Module 30 of 45Lesson 1 of 16 · Unit S22 · Bars With Meaning
Statistics & Probability · Unit S22 · Extended
Bars With Meaning
A bar chart is a picture of categories, and its gaps mean something: nothing lives between Maths and Setswana. A histogram is a picture of a measured scale, and its bars have to touch, because something lives everywhere along that scale. Once the bars touch, the outline they make has a name, a shape and a story — and this unit is about reading that story honestly, and about the one decision that can quietly rewrite it: how wide you make the classes.
Two decisions do most of the work here. Whether the bars touch is decided by the data and you have no say in it. How wide the classes are is decided by you, and the shape everybody reads off the chart depends on it. Get the first right and say the second out loud, and the rest of this unit is reading.
By the end of this unit you can
- Draw a histogram for continuous data grouped in equal fixed intervals, with the bars touching and both axes labelled
- Justify why the bars of a histogram touch and the bars of a bar chart do not, from the type of data each one shows
- Identify the shape of a distribution — symmetrical, skewed to the right, skewed to the left or bimodal — and justify the choice from the histogram itself
- Explain what each shape says about the measurements that made it
- Construct a frequency polygon by joining the midpoints of the class tops, and overlay two of them to compare two distributions
- Justify a comparison of two distributions, including why the two charts must use the same classes and the same axes
- Determine the effect of changing the class width on the shape a histogram shows, and justify a sensible choice of width
- Explain why a histogram states a modal class and why the tallest bar is not the average
The one idea behind all of it
A histogram is not a bar chart with the gaps removed. It is a different claim about a different kind of data. Bar chart: separate categories, separate bars, and the gaps say so. Histogram: one continuous measured scale cut into equal fixed intervals that meet, so the bars touch because the classes touch — a gap would claim that no measurement can land there, and that is false. Once the bars touch, the outline becomes information: its shape names how the measurements are spread, and joining the midpoints turns that outline into a line you can lay over another one. But the shape is not handed to you by the data alone. You chose the class width, and a different width can turn one hump into two, or two into none. Choosing it, and saying why you chose it, is part of the answer.
bars touch ⇔ the scale is continuous
and the shape you see is partly the width you chose
Three things a histogram can tell you exactly, and one it cannot. Exactly: how many measurements fall in each class, how many fall below any class boundary, and which class is the modal class. Not exactly: anything that needs a value from inside a class — which is why the tallest bar is a modal class and never an average.
Where this unit sits
What you already have. S17 Averages, Continued owns the grouped machinery: class boundaries, the rule at the join, midpoints, Σfx, the estimated mean and the median class. None of it is retaught here. This unit borrows exactly one piece — the midpoint of a class — because a frequency polygon is built on midpoints, and it sends you back to S17 for everything else.
Shape has been building for a while. S14 Box and Whisker started the shape language: a box pushed to one end of its whiskers is already telling you about skew. A histogram says the same thing with far more detail, and this unit gives the shapes their names.
Honesty. S12 Lies and Statistics set the rule this unit leans on hardest: a chart that changes its scale between two pictures is not comparing anything. Here that becomes concrete — two histograms may only be compared when the classes and the frequency axis are identical, and this unit refuses the comparison otherwise.
Where the numbers come from. Every dataset on this page is invented illustrative data, printed in full so you can regroup and check it yourself, and grouped by counting the raw values into classes. The 60 pumpkin masses and the 80 seedling heights are printed below in full, which is what lets you regroup them yourself and check that the shape really does change.
1
Bars that touch
the histogram for continuous data with equal fixed intervals
Continuous data is measured, not counted, so it can take any value along a scale. Grouping it into equal fixed intervals that meet — 0 ≤ x < 5, 5 ≤ x < 10, and so on — is how it becomes a table, and S17 set that up. The histogram is what that table looks like. Along the bottom runs the measured scale itself, marked at the class boundaries, not at category names. Up the side runs frequency, starting at 0. Each bar stands over its own class, from boundary to boundary, at the height of its frequency — and because the top of one class is the bottom of the next, the bars touch because the classes touch. A gap would be a claim, and the claim would be false: it would say that no pumpkin in the field weighs 5 kg, when in fact 11 of them weigh between 5 and 10 kg. Two more rules follow from the same idea. The intervals must all be the same width, or the heights stop being comparable; and the frequency axis must start at 0, or the heights stop being frequencies.
EXAMPLEThe 60 pumpkin masses, in kilograms, exactly as they were recorded. 18.7, 7.8, 4.6, 17.8, 9.5, 22.7, 12.9, 17.5, 17.2, 10.2, 21.5, 20.3, 16.1, 18.3, 5.2, 16.7, 12.4, 10.6, 13.2, 23.2, 20.5, 1.8, 9.6, 10.7, 12.0, 13.0, 14.4, 8.1, 24.8, 20.3, 8.6, 15.7, 8.4, 21.3, 28.5, 20.4, 9.5, 13.7, 16.8, 11.5, 8.0, 11.6, 2.3, 18.6, 14.8, 5.4, 25.4, 11.9, 20.7, 23.0, 8.8, 16.0, 3.9, 14.9, 16.4, 28.7, 25.6, 0.8, 24.5, 19.5. Counting them into 6 classes of width 5 gives the frequencies 5, 11, 15, 13, 12, 4, which is the table the histogram is drawn from. The list is printed once and kept, because this unit regroups it more than once and you should be able to check every regrouping yourself.
The chart the rest of the unit works from. The mass scale runs along the bottom marked at the class boundaries, the frequency axis starts at 0, and every bar meets its neighbours.
Worked example 1 · Touching bars
1
Check the data type before drawing anything.
measured in kg → histogram
2
Check the classes: equal fixed intervals that meet.
6 classes, all 5 kg wide ✓
3
0 ≤ x < 5 holds 5 pumpkins: a bar from 0 to 5.
height 5
4
5 ≤ x < 10 holds 11 pumpkins: a bar from 5 to 10.
height 11
5
10 ≤ x < 15 holds 15 pumpkins: a bar from 10 to 15.
height 15
6
15 ≤ x < 20 holds 13 pumpkins: a bar from 15 to 20.
height 13
7
20 ≤ x < 25 holds 12 pumpkins: a bar from 20 to 25.
height 12
8
25 ≤ x < 30 holds 4 pumpkins: a bar from 25 to 30.
height 4
9
Add the frequencies as a check, then label the mass axis at the boundaries 0, 5, 10, 15, 20, 25, 30 and the frequency axis from 0.
60 = n ✓
Justify the touching. The bar for 5 ≤ x < 10 runs to 10 and the bar for 10 ≤ x < 15 starts at 10, so they share an edge and there is nowhere for a gap to go. That is not a drawing convention: a gap at 10 kg would be a claim that no pumpkin in this field can weigh 10 kg, and 15 of them weigh between 10 and 15 kg. In one sentence, the bars touch because the classes touch — and because every class here is 5 kg wide, the heights may be compared directly, which is what makes the tallest bar mean anything at all.
EXAMPLEThe class 5 ≤ x < 10 holds 11 of the 60 pumpkins, so its bar is drawn from 5 to 10 along the mass axis at a height of 11. The next bar begins exactly where that one stops. A pumpkin of exactly 10 kg is counted in the class above — the inequality signs say so — and on the chart it sits on the line where the two bars meet, which is why there is no room for a gap.
Practice · Bars that touch · 10 questions
P1State the one word that describes the kind of data a histogram is drawn for.Fluency
Answercontinuous
P2The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. State the class width of this table.Fluency
Answer5 kg
P3The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. State the height of the bar drawn over the class 15 ≤ x < 20.Fluency
Answer13
P4The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. State how many classes this histogram has.Fluency
Answer6
P5Explain, in one sentence, why the bars of a histogram of continuous data touch.Applied
Answerbecause the top boundary of one class is the bottom boundary of the next, so the bars touch because the classes touch and a gap would claim that no measurement can land thereEvery class ends exactly where the next one begins: 5 ≤ x < 10 stops at 10 and 10 ≤ x < 15 starts at 10. There is no stretch of the mass axis that belongs to neither class, so there is nowhere on the axis for a gap to sit. Put one there and you have made a claim about the data — that nothing was measured at 10 kg — which the table itself contradicts, since 15 pumpkins fall in the class beginning at 10 kg.
