Corrected exercises: frequency distributions and graphs (360-300)

Relative frequencies: $\dfrac{96}{200}=48.0\%$, $\dfrac{42}{200}=21.0\%$, $\dfrac{50}{200}=25.0\%$, $\dfrac{12}{200}=6.0\%$, summing to 100%. Cumulative frequencies make no sense because the variable is NOMINAL: cumulating assumes an order, and “public transit, then walking” is not a progression. Adding these frequencies amounts to adding categories, which only an order would make legitimate. Cumulation is reserved for ordinal variables, ordered discrete variables and continuous variables.

Frequencies 104, 52, 28, 16, summing to 200. Relative frequencies: 52.0%, 26.0%, 14.0%, 8.0%. Cumulative frequencies: 104, 156, 184, 200. Cumulative relative frequencies: 52.0%, 78.0%, 92.0%, 100.0%. Two checks to make every time: the sum of the frequencies gives back $n$, and the last cumulative relative frequency is exactly 100%. If either fails, there is a rounding error or a missing value, and it must be settled before going any further.

At most one course failed: this is the cumulative relative frequency at rank 1, that is 78.0%, which corresponds to 156 students. At least two: the complement, $100-78=22.0\%$, that is 44 students. Going through the complement is almost always faster than the direct sum, and it avoids the boundary error of including or excluding the pivot category.

“3 or more” is an OPEN class, with no upper bound. What is lost is the exact value of the observations it contains, hence the possibility of computing a mean, a variance or a class midpoint for this group. It is kept all the same because it protects the anonymity of rare cases and avoids a tail of the table with frequencies of 1 or 2. It is a deliberate compromise, and it must be noted under the table rather than discovered during the analysis.

The number is NOT computable from the table alone. The first three categories give $0\times 104+1\times 52+2\times 28=108$ failures for 184 students; the 16 students of the open class contribute at least $3\times 16=48$ failures, with no upper bound. The mean is therefore at least $\dfrac{108+48}{200}=0.78$ and could be much larger. The stated value coincides exactly with the LOWER bound, which gives away a computation done by assigning 3 to the entire open class. The correct wording would be “at least 0.78 courses failed on average”, or better, a return to the raw data.

Range: $486-42=444$ dollars. Sturges' rule with $n=120$: $\log_{10}(120)\approx 2.0792$, so $k=1+3.322\times 2.0792\approx 7.91$, which is rounded to 8 classes. The rule is not compulsory: it proposes a compromise between too much detail and too much smoothing, and it grows slowly, as a logarithm, which reflects a fact: a sample four times larger deserves only two more classes.

Raw width: $\dfrac{444}{8}=55.5$ dollars, which is rounded UP to a convenient value, here 60 dollars. Limits: $\left[40, 100\right)$, $\left[100, 160\right)$, and so on up to $\left[460, 520\right)$, that is 8 classes covering 40 to 520 dollars. We check that the first contains the minimum, 42, and the last the maximum, 486. Rounding up rather than down is what guarantees that $k$ classes are enough to cover everything; rounding down would force us to add a ninth.

First rule, the classes are MUTUALLY EXCLUSIVE: an observation falls in only one, hence the convention of limits closed on the left and open on the right. Second rule, they are EXHAUSTIVE: their union covers all the observed values, with no gap. Third rule, they have EQUAL WIDTH, unless an explicit decision is made, which then forces us to reason with density. In practice we add a readability requirement: round limits, chosen for the reader and not for the software.

With 3 classes, the smoothing is so heavy that the shape disappears: a bimodal distribution, for example part-time and full-time students, will appear as a single block, and all the information about spread is lost. With 25 classes for 120 observations, there are fewer than 5 observations per class on average: the histogram becomes an irregular comb where chance in the sample creates false peaks, and the reader sees a structure that does not exist. Between the two, good practice is to try two or three divisions and keep the one whose shape stays stable.

The assumption is that the observations are spread UNIFORMLY within the class, so that their mean coincides with the midpoint. It fails as soon as the internal distribution is strongly skewed, which typically happens in the last income class, where almost everyone is near the lower limit, and in wide classes. The consequence is a slightly biased mean computed from grouped data: it is therefore presented as an approximation, and we go back to the raw data whenever they exist.

