
|---|---|---|
| 10 to 30 | 12 | 20 |
| 30 to 35 | 9 | 5 |
First interval:
Frequency density = 12 ÷ 20 = 0.6.
Second interval:
Frequency density = 9 ÷ 5 = 1.8.
The second bar is taller even though it contains fewer observations.
Why?
Because its class interval is much narrower.
Check the areas:
First bar:
0.6 × 20 = 12.
Second bar:
1.8 × 5 = 9.
The areas correctly represent the frequencies.
The examination rule
For equal-width histogram classes, bar heights can represent frequencies or relative frequencies.
For unequal-width classes, use frequency density so that bar area represents frequency.
This is particularly important for GCSE Higher and A-level Statistics questions.
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PART 8: AP STATISTICS FREE-RESPONSE ANSWER TEMPLATES
A statistical graph question often requires more than identifying a peak or calculating a median.
Your explanation should identify the variable, give relevant numerical evidence and interpret the result.
Template 1: Describe one distribution
"The distribution of [variable and units] is approximately [shape]. It is [unimodal/bimodal] with a main concentration around [value or interval]. The center is approximately [mean/median]. The values range from [minimum] to [maximum], with an IQR of [value]. [Identify any outliers and explain the evidence.]"
Template 2: Compare two distributions
"[Group A] has a [higher/lower/similar] median [variable] than [Group B], at [value] compared with [value]. However, [Group A/Group B] is more variable, with an IQR of [value] compared with [value]. Their shapes are [comparison]. [Describe unusual observations.] In context, this suggests [interpretation]."
Template 3: Explain why the median is preferred
"Because the distribution is skewed and/or contains influential outliers, the median and IQR are more appropriate summaries than the mean and standard deviation. The mean and standard deviation are more sensitive to extreme observations."
Template 4: Identify a histogram limitation
"The histogram groups observations into intervals, so the exact individual values cannot be recovered. It is possible to identify the interval containing the median, but an exact median cannot generally be determined without the original observations."
Template 5: Interpret an outlier
"The observation of [value] is above the upper fence of [value], calculated using Q₃ + 1.5 × IQR. Therefore, it is an outlier under the 1.5 × IQR rule."
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PART 9: TEN COMMON STATISTICS MISTAKES
Naming skewness according to the peak rather than the tail.
Correction: Skewness is named according to the direction of the tail.
Confusing a histogram with a bar chart.
Correction: Histograms display quantitative intervals; bar charts display categories.
Reading exact observations from a histogram.
Correction: A histogram provides grouped information, not individual values.
Forgetting the key in a stem-and-leaf plot.
Correction: Always explain what a stem and leaf represent.
Omitting empty stems.
Correction: Retain intermediate empty stems because they reveal gaps.
Forgetting to sort the observations.
Correction: Arrange the data before constructing a stem-and-leaf plot.
Calling every isolated observation an outlier.
Correction: Use the appropriate outlier rule when required.
Treating the mean and median as interchangeable.
Correction: Consider the shape and extreme observations before choosing a summary.
Comparing frequencies when sample sizes differ.
Correction: Relative frequencies may provide a fairer comparison.
Assuming equal centers mean identical distributions.
Correction: Compare shape, center, outliers and spread.
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PART 10: AP STATISTICS PRACTICE QUESTIONS WITH ANSWERS
Try the questions before reading the solutions.
QUESTION 1: DOT PLOTS AND OUTLIERS
Twelve students report the number of siblings they have:
0, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 7
Answer the following:
(a) Construct a dot plot.
(b) Describe the distribution's shape.
(c) Calculate the median.
(d) Find Q₁, Q₃ and the IQR.
(e) Determine whether 7 is an outlier using the 1.5 × IQR rule.
QUESTION 2: HISTOGRAMS AND CUMULATIVE FREQUENCY
The examination scores of 50 students are grouped as follows.
| Score interval | Frequency |
|---|---|
| 0 to under 20 | 2 |
| 20 to under 40 | 5 |
| 40 to under 60 | 10 |
| 60 to under 80 | 18 |
| 80 to under 100 | 15 |
Answer:
(a) Describe the distribution's shape.
(b) Which interval contains the median?
(c) What percentage scored at least 60?
(d) Would you expect the mean to be above or below the median?
QUESTION 3: STEM-AND-LEAF PLOT
The scores of 15 players are:
41, 43, 47, 52, 55, 55, 58, 61, 62, 64, 64, 67, 70, 71, 93
(a) Construct a stem-and-leaf plot.
(b) Include an appropriate key.
(c) Find the median.
(d) Calculate the IQR.
(e) Use the 1.5 × IQR rule to check for outliers.