P6The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. Calculate how many pumpkins weighed less than 10 kg.Applied
Answer16Two bars lie below 10 kg, so add their heights: 5 + 11 = 16. Nothing has to be estimated here — the classes meet, so every pumpkin under 10 kg is in one of those two classes and none is counted twice. This is the kind of question a histogram answers exactly, and it is why the frequency axis has to start at 0: the heights are the counts themselves.
P7The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. State the numbers that should be written along the horizontal axis of the histogram.Applied
Answer0, 5, 10, 15, 20, 25, 30The horizontal axis of a histogram carries the measured scale itself, so it is marked at the class boundaries: 0, 5, 10, 15, 20, 25, 30. It is not marked with class labels such as “5–10” sitting under the middle of each bar, because that turns a continuous scale into a row of names, and once the axis reads as names the gaps look natural and the chart has become a bar chart.
P8A learner draws the pumpkin histogram with the frequency axis starting at 4 instead of 0. Explain what goes wrong.Applied
Answerthe bar heights stop being the frequencies, so the differences between the classes are exaggeratedWith the axis starting at 4 the bar for the class holding 4 pumpkins is drawn only 0 units tall and the bar for the class holding 15 is drawn 11 units tall, so a real ratio of 4 to 15 is shown as 0 to 11. The chart still looks like a histogram, which is what makes it dangerous. A frequency axis begins at 0 for the same reason the bars touch: the picture has to say what the data says and nothing more.
P9The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. A second table groups the same masses in classes 10 kg wide. Justify whether the two histograms may be drawn on the same frequency axis and compared bar by bar.Challenge
Answerno — the classes are different widths, so a bar on one covers twice as much of the mass scale as a bar on the other and the heights are not comparableA bar’s height is the number of measurements in the stretch of scale it covers. A class 10 kg wide catches roughly twice as many as a class 5 kg wide from the same data, whatever the data is like: here the tallest bar goes from 15 to 28 simply because the classes doubled. Comparing those two heights measures the class width, not the pumpkins. The comparison becomes legitimate only when both pictures use one set of classes and one frequency axis.
P10The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. Determine the smallest and the largest possible number of pumpkins weighing more than 22 kg that this table is consistent with.Challenge
Answer4 and 16The class 25 ≤ x < 30 is entirely above 22 kg, so its 4 pumpkins certainly count. The class 20 ≤ x < 25 straddles 22 kg: none of its 12 pumpkins need be above 22 kg, and all of them could be. So the answer lies between 4 and 16. This is the honest limit of a grouped table — it can answer exactly at a boundary and only within a range in between, which is precisely what was traded away when the raw masses were grouped.
2
Bar chart or histogram?
the data type chooses the chart
Both charts draw rectangles whose heights are frequencies, and there the resemblance stops. A bar chart shows data that comes in separate lumps: categories with no order (favourite subject), or whole numbers that are counted (children in a household). Between one lump and the next there is genuinely nothing, so the bars are drawn apart and the gaps are part of the message. A histogram shows data that was measured on a scale where every value in between is possible, so the bars touch because the classes touch. The test is one question asked of the variable, before you draw anything: could a value fall between two of the labels I am about to write? Between Maths and Setswana, no — bar chart. Between 5 kg and 10 kg, yes, endlessly — histogram. Notice that counted numbers such as “2 children” look numeric but behave like categories: nobody has 2.4 children, so the gaps stay.
The same six frequencies, drawn as a bar chart and as a histogram. Only one of them is telling the truth about a measured quantity, and the difference is the gaps.
A real bar chart above — the gaps are information, because nothing exists between Maths and Setswana. Below, the same decision made for four datasets at once.
Worked example 2 · Bar chart or histogram?
1
Dataset 1 — favourite subject of 30 learners: Maths: 9; Setswana: 7; Science: 6; Art: 5; History: 3.
categorical → bar chart
2
Why: could a value fall between two labels?
no — gaps
3
Dataset 2 — number of children in 30 households: 0: 4; 1: 7; 2: 9; 3: 6; 4: 3; 5: 1.
discrete → bar chart
4
Why: is 1.5 children possible?
no — gaps
5
Dataset 3 — The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4.
continuous → histogram
6
Why: is 12.7 kg possible?
yes — bars touch
7
Dataset 4 — The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0.
continuous → histogram
8
Why: is 36.4 g possible?
yes — bars touch
Justify each verdict from the variable, not from the table. Datasets 1 and 2 are drawn as bar charts with gaps because their values come in separate lumps — there is no subject between Maths and Setswana, and no household with 1.5 children — so the gap is true information. Datasets 3 and 4 were measured on continuous scales where every intermediate value exists, so they are grouped into equal fixed intervals and drawn as histograms in which the bars touch because the classes touch. The give-away is never how the numbers look on the page: 2 and 41 are both whole numbers, and one of them was counted while the other was rounded from a measurement.
EXAMPLEThirty learners named a favourite subject: Maths: 9; Setswana: 7; Science: 6; Art: 5; History: 3. That is a bar chart — there is no scale running from Maths to History, and no value in between. The same thirty households reported 0, 1, 2, 3, 4 or 5 children: 0: 4; 1: 7; 2: 9; 3: 6; 4: 3; 5: 1. Still a bar chart, because 1.5 is impossible. But the mass of a pumpkin can be anything at all between 0.8 and 28.7 kg, so it is grouped and drawn as a histogram.
Practice · Bar chart or histogram? · 10 questions
P11Identify the chart that shows the favourite subject of 30 learners: Maths: 9; Setswana: 7; Science: 6; Art: 5; History: 3.Fluency
Answerbar chart
P12Identify the chart that shows the mass of 60 pumpkins measured to the nearest 0.1 kg.Fluency
Answerhistogram
P13State what the gaps in a bar chart tell the reader.Fluency
Answerthat the categories are separate, so no value can fall between them
P14State the one question to ask of a variable before choosing between a bar chart and a histogram.Fluency
Answercan a value fall between two of the labels?
P15The number of children in 30 households is 0: 4; 1: 7; 2: 9; 3: 6; 4: 3; 5: 1. Justify the choice of chart.Applied
Answera bar chart, because the number of children is discrete — counted in whole numbers — and no household can have a value between themThe variable is discrete: it is counted, and its possible values are 0, 1, 2, 3, 4 or 5 and nothing in between. A household with 1.5 children does not exist, so the stretch of scale between 1 and 2 is genuinely empty and the gap on the chart is true information. The numbers looking numeric is not the test — the test is whether the in-between values exist, and here they do not.
P16The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. Justify the choice of chart for these masses.Applied
Answera histogram, because mass is measured on a continuous scale where every value in between is possible, so the bars touch because the classes touchMass is continuous: a tomato can weigh anything at all between the lightest and the heaviest recorded, and rounding the record to the nearest gram does not change that. So the values are grouped into 10 equal classes of 10 g that meet, and the bars are drawn from boundary to boundary with no gaps. The fact that the recorded numbers are whole is a rounding decision, not a property of mass.
P17Explain why a bar chart of favourite subjects may have its bars rearranged, while a histogram may not.Applied
Answerbecause the subjects have no order, but the classes of a histogram sit on a measured scale that fixes their positionsThe categories of a bar chart are names. Put History first or last and no information is lost, because there is no scale saying where History belongs. The classes of a histogram are stretches of a measured scale: 10 ≤ x < 15 comes after 5 ≤ x < 10 because 10 kg comes after 5 kg. Move a bar and the axis becomes nonsense, which is another way of seeing why one chart has gaps and the other cannot.
P18A learner says shoe size should be drawn as a histogram because it is a measurement. Explain whether they are right.Applied
Answerno — shoe size is recorded in fixed steps, so it is discrete and belongs on a bar chart with gapsFeet are measured, but shoe size is not: it is issued in fixed steps, and there is no size between 7 and 8 that anybody can buy. The variable that reaches the chart is a counted label, so the bars stay apart. The general test survives: ask whether a value can fall between two labels. For foot length in centimetres, yes — histogram. For shoe size, no — bar chart.
P19A survey records the exact time, in minutes, that each of 40 learners spent on homework, and also which of four subjects they worked on. Justify which chart each of the two variables needs.Challenge
Answertime needs a histogram because it is continuous; subject needs a bar chart because the four subjects are separate categoriesOne dataset, two variables, two different charts — and the decision is made per variable, never per dataset. Time is measured on a continuous scale: any value between the shortest and longest is possible, so the times are grouped into equal fixed intervals and drawn with touching bars. Subject is a set of four names with no order and nothing in between, so it is drawn as four separate bars with gaps that mean something. Drawing both the same way would misrepresent one of them.
