S1.1 Representing data
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Statistics 1 (AS): statistical models, histograms, stem-and-leaf diagrams, box plots, outliers and skewness. Means, standard deviations, coding and interpolated medians are on the S1.2 page; here those values are given to you whenever they are needed.
What this page covers
8 points
- What a statistical model is, the modelling cycle, and the advantages and limitations of models in context
- Class boundaries, class widths and frequency density
- Histograms with unequal class widths, the modal class and estimating a frequency from part of a class
- Histogram scale problems, where a bar is given in centimetres
- Stem-and-leaf diagrams (including back-to-back) and quartiles from raw data
- Box plots, the interquartile range and outliers using the rule stated in the question
- Skewness from quartiles, from mean, median and mode, and from a coefficient given in a question
- Comparing two distributions: location, spread, skewness and outliers, always in context
In one minute
6 points
- Data are represented so that their shape, centre and spread can be seen at a glance: a histogram shows grouped data through area, a stem-and-leaf diagram keeps every value, and a box plot reduces a data set to five numbers and flags anything unusual.
- A statistical model is a simplified description of a real situation. It is quick and cheap and makes predictions, but it rests on assumptions; data are collected to test it, and the model is refined if it does not fit.
- In a histogram the area of each bar is proportional to the frequency. With frequency density on the vertical axis, frequency density = frequency ÷ class width, and area = frequency.
- For \(n\) raw values, \(Q_1, Q_2, Q_3\) sit at positions \(\tfrac n4, \tfrac n2, \tfrac{3n}4\): a whole number means average that value and the next; otherwise round up.
- An outlier is a value beyond the fences given in the question, often \(Q_1-1.5\times\text{IQR}\) and \(Q_3+1.5\times\text{IQR}\); plot it as a cross and end the whisker at the last value inside.
- Positive skew: \(Q_3-Q_2>Q_2-Q_1\), or mode < median < mean. Negative skew: the reverse. Compare distributions using a measure of location and a measure of spread, in context.
Contents
13 sections
- What is a statistical model, and why use one?
- How do you draw and read a histogram?
- How do you build a stem-and-leaf diagram and find quartiles from it?
- How do you draw a box plot and identify outliers?
- How do you compare two distributions?
- How do you describe skewness?
- How do you tackle a full data-representation question?
- Exam tips for S1.1
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What is a statistical model, and why use one?
Real situations are too complicated to describe exactly, so statistics replaces them with something simpler that still captures what matters: a model.
Examples: "each face of this dice is equally likely", "the proportion of seeds that germinate is 0.8", "heights of adult men are spread symmetrically about their mean". None is exactly true, but each is useful and makes predictions that can be checked.
Predictions are compared with real data; if the fit is poor, the model is refined and tested again: the modelling cycle.
| Advantages of a model | Limitations of a model |
|---|---|
| quick and cheap compared with experimenting on the real situation | it simplifies reality, so its predictions are only approximate |
| gives predictions, which help decision-making and planning | it rests on assumptions that may be wrong or may stop being true |
| helps us understand the situation and which factors matter | it may only work for a restricted range of values or situations |
| can be changed and re-run easily when conditions change | it needs to be tested against real data before it can be trusted |
A seed company models the probability that one of its seeds germinates as 0.8, and assumes that seeds germinate independently. (a) Give two advantages of using this model. (b) Give one limitation. (c) Use the model to predict the number of seeds that germinate when 250 are planted. (d) In a trial, 171 of 250 seeds germinate. Comment on the model and describe the next stage of the modelling process.
(a) Any two, in context: it is quick and cheap to make predictions without planting large batches; it helps the company plan how many seeds to put in each packet.
(b) The germination rate depends on soil, temperature and the age of the seed, which the single value 0.8 ignores.
(c) \(0.8\times250=200\), so 200 seeds.
(d) 171 is well below 200 (proportion \(171/250=0.684\)), so the model does not fit these data well. Refine it, for example by using a probability nearer 0.68, then collect more data to test the refined model.
Writing a generic answer such as "models are simplified". Advantages and limitations count only when they refer to the situation in the question: the seeds, the packets, the conditions they are grown in.
How do you draw and read a histogram?
A histogram is the picture for continuous grouped data (lengths, times, masses), so the bars touch. The key idea: the area of each bar is proportional to the frequency. With equal widths, area and height go together; with unequal widths they do not.