QUESTION 4: COMPARING TWO DISTRIBUTIONS
Two groups take the same examination.
| Statistic | Group X | Group Y |
|---|---|---|
| Median | 72 | 72 |
| IQR | 10 | 24 |
| Range | 35 | 60 |
| Shape | Approximately symmetric | Skewed left |
| Outliers | None | None |
Write a statistical comparison of the two groups.
QUESTION 5: FIND THE ERROR
A student writes:
"This histogram has a tail extending towards the right, so the distribution is negatively skewed. The mean must therefore be smaller than the median."
Identify the mistakes and rewrite the statement correctly.
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FULL ANSWERS AND EXPLANATIONS
ANSWER 1: DOT PLOT
(a) Frequency table:
| Number of siblings | Frequency |
|---|---|
| 0 | 1 |
| 1 | 3 |
| 2 | 4 |
| 3 | 2 |
| 4 | 1 |
| 5 | 0 |
| 6 | 0 |
| 7 | 1 |
(b) The distribution is unimodal, with a peak at 2 siblings. It has a right-hand tail and an isolated observation at 7.
(c) There are 12 observations.
Median = average of the 6th and 7th values.
Median = (2 + 2) ÷ 2 = 2.
(d) Lower half:
0, 1, 1, 1, 2, 2.
Q₁ = (1 + 1) ÷ 2 = 1.
Upper half:
2, 2, 3, 3, 4, 7.
Q₃ = (3 + 3) ÷ 2 = 3.
IQR = 3 − 1 = 2.
(e) Upper fence:
3 + 1.5 × 2 = 6.
Lower fence:
1 − 1.5 × 2 = −2.
Since 7 is greater than 6, the observation of 7 siblings is an outlier under the 1.5 × IQR rule.
ANSWER 2: HISTOGRAM
(a) The distribution is unimodal and skewed to the left. The highest frequency occurs in the 60 to under 80 interval, with a tail extending towards lower scores.
(b) There are 50 observations.
The median is between the 25th and 26th observations.
Cumulative frequencies:
2, 7, 17, 35, 50.
Both middle observations fall within the 60 to under 80 interval.
(c) Scores of at least 60:
18 + 15 = 33.
Percentage:
(33 ÷ 50) × 100 = 66%.
(d) The mean would generally be expected to be below the median because the distribution has a left-hand tail.
This is a tendency rather than a guarantee.
ANSWER 3: STEM-AND-LEAF
(a) The plot is:
Stem | Leaves
4 | 1 3 7
5 | 2 5 5 8
6 | 1 2 4 4 7
7 | 0 1
8 |
9 | 3(b) Key:
6 | 1 = 61 points.
(c) There are 15 observations.
Median = eighth observation = 61.
(d) Lower half:
41, 43, 47, 52, 55, 55, 58.
Q₁ = 52.
Upper half:
62, 64, 64, 67, 70, 71, 93.
Q₃ = 67.
IQR = 67 − 52 = 15.
(e) Upper fence:
67 + 1.5 × 15 = 89.5.
Lower fence:
52 − 1.5 × 15 = 29.5.
The observation 93 exceeds 89.5.
Therefore, 93 is an outlier under the 1.5 × IQR rule.
There are no lower outliers.
ANSWER 4: COMPARING DISTRIBUTIONS
Both groups have the same median examination score of 72, suggesting similar typical performance.
However, Group Y has substantially greater variability, with an IQR of 24 compared with 10 for Group X. Its range is also larger, at 60 compared with 35.
Group X is approximately symmetric, whereas Group Y is negatively skewed, with a tail towards lower scores.
Neither group has identified outliers.
Overall, Group X has more consistent examination results, while Group Y has a wider spread of scores.
ANSWER 5: FIND THE ERROR
There are two errors.
Error 1: A distribution with a tail towards larger values is positively skewed, or right-skewed, not negatively skewed.
Error 2: In a right-skewed distribution, the mean is generally greater than the median, not smaller.
Correct statement:
"This histogram has a tail extending towards larger values, so the distribution is positively skewed. The mean would generally be greater than the median because the high values pull the mean towards the right-hand tail."
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FREQUENTLY ASKED QUESTIONS ABOUT DOT PLOTS, HISTOGRAMS AND DISTRIBUTIONS
1. What is the difference between a dot plot and a histogram?
A dot plot represents each observation individually using a dot. A histogram groups observations into numerical intervals and displays their frequencies using adjoining bars.
Dot plots preserve individual values, while histograms provide a compact summary of larger data sets.
2. How do you describe the shape of a distribution in AP Statistics?
Describe whether the distribution is approximately symmetric, right-skewed or left-skewed. Also identify the number of peaks, such as unimodal or bimodal, and mention any unusual gaps or isolated observations.