P20Justify why a bar chart of 6 categories and a histogram of 6 classes can look almost identical on the page and still be different claims.Challenge
Answerbecause the histogram’s horizontal axis is a measured scale whose bars must touch, while the bar chart’s is a list of separate categories whose gaps are informationBoth pictures are 6 rectangles whose heights are frequencies, so the ink is nearly the same. What differs is what the horizontal axis is. On the histogram it is the mass scale itself, running unbroken from 0 to 30 kg; every point on it is a possible measurement, so every point must be covered and the bars touch. On the bar chart it is a shelf of names; the space between two names corresponds to no possible value at all, so it is left empty on purpose. Read the axis, not the rectangles.
3
Reading the shape
symmetrical, skewed, bimodal – and the story each one tells
Once the bars touch, run your eye along their tops. That outline is the shape of the distribution, and it has four names worth knowing. Symmetrical: one hump, and the two sides mirror each other — most measurements near the middle, fewer as you go either way. Skewed to the right (also called positively skewed): one hump low down the scale and a long thin tail stretching to the right — most values small, a few much larger. Skewed to the left: the mirror image, the tail running down to the low end. Bimodal: two separate humps with a dip between them, which almost always means two different groups have been mixed into one dataset. The name is only half the answer. The other half is the story: a right-skewed histogram of incomes says most people earn little and a few earn a great deal; a bimodal histogram of seedling heights says somebody planted two batches two weeks apart. And say the modal class, never a single value — a continuous measurement has no mode.
Two histograms on identical axes. Read the outline, not the bars: one of these rises and falls evenly, the other leans hard to one end and trails away.
Two more. The upper one is the mirror image of dataset B; the lower one has something no single measurement process usually produces.
Worked example 3 · Name the shape
1
Dataset A — Dataset A, 40 lengths in centimetres: 0 ≤ x < 5: 1; 5 ≤ x < 10: 3; 10 ≤ x < 15: 6; 15 ≤ x < 20: 11; 20 ≤ x < 25: 9; 25 ≤ x < 30: 6; 30 ≤ x < 35: 3; 35 ≤ x < 40: 1.
peaks: 1
2
One peak, and the tails are 3 classes left and 4 right.
symmetrical
3
Dataset B — Dataset B, 40 lengths in centimetres: 0 ≤ x < 5: 6; 5 ≤ x < 10: 12; 10 ≤ x < 15: 8; 15 ≤ x < 20: 5; 20 ≤ x < 25: 4; 25 ≤ x < 30: 3; 30 ≤ x < 35: 1; 35 ≤ x < 40: 1.
peaks: 1
4
One peak low down, tails 1 class left and 6 right.
skewed to the right
5
Dataset C — Dataset C, 40 lengths in centimetres: 0 ≤ x < 5: 1; 5 ≤ x < 10: 1; 10 ≤ x < 15: 2; 15 ≤ x < 20: 3; 20 ≤ x < 25: 4; 25 ≤ x < 30: 10; 30 ≤ x < 35: 16; 35 ≤ x < 40: 3.
peaks: 1
6
One peak high up, tails 6 classes left and 1 right.
skewed to the left
7
Dataset D — Dataset D, 40 lengths in centimetres: 0 ≤ x < 5: 1; 5 ≤ x < 10: 10; 10 ≤ x < 15: 9; 15 ≤ x < 20: 2; 20 ≤ x < 25: 6; 25 ≤ x < 30: 8; 30 ≤ x < 35: 3; 35 ≤ x < 40: 1.
peaks: 2
8
Two peaks, at 5 ≤ x < 10 and 25 ≤ x < 30, with a dip between.
bimodal
The name, then the story. A is symmetrical: most lengths cluster near the middle and the two sides fall away together. B is skewed to the right: the peak sits at 5 ≤ x < 10 and the tail runs 6 classes to the right, so most of the lengths are small and a handful are much larger. C is skewed to the left, the mirror image, with 6 classes of tail on the low side. D is bimodal, and that one is a warning rather than a description: two humps in one histogram usually means two groups have been mixed, and the right response is to find the two groups and draw them separately. In every case the answer names a class — the modal class of B is 5 ≤ x < 10 — and never a single measurement, because these lengths are continuous and have no mode.
EXAMPLEDataset B has its tallest bar at 5 ≤ x < 10 and then falls away class after class, with 6 classes of tail to the right of the peak and only 1 to the left. That is skewed to the right: most of the lengths are small and a few run far above them. Dataset D has two separate peaks, at 5 ≤ x < 10 and at 25 ≤ x < 30, with a dip between them, so it is bimodal — and the honest reading is not “the average is in the dip” but “there are probably two groups here”.
Practice · Reading the shape · 10 questions
P21Dataset A, 40 lengths in centimetres: 0 ≤ x < 5: 1; 5 ≤ x < 10: 3; 10 ≤ x < 15: 6; 15 ≤ x < 20: 11; 20 ≤ x < 25: 9; 25 ≤ x < 30: 6; 30 ≤ x < 35: 3; 35 ≤ x < 40: 1. State the shape of this distribution.Fluency
Answersymmetrical
P22Dataset B, 40 lengths in centimetres: 0 ≤ x < 5: 6; 5 ≤ x < 10: 12; 10 ≤ x < 15: 8; 15 ≤ x < 20: 5; 20 ≤ x < 25: 4; 25 ≤ x < 30: 3; 30 ≤ x < 35: 1; 35 ≤ x < 40: 1. State the shape of this distribution.Fluency
Answerskewed to the right
P23Dataset C, 40 lengths in centimetres: 0 ≤ x < 5: 1; 5 ≤ x < 10: 1; 10 ≤ x < 15: 2; 15 ≤ x < 20: 3; 20 ≤ x < 25: 4; 25 ≤ x < 30: 10; 30 ≤ x < 35: 16; 35 ≤ x < 40: 3. State the modal class.Fluency
Answer30 ≤ x < 35
P24State the word for a distribution whose histogram has two separate peaks.Fluency
Answerbimodal
P25Dataset C, 40 lengths in centimetres: 0 ≤ x < 5: 1; 5 ≤ x < 10: 1; 10 ≤ x < 15: 2; 15 ≤ x < 20: 3; 20 ≤ x < 25: 4; 25 ≤ x < 30: 10; 30 ≤ x < 35: 16; 35 ≤ x < 40: 3. Justify the shape you would give this distribution.Applied
Answerskewed to the left, because the single peak is at 30 ≤ x < 35 and the tail runs 6 classes to the left against 1 to the rightCount the peaks first: the frequencies 1, 1, 2, 3, 4, 10, 16, 3 rise once and fall once, so there is 1 peak and the distribution is not bimodal. Now look at the tails either side of the peak at 30 ≤ x < 35: 6 classes hold values below it and only 1 above it, so the long thin tail runs down the scale. That is skewed to the left — most of the lengths are large, and a few are much smaller.
P26Dataset D, 40 lengths in centimetres: 0 ≤ x < 5: 1; 5 ≤ x < 10: 10; 10 ≤ x < 15: 9; 15 ≤ x < 20: 2; 20 ≤ x < 25: 6; 25 ≤ x < 30: 8; 30 ≤ x < 35: 3; 35 ≤ x < 40: 1. Explain what this shape suggests about where the 40 measurements came from.Applied
Answerthat two different groups have been mixed into one dataset, because the histogram is bimodal with peaks at 5 ≤ x < 10 and 25 ≤ x < 30The frequencies 1, 10, 9, 2, 6, 8, 3, 1 climb to 10 at 5 ≤ x < 10, fall back to 2, and climb again to 8 at 25 ≤ x < 30. Two humps with a dip between them is bimodal, and a single population of one thing measured one way very rarely produces it. The usual explanation is that two groups have been counted together — two plantings, two age groups, two machines — and the useful response is not to average across the dip but to split the data and draw the two groups separately.
P27Dataset A, 40 lengths in centimetres: 0 ≤ x < 5: 1; 5 ≤ x < 10: 3; 10 ≤ x < 15: 6; 15 ≤ x < 20: 11; 20 ≤ x < 25: 9; 25 ≤ x < 30: 6; 30 ≤ x < 35: 3; 35 ≤ x < 40: 1. Explain why the answer to “what is the mode?” is a class and not a number.Applied
Answerbecause the lengths are continuous, so almost no value occurs twice; the table gives the modal class 15 ≤ x < 20 insteadA mode is a value that occurs most often. These lengths were measured to the nearest tenth of a centimetre, so repeats are accidents and mean nothing about the data. What the table can give is the class with the highest frequency — here 15 ≤ x < 20, holding 11 of the 40 lengths — and that is a statement about an interval of the scale, not about any single leaf, pod or learner.