Class boundaries and widths
Class boundaries are where one class ends and the next begins. They depend on how the data were recorded. For lengths recorded to the nearest mm, the class "20–24" contains every length that rounds to 20, 21, 22, 23 or 24, so its boundaries are 19.5 and 24.5. For ages in completed years, "20–24" means \(20\le \text{age}<25\), because a person aged 24 years and 11 months still says they are 24.
Class width = upper boundary − lower boundary.
The class with the tallest bar, that is the largest frequency density, is the modal class. It is not necessarily the class with the largest frequency: a wide class can collect many values without being crowded.
To estimate how many values lie in part of a class, assume the values are spread evenly across the class. Then the number in the part is the area of that part of the bar: frequency ≈ frequency density × width of the part.
The table shows the journey times, \(t\) minutes, of 100 commuters.
| Time \(t\) (min) | \(0\le t<10\) | \(10\le t<15\) | \(15\le t<20\) | \(20\le t<30\) | \(30\le t<50\) |
|---|---|---|---|---|---|
| Frequency | 12 | 18 | 25 | 30 | 15 |
(a) Calculate the frequency density of each class. (b) State the modal class. (c) Estimate the number of commuters whose journeys took between 12 and 25 minutes.
(a) Divide each frequency by its class width: \(12\div10=1.2\), \(18\div5=3.6\), \(25\div5=5\), \(30\div10=3\), \(15\div20=0.75\).
(b) The largest frequency density is 5, so the modal class is \(15\le t<20\), even though \(20\le t<30\) has the largest frequency.
(c) Add the areas between 12 and 25: \((15-12)\times3.6+25+(25-20)\times3=10.8+25+15=50.8\), so about 51 commuters.
Draw the same data with frequency as the height and the picture lies: the class \(30\le t<50\) holds only 15 commuters but gets one of the largest areas.
The lengths of seedlings were measured to the nearest millimetre and grouped. On a histogram, the class 20–24 mm, which has frequency 15, is drawn as a bar 2 cm wide and 6 cm tall. The class 25–39 mm has frequency 9. (a) Write down the class boundaries of the class 25–39. (b) Find the width and the height of the bar for the class 25–39.
(a) Lengths are rounded to the nearest mm, so the boundaries are 24.5 mm and 39.5 mm (width 15 mm).
(b) Horizontal scale from the known bar: 20–24 has width \(24.5-19.5=5\) mm, drawn as 2 cm, so 1 mm ↔ 0.4 cm. The new width is \(15\times0.4=6\) cm.
Area scale from the known bar: \(2\times6=12\text{ cm}^2\) represents 15 seedlings, so 1 seedling ↔ 0.8 cm². The new area is \(9\times0.8=7.2\text{ cm}^2\), so the height is \(7.2\div6=1.2\) cm.
Width 6 cm, height 1.2 cm.
In every scale problem, work out two separate scales from the bar you know: width (cm per unit of the variable) and area (cm² per unit of frequency). Height then comes from area ÷ width.
How do you build a stem-and-leaf diagram and find quartiles from it?
A stem-and-leaf diagram keeps every value. Each value splits into a stem (leading digits) and a leaf (last digit), with leaves in order, smallest next to the stem: a sorted list that also shows the shape.
Every diagram needs a key, for example "2 | 3 means 23 seconds". Without it, 2 | 3 could be 2.3 or 230. In a back-to-back diagram two data sets share the stems: leaves on the right are read left to right as usual, leaves on the left are read from right to left (smallest nearest the stem), and the key must say how to read each side.
Because the values are in order, the median and quartiles can be read off directly. S1 uses a fixed rule for raw (ungrouped) data.
For grouped data the quartiles are estimated by interpolation instead; that method is on the S1.2 page.
The times, in seconds, taken by 19 people to solve a puzzle were
41, 27, 53, 35, 64, 23, 45, 38, 91, 50, 31, 44, 56, 36, 42, 58, 34, 47, 39
(a) Draw an ordered stem-and-leaf diagram. (b) Find the median and the quartiles.
(a) Order the values, then split into stems (tens) and leaves (units).
| Stem | Leaf |
|---|---|
| 2 | 3 7 |
| 3 | 1 4 5 6 8 9 |
| 4 | 1 2 4 5 7 |
| 5 | 0 3 6 8 |
| 6 | 4 |
| 7 | |
| 8 | |
| 9 | 1 |
Key: 2 | 3 means 23 seconds. Empty stems 7 and 8 must be shown, or the gap before 91 disappears.
(b) Positions for \(n=19\): \(19/4=4.75\to\) 5th value; \(19/2=9.5\to\) 10th; \(3\times19/4=14.25\to\) 15th.
\(Q_1=35,\ Q_2=42,\ Q_3=53\) seconds.