The frequencies read 8, 18, 26, 20, 10 and 4, with sum $8+18+26+20+10+4=86$. This check is not a formality: on a histogram read by eye, an error of one gradation is common, and the sum of the frequencies is the only check available when the original table does not accompany the figure.

Because the variable is CONTINUOUS: between 9.9 and 10.1 hours there is no gap, the value 10 exists, and the classes follow each other edge to edge, covering the whole interval. A bar chart represents a discrete or qualitative variable, where the categories are separated by emptiness: between 2 and 3 courses failed, no value exists, and leaving a space says so visually. The difference in the drawing therefore reflects a difference in nature, and it is exactly the kind of detail on which a marker takes off points.

A point is placed at the MIDDLE of the top of each bar, at the class midpoint on the horizontal axis and at the frequency on the vertical axis: here $(2.5, 8)$, $(7.5, 18)$, $(12.5, 26)$, $(17.5, 20)$, $(22.5, 10)$ and $(27.5, 4)$. These points are joined by segments. At both ends, an empty fictitious class is added on each side, here with midpoints $-2.5$ and $32.5$, and the polygon is brought back to zero: this closes the line and gives the area under the polygon the same value as the area of the histogram. Many students forget this closure, and the graph then floats with no reference.

The MODAL class is $\left[10, 15\right)$, the one with the highest frequency, 26 students. For the median, we cumulate: 8, then 26, then 52; the 43rd and 44th observations, which surround the middle of 86, fall in the class $\left[10, 15\right)$, which is therefore also the median class. The distribution is unimodal and SLIGHTLY SKEWED TO THE RIGHT: the right tail, from 20 to 30 hours, is longer than the left one, which is the usual shape for the time spent on an activity, bounded by zero with no upper limit.

We must switch from FREQUENCIES to RELATIVE FREQUENCIES, as percentages or proportions. Without that, the histogram of the second CEGEP would swamp the first by its size alone, and we would be comparing populations rather than shapes. Once both distributions are expressed as percentages, the two POLYGONS are superimposed rather than the two histograms, precisely because two overlapping sets of bars are unreadable while two lines are easy to tell apart.

Densities: $\dfrac{20}{10}=2.0$, $\dfrac{30}{10}=3.0$, $\dfrac{40}{20}=2.0$ and $\dfrac{40}{40}=1.0$ households per thousand dollars. Density is interpreted directly: the class $\left[10, 20\right)$ concentrates three households per thousand dollars, the last class only one.

On the left histogram, drawn at the height of the frequency, the last two classes look the largest with 40 each, and the last seems overwhelming since it is also the widest: the eye gives it an area more than five times that of the class $\left[10, 20\right)$ ($40\times 40$ against $30\times 10$), although it contains only 33% more households. On the right histogram, the DENSEST class is $\left[10, 20\right)$, with 3.0, and that is indeed where the bulk of the population lies: the corrected drawing tells the opposite story from the first.

Areas: $10\times 2.0=20$, $10\times 3.0=30$, $20\times 2.0=40$, $40\times 1.0=40$. We recover exactly the four frequencies, and their sum, 130, is the total area. This is the property that defines the correct histogram: the area, not the height, represents the frequency. On a histogram with equal classes, the two coincide up to a factor, which is why it is so easily forgotten.

Less than 20,000 dollars: $\dfrac{20+30}{130}=\dfrac{50}{130}\approx 38.5\%$. Between 20,000 and 30,000: the class $\left[20, 40\right)$ contains 40 households over 20 units of width, so under the uniformity assumption half of the class, that is 20 households, falls in the first half, which gives $\dfrac{20}{130}\approx 15.4\%$. We could have gone directly through the density: $10\times 2.0=20$ households. It should be recalled here that uniformity is a convenient and probably false assumption for incomes, where the concentration is stronger near the lower limit.

Because the income distribution is strongly skewed to the right: the great majority of households are concentrated in a narrow interval, while a small minority is spread over a very wide range. Equal classes fine enough to describe the bottom of the distribution would require hundreds of classes to reach the top, almost all empty. So we tighten the classes where the people are and widen them where there are none, even if it means reasoning with density. This is also why the last class is almost always open, of the type “200,000 and over”: its upper limit would be both useless and indiscreet.