P28Dataset B, 40 lengths in centimetres: 0 ≤ x < 5: 6; 5 ≤ x < 10: 12; 10 ≤ x < 15: 8; 15 ≤ x < 20: 5; 20 ≤ x < 25: 4; 25 ≤ x < 30: 3; 30 ≤ x < 35: 1; 35 ≤ x < 40: 1. Calculate how many of the 40 lengths were 20 cm or more.Applied
Answer9Add the heights of every bar from 20 ≤ x < 25 upwards: 4 + 3 + 1 + 1 = 9. That is 9 of the 40 lengths, which is the thin right-hand tail this distribution is named for — a small number of values spread over a long stretch of the scale, while 31 of the lengths sit below 20 cm.
P29Dataset B, 40 lengths in centimetres: 0 ≤ x < 5: 6; 5 ≤ x < 10: 12; 10 ≤ x < 15: 8; 15 ≤ x < 20: 5; 20 ≤ x < 25: 4; 25 ≤ x < 30: 3; 30 ≤ x < 35: 1; 35 ≤ x < 40: 1. Taking each class at its midpoint, as in S17, this table estimates the mean of the 40 lengths at 13.4 cm. Justify why that mean lies outside the modal class.Challenge
Answerbecause the distribution is skewed to the right: the 6 classes of tail on the right pull the mean up, while the tallest bar stays at 5 ≤ x < 10First check the figure you have been given, because every number in it is on the table: Σfx = 535 over 40 lengths, so the estimated mean is 535 ÷ 40 = 13.375, which is 13.4 cm to 3 significant figures. Now the argument. The modal class 5 ≤ x < 10 answers “where do most of the lengths sit?”. The mean answers “where is the balance point?” and every value pulls on it in proportion to how far out it is. Here 9 lengths lie above 20 cm, some of them far above, and each of those drags the balance point up the scale; nothing on the left can pull back as hard, because the scale stops at 0. So the mean, 13.4 cm, ends up above the modal class. In a symmetrical distribution the two would very nearly coincide, which is exactly why this slip is invisible until the shape is skewed.
P30Two histograms of 40 measurements each have the same modal class and the same range. Justify whether they must have the same shape.Challenge
Answerno — one could be symmetrical and the other bimodal, with the same tallest class and the same rangeThe modal class fixes one bar and the range fixes the two ends; everything between is still free. Datasets A and D make the point: both are 40 lengths in 8 classes of 5 cm, but A is symmetrical while D is bimodal, with peaks at 5 ≤ x < 10 and 25 ≤ x < 30. Shape is a statement about the whole outline, and it cannot be recovered from any one summary number — which is the case for drawing the histogram rather than quoting statistics about it.
4
Joining the midpoints
the frequency polygon, and two distributions on one axes
A frequency polygon is the same table drawn as a line instead of a block of bars. One point per class, placed at the midpoint of the class — S17’s midpoint, (lower + upper) ÷ 2 — at a height equal to that class’s frequency. Join the points in order with straight lines and stop at the first and last points you plotted. The midpoint is used because it is the one value in the class with nothing to favour. What a polygon buys you is the whole point of this concept: two sets of bars laid over each other fight, but two lines do not. Draw both distributions as polygons on one pair of axes and the comparison — which one sits further right, which is more spread out, which has two peaks — is there in a single picture. One warning to carry into the next concept: a polygon must not be dragged down to the axis at each end unless there really are empty classes there, because a line reaching the axis claims a frequency of zero, and a claim about a class that does not exist is an invented measurement.
One point per class, at the class midpoint, at the height of the bar. Joined in order, and stopped at the last point the table gives.
Two distributions, one pair of axes, and the comparison readable at a glance. Two sets of bars could not have done this — the taller ones would have hidden the rest.
Worked example 4 · Two seasons, one pair of axes
1
Count both plantings into the SAME classes.
10 classes of 10 g
2
Early planting frequencies.
2, 4, 12, 16, 14, 10, 2, 0, 0, 0
3
Late planting frequencies.
0, 0, 0, 2, 8, 16, 13, 12, 8, 1
4
Write the midpoints once — both polygons use them.
5, 15, 25, 35, 45, 55, 65, 75, 85, 95
5
Plot the early points and join them.
(5, 2), (15, 4), (25, 12), (35, 16), …
6
Plot the late points on the same axes and join them.
(5, 0), (15, 0), (25, 0), (35, 2), …
7
Compare the centres.
modal class 30 ≤ x < 40 vs 50 ≤ x < 60
8
Compare the spreads — from the classes each table uses.
70 g of scale vs 70 g
9
Compare the shapes.
symmetrical vs single peak, slightly skewed to the right
Three sentences, each defended from the picture. Centre: the late planting is heavier — its polygon sits to the right of the early one all the way along, and its modal class is 50 ≤ x < 60 against 30 ≤ x < 40. Spread: the two are equally spread, 70 g of the scale for the late planting and 70 g for the early one, so the late crop is not more variable, just heavier. Shape: both are single humps; the early one is symmetrical, 3 classes either side of its peak, while the late one has 2 classes below its peak and 4 above, so it is slightly skewed to the right — the shift is the main difference, with a little extra tail on the heavier side. Two histograms drawn on top of each other could not have shown this — the bars would have hidden one another — and that is what the polygon is for. Note also where each line stops: at the first and last classes the table has, and not dragged down to the axis beyond them.
EXAMPLEThe early planting has 16 tomatoes in the class 30 ≤ x < 40, so the polygon carries a point at (35, 16) — the midpoint, 35 g, at a height of 16. The full set of points is (5, 2), (15, 4), (25, 12), (35, 16), (45, 14), (55, 10), (65, 2), (75, 0), (85, 0), (95, 0). The line joins them in order and stops at the last one; it is not dragged down to the axis beyond the classes the table actually has.
Practice · Joining the midpoints · 10 questions
P31State the value in each class that a frequency polygon is plotted at.Fluency
Answerthe midpoint of the class
P32Calculate the midpoint of the class 40 ≤ x < 50.Fluency
Answer45
P33The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. State the coordinates of the polygon point for the class 20 ≤ x < 30.Fluency
Answer(25, 12)
P34State how many points a frequency polygon for a table of 10 classes has.Fluency
Answer10
P35The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. Calculate the coordinates of all 10 points of the frequency polygon.Applied
Answer(5, 2), (15, 4), (25, 12), (35, 16), (45, 14), (55, 10), (65, 2), (75, 0), (85, 0), (95, 0)Take the midpoints first — (lower + upper) ÷ 2 for each class, giving 5, 15, 25, 35, 45, 55, 65, 75, 85, 95 — and pair each one with its own frequency, 2, 4, 12, 16, 14, 10, 2, 0, 0, 0. That gives (5, 2), (15, 4), (25, 12), (35, 16), (45, 14), (55, 10), (65, 2), (75, 0), (85, 0), (95, 0). The points are joined in order and the line stops at the last one; the three zero heights at the top of the scale are real, because those classes genuinely hold no tomatoes.
P36Explain why a frequency polygon is easier than two histograms when two distributions are being compared.Applied
Answerbecause two lines can be drawn on one pair of axes without hiding each other, while two sets of bars overlapBars are solid. Draw one histogram over another and the taller bars hide the shorter ones, so the reader has to look at two pictures and hold one in their head while reading the other. A polygon is a single line, so both distributions fit on one pair of axes and the comparison — which sits further right, which is more spread out, which has two peaks — is visible in one glance. The information is identical; only the reading is easier.
P37The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. Explain why the polygon for the early planting is not drawn down to the axis to the right of the point (95, 0).Applied
Answerbecause there is no class beyond it, so a line drawn there would invent a frequency for measurements nobody madeThe polygon plots one point per class in the table, and the table stops at 90 ≤ x < 100. Continuing the line past the last point means plotting a point for a class that does not exist, giving it a height, and joining it up — three claims about data that was never collected. Note the honest exception here: this polygon does reach the axis, at (75, 0), (85, 0) and (95, 0), and it is entitled to, because those classes really are in the table and really do hold nothing.