Averaging two values when the position is not a whole number: for \(19/4=4.75\), taking the mean of the 4th and 5th values instead of the 5th value. Average only when the position is a whole number; otherwise round up, and write the position down so the method is clear.
How do you draw a box plot and identify outliers?
A box plot is drawn against a labelled scale: a box from \(Q_1\) to \(Q_3\), a line across it at the median, and whiskers out to the smallest and largest values that are not outliers. The box holds the middle half of the data, so its length is the IQR.
An outlier is an extreme value that lies outside a stated range. The question always gives the rule: most often outside \(Q_1-1.5\times\text{IQR}\) to \(Q_3+1.5\times\text{IQR}\), sometimes outside mean ± 2 × standard deviation. Use exactly the rule and multiplier you are given.
For the puzzle times in Worked example 4, an outlier is a value less than \(Q_1-1.5\times\text{IQR}\) or greater than \(Q_3+1.5\times\text{IQR}\). (c) Show that there is exactly one outlier. (d) Draw a box plot. (e) Describe the skewness of the data, giving a reason.
(c) \(\text{IQR}=53-35=18\) and \(1.5\times18=27\). Lower fence \(35-27=8\); upper fence \(53+27=80\). No value is below 8; only \(91>80\), so 91 is the one outlier.
(d) The whiskers go to the last values inside the fences: lower whisker to 23, upper whisker to 64, and a cross at 91 (figure below).
(e) \(Q_3-Q_2=11>Q_2-Q_1=7\), so the data are positively skewed.
Drawing the upper whisker all the way to the outlier, or stopping it at the fence (80). The whisker ends at the largest value that is not an outlier, here 64.
How do you compare two distributions?
Make at least two comparisons: one of location (median or mean) and one of spread (IQR, range or standard deviation). Add skewness or outliers if they differ between the groups.
Compare like with like: medians with IQRs, means with standard deviations; with skew or outliers prefer the median and IQR, which extreme values do not affect. Every comparison must be in context: not "B has a higher median" but "group B's reaction times were generally slower".
The reaction times, in hundredths of a second, of 16 students who play a racket sport (group A) and 16 who do not (group B) are shown in the back-to-back stem-and-leaf diagram.
| Group A | Stem | Group B |
|---|---|---|
| 9 | 1 | |
| 8 8 7 5 4 2 | 2 | 6 9 |
| 8 6 4 3 1 0 | 3 | 1 3 4 5 6 7 7 8 9 |
| 9 5 1 | 4 | 0 1 2 4 |
| 5 | ||
| 6 | 1 |
Key: 9 | 1 | 2 means 19 for group A and 12 for group B (hundredths of a second).
(a) Find the median and interquartile range of each group. (b) An outlier is a value more than \(1.5\times\text{IQR}\) beyond the nearer quartile. Identify any outliers. (c) Compare the reaction times of the two groups.
(a) With \(n=16\): \(Q_1\) = mean of the 4th and 5th values, \(Q_2\) = mean of the 8th and 9th, \(Q_3\) = mean of the 12th and 13th.
A: \(Q_1=26,\ Q_2=30.5,\ Q_3=37\), so median 30.5, IQR 11.
B: \(Q_1=33.5,\ Q_2=37,\ Q_3=40.5\), so median 37, IQR 7.
(b) A: fences \(26-16.5=9.5\) and \(37+16.5=53.5\); values run from 19 to 49, so no outliers. B: fences \(33.5-10.5=23\) and \(40.5+10.5=51\); 61 is an outlier.
(c) Median A < median B: group A generally reacted faster. IQR A > IQR B: group A's times were more variable; group B was more consistent (apart from one slow time). A is positively skewed (\(6.5>4.5\)); B is symmetrical between the quartiles (\(3.5=3.5\)).
Draw comparative box plots on the same scale, one above the other, and label each. A scale that differs between the two plots makes the picture misleading even if every number is right.
How do you describe skewness?
Skewness describes which way a distribution leans. The key rule is that skew is named after the long tail, not after the peak. A long tail of high values pulls the mean upwards more than the median, and the median more than the mode.
| Shape | Tail | Quartiles | Averages |
|---|---|---|---|
| Symmetrical | none | \(Q_3-Q_2=Q_2-Q_1\) | mode = median = mean |
| Positive skew | long tail to the right (high values) | \(Q_3-Q_2>Q_2-Q_1\) | mode < median < mean |
| Negative skew | long tail to the left (low values) | \(Q_3-Q_2<Q_2-Q_1\) | mean < median < mode |
A question may define a skewness coefficient, such as \(\dfrac{3(\text{mean}-\text{median})}{\text{standard deviation}}\). Evaluate it exactly as given: a positive value means positive skew, a negative value means negative skew, and a value near 0 means roughly symmetrical.