Language spoken at home, a nominal variable: BAR CHART, or pie chart if the categories are few and form a whole. Satisfaction level from 1 to 5, ordinal: bar chart in which the ORDER of the categories is respected, or 100% stacked bars if groups are compared. Number of children, discrete quantitative: also a bar chart, with separated bars. Annual income, continuous quantitative: HISTOGRAM, possibly with density. Change in the unemployment rate over twelve years: LINE GRAPH, because time is continuous and the line rightly suggests the intermediate values.

Beyond five or six sectors, the human eye compares neighbouring angles very poorly: it easily tells two bars that differ by 3% apart, but hardly two sectors. The pie chart is therefore reserved for the case where we want to show that one share dominates, not where we want to compare shares with each other. It is flatly forbidden for multiple-response questions, where a respondent may tick several categories: the total then exceeds 100% and the disk, which represents a WHOLE that is shared out, becomes false by construction. In that case we use bars, each read on the basis of the respondents.

The variable is NOMINAL: joining its categories with a line suggests a continuous passage from one to the next, as if there were intermediate states between married and separated, and as if the chosen order were the right one. But this order is arbitrary: by permuting the categories, we get an entirely different curve, which is enough to show that the shape means nothing. The technical objection fits in one sentence: a line encodes continuity and order, and a nominal variable has neither.

A POPULATION PYRAMID, that is, two horizontal histograms back to back sharing the axis of age classes, or failing that two superimposed frequency polygons. The important point is to express the frequencies as a PERCENTAGE of each population, otherwise the larger one swamps the other. The pyramid has the advantage of making generation gaps and imbalances immediately readable, which two tables side by side would never reveal as quickly.

The BOX PLOT, one box per program, aligned on a common axis. It shows at once the median, the central line, the quartiles, the edges of the box, the range of the non-outlying data by the whiskers, and the outliers by isolated points. It is the only compact graph that gives both level and spread, and it lets us see immediately that a program with a high but very spread-out median is not comparable to a homogeneous program with the same median. What it does NOT show, on the other hand, is the fine shape of the distribution: a bimodal distribution and a flat one can give exactly the same box.

Change in points: $42.4-41.2=1.2$ points. Relative change: $\dfrac{1.2}{41.2}\approx 0.0291$, that is about 2.9%. This is a real but modest rise, of the kind that sampling noise can produce by itself if the values come from successive polls of a thousand respondents, where the margin is three points.

On the left graph, the axis starts at 41: the first bar measures $41.2-41=0.2$ units of height, the last $42.4-41=1.4$, a ratio of 7. The eye therefore reads a multiplication by SEVEN where the values change by 2.9%. On the right graph, with a full axis, the ratio of the heights is $\dfrac{42.4}{41.2}\approx 1.03$, exactly the ratio of the values. The distortion is thus enormous: the rise seen on the left (from 1 to 7, that is +600%) is about 200 times the real rise of 2.9%.

No, not always. It is LEGITIMATE when the variable has no meaningful zero or when a zero would make reading impossible: a body temperature, a stock index, a pH, a year. It becomes DISHONEST when the variable is a quantity or a proportion whose zero has a meaning and the reader will interpret the height of the bars as a quantity, which is exactly the case for a percentage of support or a sales figure. Practical rule: on BARS, the axis must start at zero, because a bar encodes a length, hence a quantity; on a LINE GRAPH, truncation is acceptable provided it is visible and flagged.

By choosing the BOUNDS of the time window. A rate that fluctuates around a stable value periodically reaches troughs and peaks: starting from a trough and stopping at a peak shows a spectacular rise, doing the opposite shows a collapse, and both graphs are accurate. This is the temporal variant of the same device, and it is harder to spot because nothing is visibly truncated. The defence is to demand the full series available, and to be wary of any window whose bounds fall exactly on a local minimum or maximum.

First question: where does the vertical axis start, and does the variable have a zero that means something? Second question: what is the unit, and are these frequencies or percentages, computed on what base? Third question: what data were left outside the frame, before or after the period shown, and why that one? A fourth question can be added, often decisive: where do the data come from and who paid for the figure?