P38The 60 late-planting tomato masses, in grams, grouped in the same classes: 0 ≤ x < 10: 0; 10 ≤ x < 20: 0; 20 ≤ x < 30: 0; 30 ≤ x < 40: 2; 40 ≤ x < 50: 8; 50 ≤ x < 60: 16; 60 ≤ x < 70: 13; 70 ≤ x < 80: 12; 80 ≤ x < 90: 8; 90 ≤ x < 100: 1. Calculate the total frequency represented by the polygon and state what it must equal.Applied
Answer60, the number of tomatoes in the datasetAdd the heights of all 10 points: 0 + 0 + 0 + 2 + 8 + 16 + 13 + 12 + 8 + 1 = 60. A frequency polygon is the same table as the histogram, drawn differently, so the heights must add to n exactly as the bar heights do. It is a free check — if the points do not add to 60, either a class has been missed or a height has been misread.
P39The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. The 60 late-planting tomato masses, in grams, grouped in the same classes: 0 ≤ x < 10: 0; 10 ≤ x < 20: 0; 20 ≤ x < 30: 0; 30 ≤ x < 40: 2; 40 ≤ x < 50: 8; 50 ≤ x < 60: 16; 60 ≤ x < 70: 13; 70 ≤ x < 80: 12; 80 ≤ x < 90: 8; 90 ≤ x < 100: 1. Justify a comparison of the two plantings in terms of centre, spread and shape.Challenge
Answerthe late planting is heavier — modal class 50 ≤ x < 60 against 30 ≤ x < 40 — the two are equally spread, 70 g of the scale for the late planting and 70 g for the early one — and both have a single peak, the early one symmetrical and the late one slightly skewed to the rightCentre. The late polygon lies to the right of the early one along its whole length; the modal class moves from 30 ≤ x < 40 to 50 ≤ x < 60, a shift of about 20 g. Spread. The early masses run over 70 g of the scale and the late ones over 70 g, so the two are equally variable — the late crop is heavier, not more inconsistent. Shape. Both polygons are single humps. The early one has 3 classes either side of its peak, so it is symmetrical; the late one has 2 classes below its peak at 50 ≤ x < 60 and 4 above, so by this unit’s rule it is slightly skewed to the right. Most of the difference is a shift along the scale, with a slightly longer tail on the heavy side. All three statements are legitimate only because both plantings were counted into the same 10 classes and drawn against the same frequency axis.
P40A frequency polygon is drawn from a table of 6 classes of width 5. Determine what happens to the polygon if every class is split in two and the data regrouped, and justify your answer.Challenge
Answerit gains points and follows the data more closely, but each height is based on fewer measurements, so the line becomes less smooth and eventually shows chance rather than shapeHalving the width doubles the number of points, so the line has twice as many places to bend and can follow a genuine feature the coarser table smoothed over. The price is that each height now counts about half as many measurements, so each is more affected by chance. The 80 seedling heights show both ends of that trade: at width 5 the polygon has 8 points and one clean hump, and at width 2 it has 20 points and 6 peaks, none of which is a feature of the seedlings. Splitting classes is worth doing until the wobble stops being data.
5
Comparing honestly
same classes, same axes, or no comparison at all
Two histograms may be compared only when they are drawn on the same classes and the same axes. That is not a style rule, it is the whole basis of the comparison: if one chart uses classes 5 wide and the other uses classes 10 wide, a bar on one is not the same kind of object as a bar on the other, and the taller bar means nothing. Frequency axes that stop at different heights do the same damage more quietly — both shapes are drawn correctly, and the eye still reads the one on the shorter axis as bigger. So the honest procedure is fixed: choose one set of classes that suits both datasets, count both into it, and draw both against one frequency scale, either as a pair of histograms or as two polygons on one axes. Then, and only then, write the comparison — and write it in three parts, centre, spread and shape, each backed by something you can point at on the chart. Where the two datasets are different sizes, say so, because a taller bar may just mean more measurements.
The same two plantings as histograms. Identical classes and an identical frequency axis are what make the comparison legitimate; without them there is nothing to compare.
EXAMPLEBoth plantings were counted into the same 10 classes, each 10 g wide, and drawn against the same frequency axis. Now the comparison can be written and defended: the late planting is heavier — its modal class is 50 ≤ x < 60 against 30 ≤ x < 40 — while the two are equally spread, 70 g of the scale for the late planting and 70 g for the early one, and both have a single-peaked shape — the early planting symmetrical, the late one slightly skewed to the right.
Practice · Comparing honestly · 10 questions
P41State the two things that must be identical before two histograms may be compared.Fluency
Answerthe classes and the frequency axis
P42The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. The 60 late-planting tomato masses, in grams, grouped in the same classes: 0 ≤ x < 10: 0; 10 ≤ x < 20: 0; 20 ≤ x < 30: 0; 30 ≤ x < 40: 2; 40 ≤ x < 50: 8; 50 ≤ x < 60: 16; 60 ≤ x < 70: 13; 70 ≤ x < 80: 12; 80 ≤ x < 90: 8; 90 ≤ x < 100: 1. State the modal class of the late planting.Fluency
Answer50 ≤ x < 60
P43State the three headings a written comparison of two distributions should cover.Fluency
Answercentre, spread and shape
P44The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. Calculate how many of the early-planting tomatoes weighed less than 30 g.Fluency
Answer18
P45The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. The 60 late-planting tomato masses, in grams, grouped in the same classes: 0 ≤ x < 10: 0; 10 ≤ x < 20: 0; 20 ≤ x < 30: 0; 30 ≤ x < 40: 2; 40 ≤ x < 50: 8; 50 ≤ x < 60: 16; 60 ≤ x < 70: 13; 70 ≤ x < 80: 12; 80 ≤ x < 90: 8; 90 ≤ x < 100: 1. Compare the centres of the two plantings, quoting evidence from the tables.Applied
Answerthe late planting is heavier: its modal class is 50 ≤ x < 60 against 30 ≤ x < 40 for the early plantingThe tallest bar of the early planting stands over 30 ≤ x < 40 and the tallest bar of the late planting stands over 50 ≤ x < 60, so the whole distribution has moved about 20 g up the scale. The supporting evidence is the tails: 2 early tomatoes weigh 60 g or more, against 34 late ones. Because both tables use the same 10 classes, those two counts mean the same thing and may be set beside each other.
P46A magazine prints the early planting as 10 bars of width 10 g beside the late planting as 5 bars of width 20 g. Explain why the two pictures cannot be compared.Applied
Answerbecause a bar covering 20 g of the scale collects far more tomatoes than a bar covering 10 g, so the heights measure the class width and not the cropEach bar height is a count of everything inside its own stretch of the mass scale. Double the stretch and you roughly double the count, whatever the crop is like — the same effect that turns the 60 pumpkin masses from a tallest bar of 15 at width 5 into a tallest bar of 28 at width 10. Set the two pictures side by side and a reader compares heights that were never measuring the same thing. The fix is not a caption; it is to recount both datasets into one set of classes.
P47Two histograms use identical classes, but one frequency axis runs to 20 and the other to 40. Explain what a reader is likely to get wrong.Applied
Answerthe distribution on the shorter axis looks taller and more concentrated than it is, because the eye compares the heights on the page and not the numbersBoth charts are drawn correctly, which is what makes this harder to spot than a missing gap. A frequency of 10 fills half the height of an axis that stops at 20 and a quarter of one that stops at 40, so the same count looks twice as impressive on one page as on the other. A reader who is comparing shapes rather than reading numbers off the axis will draw the wrong conclusion every time. One axis, one scale, both charts — and if the two datasets are different sizes, say so, because a taller bar may only mean more measurements.
P48The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. The 60 late-planting tomato masses, in grams, grouped in the same classes: 0 ≤ x < 10: 0; 10 ≤ x < 20: 0; 20 ≤ x < 30: 0; 30 ≤ x < 40: 2; 40 ≤ x < 50: 8; 50 ≤ x < 60: 16; 60 ≤ x < 70: 13; 70 ≤ x < 80: 12; 80 ≤ x < 90: 8; 90 ≤ x < 100: 1. Each table runs from the bottom of its lowest occupied class to the top of its highest, so it can support a range of 70 g for the early planting and 70 g for the late one. Compare the spreads of the two plantings.Applied
Answerthey are equally spread: 70 g of the scale for the early planting and 70 g for the late oneThe two tables span 70 g and 70 g of the mass scale respectively, so the late crop is not more variable — it is simply heavier. The counts say the same thing: each planting spreads its 60 tomatoes over 7 non-empty classes of the same width. Separating centre from spread matters here, because “bigger” and “more variable” are different claims and only one of them is true.