For a sample of weekly rents, in pounds, the mean is 186, the median is 172, the mode is 160, the standard deviation is 42, \(Q_1=148\) and \(Q_3=214\). (a) Show, in two different ways, that the rents are positively skewed. (b) One measure of skewness is \(\dfrac{3(\text{mean}-\text{median})}{\text{standard deviation}}\). Evaluate it. (c) A rent is an outlier if it is more than 2 standard deviations from the mean. Determine whether a rent of £295 is an outlier. (d) State, with a reason, whether the mean or the median is the better average for these rents.
(a) Quartiles: \(Q_3-Q_2=42>Q_2-Q_1=24\). Averages: mode 160 < median 172 < mean 186. Both show positive skew.
(b) \(\dfrac{3(186-172)}{42}=\dfrac{42}{42}=\) 1, positive, confirming positive skew.
(c) Upper limit \(186+2\times42=270\). Since \(295>270\), £295 is an outlier.
(d) The median: the data are skewed (with a high outlier), which pulls the mean upwards but hardly affects the median.
Describing skew from where the peak is: "the peak is on the left, so it is negatively skewed". The skew is named after the long tail. A peak on the left with a tail to the right is positive skew.
How do you tackle a full data-representation question?
Longer questions join several skills: recover a frequency from a histogram, rescale a bar, estimate part of a class, then judge skew and outliers. Label each step.
The lengths, \(t\) minutes, of the 136 calls made from an office in one week are summarised below.
| Length \(t\) (min) | \(0\le t<2\) | \(2\le t<4\) | \(4\le t<6\) | \(6\le t<10\) | \(10\le t<20\) |
|---|---|---|---|---|---|
| Frequency | 26 | \(x\) | 30 | 28 | 16 |
On a histogram of these data the bar for \(0\le t<2\) is 6.5 cm tall and the bar for \(2\le t<4\) is 9 cm tall. (a) Find \(x\). (b) Find the height of the bar for \(10\le t<20\). (c) Estimate the number of calls longer than 8 minutes. (d) Estimates of the quartiles are \(Q_1=2.44\), \(Q_2=4.40\) and \(Q_3=7.43\). Describe, with a reason, the skewness of the call lengths. (e) The shortest call lasted 0.3 min, and the two longest lasted 13.6 min and 19.2 min. An outlier is a value greater than \(Q_3+1.5\times\text{IQR}\) or less than \(Q_1-1.5\times\text{IQR}\). Show that exactly one call is an outlier. (f) Describe the box plot of these data.
(a) The two bars have equal widths, so frequency ∝ height: \(\dfrac{x}{26}=\dfrac{9}{6.5}\Rightarrow\) \(x=36\). (Check: \(26+36+30+28+16=136\).)
(b) First bar: fd \(=26\div2=13\) ↔ 6.5 cm, so 1 unit of fd ↔ 0.5 cm. Last bar: fd \(=16\div10=1.6\), height \(=1.6\times0.5=\) 0.8 cm.
(c) Part of the 6–10 class (fd \(28\div4=7\)) plus the whole 10–20 class: \((10-8)\times7+16=14+16=\) 30 calls.
(d) \(Q_3-Q_2=3.03>Q_2-Q_1=1.96\), so positive skew (a long tail of long calls).
(e) \(\text{IQR}=7.43-2.44=4.99\), \(1.5\times4.99=7.485\). Upper fence \(7.43+7.485=14.915\); lower fence \(2.44-7.485=-5.045\). \(19.2>14.915\), an outlier; \(13.6<14.915\) and no call is shorter than 0, so 19.2 is the only outlier.
(f) A box from 2.44 to 7.43 with the median at 4.40; whiskers to 0.3 and 13.6; a cross at 19.2.
When a histogram is given with a scale in centimetres, use one bar you know to fix the link between height and frequency density (or between area and frequency), and state that link in a line of working.
Exam tips for S1.1
Before any histogram calculation, write the class boundaries and widths in a column. "10–14" for lengths to the nearest cm is 9.5 to 14.5; for ages it is 10 to 15. Most wrong frequency densities start with a wrong width.
For quartiles from a list or a stem-and-leaf diagram, write "\(n/4=4.75\to\) 5th value" before the answer. The rule is easy to apply and the position line makes your method visible.