3. What does SOCS mean in AP Statistics?
SOCS stands for:
S = Shape.
O = Outliers.
C = Center.
S = Spread.
It is a useful checklist for describing quantitative distributions.
4. How do you identify a right-skewed histogram?
Look for a longer tail extending towards larger numerical values.
The main concentration of observations will generally be towards the lower end of the scale.
5. How do you find the median from a histogram?
Use the cumulative frequencies to identify the class interval containing the middle observation.
The exact median generally cannot be determined from grouped data alone.
Interpolation can provide an estimate if an assumption about the distribution within the median class is made.
6. What is the difference between frequency and frequency density?
Frequency is the number of observations in a class.
Frequency density is frequency divided by class width.
Frequency density is needed when histogram intervals have unequal widths, because the bar area must represent frequency.
7. Why is a stem-and-leaf plot useful?
It organizes numerical data while preserving individual observations.
It also makes it easier to identify clusters, gaps, repeated values and potential unusual observations.
8. What is a split stem-and-leaf plot?
A split stem-and-leaf plot divides each stem into multiple rows to make a crowded distribution easier to read.
A common approach is to use one row for leaves 0–4 and another for leaves 5–9.
9. What is a back-to-back stem-and-leaf diagram?
It is a graphical method for comparing two numerical data sets.
Both groups share the same stems, but their leaves are displayed on opposite sides.
10. How do you calculate an outlier using the 1.5 × IQR rule?
Calculate:
IQR = Q₃ − Q₁.
Lower fence = Q₁ − 1.5 × IQR.
Upper fence = Q₃ + 1.5 × IQR.
Values outside these fences are outliers under this rule.
11. Should you use the mean or median for skewed data?
The median is generally more appropriate because it is less sensitive to extreme observations.
The IQR is a useful measure of spread to use alongside the median.
12. Can two distributions have the same median but different shapes?
Yes.
Two groups can have identical medians but different ranges, IQRs, standard deviations, skewness and outliers.
A center measure alone never provides a complete description of a distribution.
13. Are dot plots and stem-and-leaf diagrams included in UK statistics?
These graphical methods appear in UK mathematics and statistics education, although the precise requirements vary by qualification and examination board.
GCSE and A-level students should also understand histograms, class intervals and, where applicable, frequency density.
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THE ONE-PAGE AP STATISTICS REVISION SHEET
PILE
Find where observations concentrate and where they become sparse.
SLICE
Understand how the graph organizes the data.
Dot plot = individual observations.
Stem-and-leaf = stems and leaves.
Histogram = numerical intervals.
SAY
Use SOCS:
Shape.
Outliers.
Center.
Spread.
SHAPE
Right-skewed = tail towards larger values.
Left-skewed = tail towards smaller values.
Symmetric = approximately balanced.
Unimodal = one main peak.
Bimodal = two main peaks.
Uniform = roughly equal frequencies.
CENTER AND SPREAD
Symmetric, without influential outliers: mean and standard deviation.
Skewed or containing influential outliers: median and IQR.
OUTLIERS
IQR = Q₃ − Q₁.
Lower fence = Q₁ − 1.5 × IQR.
Upper fence = Q₃ + 1.5 × IQR.
HISTOGRAMS
Bars touch.
Bin width affects appearance.
Exact individual observations cannot be recovered.
Unequal-width classes require frequency density.
STEM-AND-LEAF
Sort observations.
Keep necessary empty stems.
Add a key.
Check the number of leaves.
COMPARISON
Use the same scale.
Compare shape, outliers, center and spread.
Name both groups.
Include numerical evidence.
Explain the conclusion in context.
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FINAL THOUGHT: A GRAPH IS MORE THAN A PICTURE
Return to the two pizza delivery apps.
Their median delivery times were almost identical.
But the graphical evidence revealed a much more useful difference.
App A had a compact cluster of delivery times, with one high outlier.
App B had a much wider spread, ranging from 15 to 50 minutes.
The difference was not obvious from the medians alone.
That is the purpose of statistical graphs: to reveal patterns that individual numbers can conceal.
Once you understand how to read a dot plot, construct a histogram, interpret a stem-and-leaf diagram and compare distributions using SOCS, you have a practical foundation for describing quantitative data.
The next time an AP Statistics question asks you to describe a distribution, do not start by guessing the answer.
Pile. Slice. Say.
First identify where the observations collect. Then understand how the graph has organized them. Finally, describe the shape, outliers, center and spread using numerical evidence and the context of the question.
That simple process turns an unfamiliar statistical graph into a structured problem you can explain confidently.