P49The 60 early-planting tomato masses, in grams, grouped in tens: 0 ≤ x < 10: 2; 10 ≤ x < 20: 4; 20 ≤ x < 30: 12; 30 ≤ x < 40: 16; 40 ≤ x < 50: 14; 50 ≤ x < 60: 10; 60 ≤ x < 70: 2; 70 ≤ x < 80: 0; 80 ≤ x < 90: 0; 90 ≤ x < 100: 0. The 60 late-planting tomato masses, in grams, grouped in the same classes: 0 ≤ x < 10: 0; 10 ≤ x < 20: 0; 20 ≤ x < 30: 0; 30 ≤ x < 40: 2; 40 ≤ x < 50: 8; 50 ≤ x < 60: 16; 60 ≤ x < 70: 13; 70 ≤ x < 80: 12; 80 ≤ x < 90: 8; 90 ≤ x < 100: 1. Justify a full comparison of the two plantings under the headings centre, spread and shape.Challenge
Answercentre: the late planting is heavier, modal class 50 ≤ x < 60 against 30 ≤ x < 40; spread: equal, 70 g of the scale for the late planting and 70 g for the early one; shape: both single-peaked, the early one symmetrical and the late one slightly skewed to the rightCentre. The modal class moves from 30 ≤ x < 40 to 50 ≤ x < 60 and every one of the 34 late tomatoes weighing 60 g or more has only 2 early counterparts, so the late crop sits higher up the scale by about 20 g. Spread. The tables span 70 g and 70 g of the scale, so the two are equally variable; the shift is not accompanied by any extra inconsistency. Shape. Each table has 1 peak. The early table has 3 classes either side of its peak, so it is symmetrical; the late table has 2 classes below its peak and 4 above, so it is slightly skewed to the right. The licence for all three. Both plantings were counted into the same 10 classes of 10 g and drawn against the same frequency axis; without that, none of the three sentences could be defended.
P50One dataset has 40 measurements and another has 120, and they are drawn as two histograms on identical classes and one frequency axis. Justify what may and may not be concluded from the fact that every bar of the second is taller.Challenge
Answernothing about shape or centre — the second dataset has three times as many measurements, so its bars are taller everywhere; only the proportions in each class may be comparedFrequency is a count, so a dataset three times the size produces bars about three times as tall in every class, whatever its shape. Reading “taller everywhere” as “bigger measurements” confuses how many with how much. Two honest routes are open. Compare the shapes, which are unaffected by n — where the peak sits, which way the tail runs, how many humps. Or convert each frequency to a proportion of its own total, so that 11 out of 40 and 33 out of 120 become directly comparable. Both routes require what this concept insists on: identical classes.
6
Change the width, change the picture
what survives a regrouping and what blurs away
You chose the class width, and the histogram you get is partly your decision. Group the 60 pumpkin masses in 5s and you get 6 bars with a clear single hump; group the very same masses in 10s and you get 3 bars, and the detail is gone — too few bars to show a shape. Go the other way and the opposite failure appears: group them in 2s and each of the 15 classes holds so few pumpkins that the tops jump up and down 6 times from pure chance, and the chart is now showing which classes happened to catch an extra pumpkin. Neither picture is wrong arithmetically; both are misleading to read. Somewhere between them is a width that shows the shape without showing the chance, and finding it is a judgement you have to make and then defend. Two habits follow. First, whatever width you choose, say what it is and why you chose it. Second, when a feature appears at only one width, treat it with suspicion until you have looked at the data another way — the Criterion B task below is exactly that investigation.
Nothing in the data changed between these two pictures. The class width did, and it took most of the shape with it.
And the failure at the other end. Sixty measurements cut into fifteen classes leave about four in each, and four is far too few for a bar height to be stable.
Worked example 5 · Same data, two widths
1
Start from the 60 raw masses, not from a table.
60 values
2
Count them into classes 5 kg wide.
5, 11, 15, 13, 12, 4
3
Read that histogram.
6 bars, symmetrical
4
State its modal class.
10 ≤ x < 15
5
Now count the same masses into classes 10 kg wide.
16, 28, 16
6
Read that histogram.
3 bars, too few bars to show a shape
7
State its modal class.
10 ≤ x < 20
8
Say what survived and what did not.
centre survived, detail blurred
What survives and what blurs. The centre survives: both pictures put the bulk of the pumpkins in the middle of the scale, and the wider modal class 10 ≤ x < 20 simply contains the narrower one 10 ≤ x < 15. What blurs is everything the shape is made of. At 5 kg wide there are 6 bars, the outline rises and falls and the distribution reads as symmetrical; at 10 kg wide there are 3 bars, and with 3 of them a histogram cannot show a shape at all — the fall-away at the top of the scale, clearly visible in the 5-kg picture as the drop from 12 to 4, has been absorbed into one wide block. Neither chart is wrong. The 5-kg chart is more useful, and an answer that shows one of them has to say which width it used.
EXAMPLEThe same 60 masses, three times over. At width 5 there are 6 bars, the shape reads as symmetrical and the modal class is 10 ≤ x < 15. At width 10 there are only 3 bars: too few bars to show a shape, and the fall-away at the top of the scale — the drop from 12 to 4 in the wider picture — has been absorbed into one block. At width 2 there are 15 bars and 6 peaks, which is chance dressed up as detail. Same 60 pumpkins every time.
Beyond this course · one sentence only. If the classes are not all the same width, bar heights stop being comparable and the chart has to plot frequency density — frequency divided by class width — so that it is the area of each bar, not its height, that gives the frequency. That tool is named here so you know it exists and know its name; it is not part of this course, there is no practice on it in this unit, and every histogram you meet here has equal fixed intervals.
Practice · Change the width, change the picture · 10 questions
P51The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. State the modal class.Fluency
Answer10 ≤ x < 15
P52The same 60 masses grouped in tens: 0 ≤ x < 10: 16; 10 ≤ x < 20: 28; 20 ≤ x < 30: 16. State the modal class now.Fluency
Answer10 ≤ x < 20
P53The same 60 masses grouped in tens: 0 ≤ x < 10: 16; 10 ≤ x < 20: 28; 20 ≤ x < 30: 16. State how many classes this table has.Fluency
Answer3
P54State what happens to a histogram when the classes are made very narrow.Fluency
Answereach class holds very few values, so the tops rise and fall from chance and the chart shows noise instead of shape
P55The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. The same 60 masses grouped in tens: 0 ≤ x < 10: 16; 10 ≤ x < 20: 28; 20 ≤ x < 30: 16. Explain how the same 60 masses give both of these tables.Applied
Answerthe classes of the second table are pairs of the first, so its frequencies are the first table’s added in twos: 5 + 11 = 16 and 15 + 13 = 28 and 12 + 4 = 16Nothing was measured again. The class 0 ≤ x < 10 is exactly 0 ≤ x < 5 and 5 ≤ x < 10 joined together, so its frequency is 5 + 11 = 16, and the same happens down the table: 16, 28, 16. Σf is unchanged at 60, because no pumpkin was added or removed. What changed is how much of the mass scale each bar is responsible for, and that is enough to change the picture completely.
P56The same 60 masses grouped in tens: 0 ≤ x < 10: 16; 10 ≤ x < 20: 28; 20 ≤ x < 30: 16. Explain why no shape can honestly be named from this table.Applied
Answerbecause 3 classes is too few bars to show a shape: an outline needs several bars before it can rise, fall and have tailsA shape is a statement about how the frequencies change across the scale. With 3 bars, too few bars to show a shape, there is one rise and one fall and nothing else to see: the same three heights 16, 28, 16 would be produced by a great many different distributions, including the clearly symmetrical one the width-5 table shows. Naming a shape from it would be claiming detail the chart does not carry.
P57The same 60 masses grouped in classes 2 kg wide give 15 classes with 6 peaks. Explain why those peaks should not be reported.Applied
Answerbecause each class holds only a handful of pumpkins, so one or two extra values move a bar noticeably and the ups and downs are chance, not a feature of the fieldWith 60 pumpkins spread over 15 classes the average class holds about 4, and the frequencies run 2, 2, 3, 1, 8, 6, 6, 4, 8, 4, 7, 3, 4, 0, 2. At that size a single extra pumpkin is a visible step. Weigh a second field of 60 pumpkins from the same crop and the peaks would land in different places, which is the test of whether a feature is real. Reporting 6 peaks here would be reporting which classes happened to catch an extra value.