Show the fence calculation, list which values lie beyond it, and then end each whisker at the last value inside. Plot each outlier as its own cross.
"Compare" needs a measure of location and a measure of spread, each with a sentence about the real situation: "the median time for A is lower, so A generally reacted faster".
"Positive skew" alone is rarely enough. Add the reason: the quartile comparison with numbers, or the order of mode, median and mean.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The height of a histogram bar is the frequency. | With unequal widths a wide class would look too important. | Area is proportional to frequency; height is frequency density. |
| The modal class is the class with the largest frequency. | A wide class collects more values without being more crowded. | The modal class has the largest frequency density. |
| The whisker goes to the outlier. | Outliers are plotted separately. | The whisker ends at the most extreme value that is not an outlier. |
| A peak on the left means negative skew. | Skew is named after the tail, not the peak. | Peak on the left with a long right tail is positive skew. |
| An outlier is an error and should be removed. | Many outliers are genuine, just unusual. | Remove it only if there is a reason to think it is a mistake; otherwise keep it and say so. |
| A model that fails one test is useless. | Refining is a normal stage of modelling. | Use the data to adjust the assumptions, then test again. |
How this topic connects
S1.2 Measures of location and dispersion
Means, standard deviations, coding and medians of grouped data by interpolation supply the numbers that this page draws and interprets.
S1.6 The Normal distribution
A Normal model is symmetrical, with mean = median = mode; checking a data set's skew is one way to decide whether a Normal model is reasonable. The "mean ± 2 standard deviations" outlier rule comes from the same idea.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| class width = upper boundary − lower boundary | widths of histogram bars | No – learn it |
| frequency density = frequency ÷ class width | heights of histogram bars | No – learn it |
| area of bar = \(k\) × frequency | histogram scale problems; estimating part of a class | No – learn it |
| \(Q_1, Q_2, Q_3\) at positions \(\tfrac n4, \tfrac n2, \tfrac{3n}4\) (whole number: average with the next; otherwise round up) | quartiles of raw data, e.g. from a stem-and-leaf diagram | No – learn it |
| \(\text{IQR}=Q_3-Q_1\) | spread, unaffected by extreme values | No – learn it |
| \(Q_1-1.5\times\text{IQR}\), \(Q_3+1.5\times\text{IQR}\) | outlier fences | No – learn it |
| \(Q_3-Q_2>Q_2-Q_1\) ⇔ positive skew | skewness from quartiles | No – learn it |
| mode < median < mean ⇔ positive skew | skewness from averages | No – learn it |
Check your understanding
1. Ages are recorded in completed years. Give the class boundaries and width of the class "40–49".
40 and 50 (someone aged 49 years 11 months is still "49"), so the width is 10.
2. The class \(2.5\le x<4\) has frequency 21. Find its frequency density.
Width \(4-2.5=1.5\), so frequency density \(=21\div1.5=14\).
3. A stem-and-leaf diagram shows 27 values. Which values give \(Q_1\), \(Q_2\) and \(Q_3\)?
\(27/4=6.75\to\) 7th; \(27/2=13.5\to\) 14th; \(3\times27/4=20.25\to\) 21st.
4. \(Q_1=12\), \(Q_2=15\), \(Q_3=22\). Using the \(1.5\times\text{IQR}\) rule, is 40 an outlier? Describe the skew.
\(\text{IQR}=10\), fences \(12-15=-3\) and \(22+15=37\); \(40>37\), so yes. \(Q_3-Q_2=7>Q_2-Q_1=3\), so positive skew.
5. A data set has mean 52, median 55 and mode 58. Describe its skewness.
mean < median < mode, so negative skew (a long tail of low values).
Specification checklist
- (S1 1.1) I can explain what a statistical model is, describe the stages of the modelling cycle, and give advantages and limitations of a model in context.
- (S1 1.1) I can use a model to make a prediction, compare it with observed data, and suggest how the model could be refined.
- (S1 2.1) I can find class boundaries and widths, calculate frequency densities, draw histograms with unequal class widths and identify the modal class.
- (S1 2.1) I can solve histogram scale problems and estimate the number of values in part of a class.
- (S1 2.1) I can draw and read ordered stem-and-leaf diagrams, including back-to-back diagrams, with a key, and find quartiles from them.
- (S1 2.1) I can draw box plots and use them, with other diagrams, to compare two distributions in context.
- (S1 2.4) I can describe skewness using quartiles, using the mean, median and mode, or using a coefficient given in a question.
- (S1 2.4) I can identify outliers with the rule given in a question and show them on a box plot.