P58The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. Calculate how many pumpkins weighed 15 kg or more, and explain why the width-10 table cannot answer the same question.Applied
Answer29 from the width-5 table; the width-10 table has no boundary at 15 kgIn the width-5 table 15 kg is a boundary, so add the bars above it: 13 + 12 + 4 = 29, exactly. In the width-10 table 15 kg falls in the middle of the class 10 ≤ x < 20, which holds 28 pumpkins and says nothing about how they sit inside it — the answer could be anything from 16 to 44. Wider classes do not just blur the shape; they destroy the questions the table can answer exactly.
P59The 60 pumpkin masses, in kilograms, grouped in fives: 0 ≤ x < 5: 5; 5 ≤ x < 10: 11; 10 ≤ x < 15: 15; 15 ≤ x < 20: 13; 20 ≤ x < 25: 12; 25 ≤ x < 30: 4. The same 60 masses grouped in tens: 0 ≤ x < 10: 16; 10 ≤ x < 20: 28; 20 ≤ x < 30: 16. Justify which of the two widths you would publish, including what each one costs.Challenge
Answerwidth 5, because 6 bars show the single symmetrical hump and the fall-away at the top of the scale, while width 10 gives only 3 bars and too few bars to show a shapeWidth 5 buys detail at the cost of smaller counts per class: 6 bars, frequencies 5, 11, 15, 13, 12, 4, every one of them big enough to be stable, and the outline reads as symmetrical with a clear drop from 12 to 4 at the top of the scale. Width 10 buys stability at the cost of nearly everything else: 3 bars, frequencies 16, 28, 16, and too few bars to show a shape — the drop at the top has disappeared inside the last block. Neither is arithmetically wrong; the width-5 chart answers more questions, so it is the one to publish, and the sentence that publishes it must say which width was used.
P60A learner regroups a dataset until a second peak appears and reports it as a discovery. Justify what they should do before publishing it.Challenge
Answertest whether the peak survives at other widths and in other samples; a feature that exists at only one width is chance, not shapeSearching through widths until something appears is a way of guaranteeing that something appears: with enough narrow classes every dataset grows extra peaks, as the 60 pumpkin masses do at width 2, where 6 peaks turn up in 15 classes and none of them mean anything. The honest test has two parts. Redraw at two or three other widths and see whether the second peak is still there — a real one, like the two humps of dataset D, survives regrouping. Then ask whether there is a reason for it in the world: two plantings, two batches, two age groups. A peak that survives no width but its own and has no story behind it should be reported as what it is, an artefact of the class width.
The histogram routine
Four moves, whatever the measurements turn out to be.
1
Ask what kind of data it is, first.Measured on a scale means a histogram and touching bars. Categories or counted whole numbers mean a bar chart and gaps. Nothing else in the routine works if this step is wrong.
2
Fix equal classes, and label both axes.Equal fixed intervals that meet, boundaries marked along the measured scale, frequency up the side starting at 0, and a title. A bare axis is not a chart.
3
Read the outline, then name the shape.Count the peaks first: three or more is noise, two is bimodal, one sends you to the tails. Longer tail on the right means skewed to the right. Then say the modal class – a class, never a value.
4
Say the width you chose, and check the shape survives it.Regroup once at a different width. A feature that only exists at one width is not a feature of the data, and a comparison drawn on two different widths is not a comparison at all.
The mistakes that cost the most marks
1Leaving gaps between the bars of continuous data
WrongDrawing the 60 pumpkin masses as 6 separate bars with gaps between them, labelled “0–5”, “5–10”, “10–15” and so on, exactly as if the classes were subjects on a timetable.
RightDraw the mass itself along the bottom axis, marked at the boundaries 0, 5, 10, 15, 20, 25, 30, and let each bar run from its lower boundary to its upper one, so that the bars touch because the classes touch.
The gap is not a decoration, it is a statement, and here the statement is false. A gap between 5 ≤ x < 10 and 10 ≤ x < 15 claims that no pumpkin weighs 10 kg — when the very table being drawn puts 15 pumpkins in the class that starts at 10 kg. The same slip has a twin worth naming: writing class labels such as “5–10” along the axis instead of the boundary numbers. Labels turn a measured scale into a row of categories, and once that has happened the gaps look natural and the chart has quietly become a bar chart.
2Reading the tallest bar of a skewed histogram as “the average”
Wrong“Dataset B is skewed to the right, so the average length is about 7.5 cm.”
Right“The modal class of dataset B is 5 ≤ x < 10. Because the distribution is skewed to the right, the mean is pulled above that class by the 6 classes of tail on the right, and the table puts the mean of the 40 lengths at 13.4 cm.”
The tallest bar answers the question “where do most of the measurements sit?” and nothing else. In a symmetrical distribution that is close enough to the middle that the slip does no harm, which is exactly why it goes unnoticed — and then costs marks the moment the shape is skewed. Here the modal class runs from 5 to 10 cm while the mean the table gives is 13.4 cm, well outside it, because a long right tail drags the mean up and leaves the tallest bar behind. Say modal class, say it is a class and not an average, and if the question wants a typical value, name which average you are giving.
3Comparisons the two charts cannot support
WrongPutting the 5-kg-wide picture of the pumpkins beside the 10-kg-wide one and concluding “the second field has far more heavy pumpkins, look at the height of that bar” — and, on the polygon, running the line down to the axis at each end so that it closes neatly on the frequency scale.
RightCount both datasets into one set of classes and draw them against one frequency axis before comparing anything — as the two plantings are, in 10 identical classes of 10 g. Stop each polygon at the first and last points the table actually gives: (5, 2) and (95, 0) for the early planting.
Both halves of this trap are the same mistake — the chart claiming something the data never said. A bar 10 kg wide holds about twice as many pumpkins as a bar 5 kg wide whatever the field is like, so comparing their heights measures the class width and not the harvest; the same 60 masses give a tallest bar of 28 at one width and 15 at the other. And a polygon dragged down to the axis beyond the last class invents a class, gives it a frequency of zero and joins it up, which is a measurement nobody made. If the end classes really are empty — as 70 ≤ x < 80 and 80 ≤ x < 90 honestly are for the early planting — then the line reaching the axis is true, and that is the only case in which it may.
Tasks to submit
Criterion B · Investigating patterns
Bin it again
Eighty sunflower seedlings were measured four weeks after planting, to the nearest tenth of a centimetre. Every one of the eighty heights is printed below, so nothing here has been decided for you: you group them yourself, as many ways as the task asks. The question is not what the average height is. It is this. The shape a histogram shows is supposed to be a property of the seedlings — but the class width is a property of your decision. How much of the shape you see is theirs, and how much is yours?
The 80 seedling heights, in centimetres, in full
17.6, 9.4, 20.3, 18.6, 10.5, 22.9, 25.0, 22.8, 20.8, 18.5, 34.2, 25.9, 20.7, 8.8, 27.6, 20.2, 31.0, 19.4, 13.1, 11.7, 20.5, 30.8, 10.4, 20.0, 16.0, 15.1, 27.3, 21.6, 17.8, 22.5, 27.1, 17.4, 3.4, 31.4, 27.9, 20.5, 22.5, 28.9, 5.6, 20.8, 28.7, 13.3, 28.4, 16.4, 22.5, 4.3, 22.9, 21.8, 35.6, 11.3, 27.9, 18.6, 8.7, 19.7, 12.8, 21.4, 26.1, 7.8, 32.9, 24.9, 16.7, 27.2, 16.8, 17.8, 31.7, 27.9, 25.0, 12.9, 27.2, 23.3, 12.0, 15.3, 18.0, 21.7, 29.1, 29.1, 13.8, 19.0, 35.9, 21.8.
The second dataset · 60 heights from a different bed, in full
31.8, 13.7, 11.5, 5.7, 31.8, 28.8, 14.5, 18.6, 32.5, 15.9, 10.8, 8.8, 8.9, 28.7, 9.9, 28.4, 9.5, 27.4, 15.0, 28.2, 11.1, 12.6, 32.2, 27.4, 30.0, 12.9, 11.5, 31.1, 7.0, 24.4, 16.6, 27.1, 9.1, 4.0, 11.7, 14.1, 13.9, 12.8, 13.3, 30.7, 23.6, 24.6, 9.6, 25.2, 36.2, 5.1, 10.4, 12.1, 30.6, 24.8, 7.8, 25.5, 13.4, 14.7, 28.4, 28.4, 28.3, 12.2, 30.5, 7.4.
Part 1 · the structured investigation
- Predict, before you group anything, what you expect to happen to the shape of the histogram as the classes get narrower. Write the prediction down and do not change it afterwards.
- Tabulate the eighty heights at class widths 20, 10, 5 and 2, all starting at 0. For each width record the number of classes, the frequencies, the modal class, the number of peaks in the outline, and a one-phrase shape verdict.
- Draw the histogram at each of the four widths, on axes whose frequency scale starts at 0, so that the four shape verdicts can be seen and not just asserted.
- Identify the pattern in your table: say what improves as the classes narrow, at which width it stops improving, and what starts happening instead.
- Predict, from that pattern and before you build it, what the table at width 1 would look like — how many classes, roughly what frequencies, and how many peaks. Then build it and say whether you were right.
- Determine a general statement about choosing a class width, written as a sentence a classmate could apply to a dataset they have never seen. It must mention the number of measurements, because a rule that ignores n cannot be right.
- Test and verify your general statement on the second dataset printed below — 60 heights from a different bed. Apply your rule to choose a width, group the data at that width and at one wider and one narrower, and say whether the width your rule chose is the one that shows the data best. Be honest if it is not.
- Justify the trade-off from first principles: explain what a class width that is too wide destroys, explain what a class width that is too narrow manufactures, and explain why the second failure gets worse as the classes get narrower for a fixed n.
The table to fill in
| Class width | Number of classes | Frequencies | Modal class | Peaks | Shape verdict |
|---|---|---|---|---|---|
| 20 | |||||
| 10 | |||||
| 5 | |||||
| 2 | |||||
| 1 — predict first |
Every entry in this table is yours to compute from the printed heights. Nothing in it has been given to you.
Part 2 · your own investigation
A general statement is only worth as much as the tests it survives. Construct your own rule for choosing a class width — it may be the one you determined above, sharpened, or a completely different one, and it may use n, the range, the smallest acceptable class frequency, or anything else you can defend. State it precisely enough that two people applying it to the same dataset would choose the same width. Then stress-test it. Find, or construct, a dataset on which your rule gives a width you would not defend, and say why it fails there. Finally, justify the rule’s core claim by argument and not by example: show, in general, why a dataset of n measurements split into k classes puts on average n/k measurements in each class, and use that to argue why the number of classes cannot sensibly grow as fast as n does. You choose the rule, the tests and the way you present the argument — but a reader must be able to check every number and follow every step.
Criterion B · Investigating patterns. Levels 1–2: groups the eighty heights correctly at one or two widths and draws at least one correct histogram. Levels 3–4: all four tables complete and correct, each with its modal class and a shape verdict, the four histograms drawn on honest axes, and the pattern described in words. Levels 5–6: states a correct general statement about class width that refers to the number of measurements, predicts the width-1 table in advance, and tests the statement on the second dataset, reporting the result whichever way it falls. Levels 7–8: in addition, justifies the trade-off from first principles rather than from the two datasets — a wider class destroys information that no later step can recover, while a narrower class leaves about n/k measurements per class, so the chance wobble in each bar grows as k grows — and carries Part 2 through to a precisely stated rule, a dataset that breaks it, and a general argument rather than a worked example.
Remember, you should communicate in an organized and coherent manner.
Criterion D · Applying mathematics in real-life contexts
Was the second month really more changeable?
A school weather-watch group recorded the maximum temperature every day for two months. At the end of the second month one of them wrote in the report: “The two months were about as hot as each other, but the second month was more variable.” The group has the sheet below and has been asked to defend that sentence — or withdraw it — with a chart and a calculation, in a report the whole school will read.
Weather-watch record sheet
| Entry | Recorded |
|---|---|
| Recorded by | the Form 4 weather-watch group, one reading a day |
| Instrument | a maximum-minimum thermometer in a screened box |
| Reading taken | 17:00 each day, to the nearest 0.1 °C |
| Month 1 | 30 days, no readings missed |
| Month 2 | 31 days, no readings missed |
| Also recorded each day | minimum temperature, rainfall, wind direction, cloud cover |
| Rainfall, month 1 | 4 mm on 2 days |
| Rainfall, month 2 | 61 mm on 9 days |
| Thermometer checked against a second one | start of month 1 only |
| Month | Daily maximum temperature (°C), day 1 onwards |
|---|---|
| Month 1 | 27.2, 32.9, 26.0, 30.6, 27.2, 29.0, 36.2, 31.4, 31.7, 33.7, 23.8, 30.2, 31.8, 33.0, 34.0, 31.3, 33.5, 34.1, 29.6, 31.8, 32.6, 31.0, 33.2, 29.1, 25.3, 34.2, 30.5, 27.1, 30.7, 29.8 |
| Month 2 | 27.0, 29.2, 28.1, 36.1, 29.9, 26.0, 28.8, 34.4, 26.6, 23.8, 30.0, 26.3, 23.8, 31.6, 31.2, 33.4, 31.0, 41.8, 37.6, 26.9, 35.9, 22.6, 24.3, 30.9, 28.8, 27.0, 30.4, 35.1, 27.5, 33.9, 37.8 |
This record sheet is an invented illustrative example written for the task. It is not taken from any business, price list or publication, and no organisation, person or source is named or implied.
Blank axes for the Criterion D histograms. Choose your class interval first, use the same one for both months, and start the frequency axis at 0.
- Interpreting the situation. Justify which entries on the record sheet the claim actually needs and which are surplus to it. For two of the surplus entries, say what question they would be the right data for instead, and justify why one of them — the note about the thermometer check — still matters even though it is not used in any calculation.
- Selecting the mathematics. Justify the choice of a histogram for each month and a frequency polygon overlay for the comparison, rather than two bar charts or a single list of numbers. Justify the class interval you will use, saying why it must be the same for both months and what a wider or a narrower interval would cost you.
- Applying the mathematics. Draw a histogram of each month’s maximum temperatures on the printed axes, using your chosen interval and a frequency axis starting at 0, with the bars touching. Then draw both frequency polygons on one pair of axes. State the modal class and the shape of each month, and calculate the range and the interquartile range of each month from the full lists of daily maxima.
- Accuracy. Justify how far your answers can be trusted: say which of them come exactly from the raw lists and which depend on the interval you chose, explain what the readings being taken to the nearest 0.1 °C at one fixed time of day does and does not guarantee, and discuss what the unchecked thermometer means for a comparison of two months rather than for a single reading.
- Sense in context. Justify a verdict on the sentence in the report. Does the evidence support “about as hot as each other”, and does it support “more variable”? Say which measure carries most of the weight of your verdict and why, discuss whether two months of readings from one thermometer at one school can support a sentence the whole school will read as a statement about the weather, and rewrite the sentence so that it claims exactly what the data supports and no more.
Criterion D · Applying mathematics in real-life contexts. Levels 1–2: identifies some relevant information and begins a grouped table for at least one month. Levels 3–4: separates the needed data from the surplus, groups both months into the same classes, and draws at least one correct histogram with touching bars and a frequency axis starting at 0. Levels 5–6: both histograms and both polygons correct on shared axes, modal classes and shapes stated, range and interquartile range correct for both months, the interval choice explained, and a defensible verdict on the report sentence. Levels 7–8: in addition, justifies the interval choice against what a wider and a narrower one would cost, separates the answers that are exact from those that depend on the grouping, reaches a verdict that treats “as hot as” and “more variable” as two separate claims with separate evidence, weighs the unchecked thermometer as a threat to a comparison rather than to a single reading, and rewrites the sentence so that it claims exactly what two months of one school’s readings can support.
When you are ready
The mastery quiz is 10 questions drawn at random from a bank of 25. You need 70% to pass and you have three attempts. Type a frequency or a count as digits only, with no unit — 15, not “15 pumpkins”. Type a class as its two boundaries with the word to between them, 10 to 15. Type a shape in lower case exactly as this unit names it: symmetrical, skewed to the right, skewed to the left or bimodal. A multi-gap answer separates its gaps with the bar character, in the order the gaps appear.
Printable version
Download Unit S22 as a PDF — the same lesson, laid out for printing and for working offline.
Download Unit S22 as a PDF — the same lesson, laid out for printing and for working offline.