No: doubling the height of a figure while keeping its proportions also doubles its width, so the area is multiplied by $2\times 2=4$. The reader, who spontaneously perceives the amount of drawing, that is the area, reads a multiplication by four where the budget doubled. The reader sees an increase of 300% where the real increase is 100%, and the exaggeration is all the more effective because it rests on no false number.

In three dimensions, the perceived ratio becomes $2^{3}=8$ if the reader interprets the volume, since all three dimensions are multiplied by two. Depending on whether the reader looks at the height, the face or the volume, the same figure suggests a ratio of 2, 4 or 8. A graph whose reading depends this much on the reader's eye is no longer a measuring instrument, which is why visualization textbooks forbid relief effects.

For the AREA to double, each dimension must be multiplied by $\sqrt{2}\approx 1.414$, not by 2. The truly robust solution, however, is not to resize at all: the pictograph is repeated, two identical silhouettes instead of one, which encodes the quantity by a COUNT rather than a size. This is the form used by serious infographics, and it has the advantage of staying readable even when the ratio is not a whole number, the last silhouette then being cut off.

The tilt squashes the disk into an ellipse and breaks the proportionality between angle and visible area: the sectors at the FRONT look clearly larger than those at the back, for the same angle. Two sectors of 20% will occupy different areas depending on their position, and the order of the shares itself may seem reversed. Since the position in the disk is a choice of the designer, perspective gives a convenient way of putting forward the share one wants to highlight.

The ratio of the populations is $\dfrac{32}{8}=4$. Since the area of a disk is $\pi r^{2}$, we need $\dfrac{r_{2}^{2}}{r_{1}^{2}}=4$, hence $\dfrac{r_{2}}{r_{1}}=2$. The second circle must have a DOUBLE radius, not a quadruple one. Quadrupling the radius would produce an area sixteen times larger, four times the real gap; this is one of the most frequent errors in infographics, and it is spotted by checking that the ratio of the radii is the square root of the ratio of the values.

FALSE. Spacing is not a layout choice, it is a sign that encodes the nature of the variable: touching bars for a continuous variable, separated bars for a discrete or qualitative variable. Separating the bars of a histogram visually asserts that impossible values exist between the classes, which is false for an income or a duration. Correct statement: in a histogram, the bars touch, because the classes follow each other edge to edge and cover the whole interval.

FALSE. The fact that the cumulation reaches 100% proves nothing: it will always reach it, whatever the order of the categories, since all the frequencies are being added. The problem is that the variable is NOMINAL: the cumulation depends on an arbitrary order, and swapping two categories changes all the intermediate values. Correct statement: we only cumulate on an ordered variable; for means of transportation, only the simple frequencies make sense.

FALSE, and the effect obtained is exactly the opposite of the one intended. A bar twice as wide AND twice as tall has a QUADRUPLE area, although the class may contain no more observations than the others. Correct statement: when classes have unequal widths, the height must be the DENSITY, frequency divided by class width, so that the area of the rectangle represents the frequency.

FALSE. Forty classes for 400 observations give 10 observations per class on average, with purely random irregularities that will make peaks and troughs appear that do not exist in the population. The histogram then stops summarizing and starts reproducing noise. Correct statement: Sturges' rule suggests here $1+3.322\log_{10}(400)\approx 9.6$, so about ten classes; we may go a little above if the distribution is very structured, never up to forty.

FALSE. The explanation for the total exceeding 100% is correct, but it disqualifies precisely the choice of graph: a disk represents a whole divided into exclusive shares, and it cannot display 168%. Correct statement: for a multiple-response question, we use a bar chart in which each category is read independently, as a percentage of the RESPONDENTS, and we state under the figure that the total exceeds 100% because several answers were allowed.

A dual axis lets you superimpose two series with different units, here dollars and visits, on a single drawing. The freedom it leaves is total: the designer chooses the bounds of EACH axis independently, and can therefore make the two curves cross wherever they like, or make them coincide almost perfectly, without changing a single data value. It is the only figure whose visual conclusion can be reversed by changing two numbers that appear nowhere.

None. The crossing point depends solely on the choice of the two scales: by moving the lower bound of the right axis by a few tenths, it moves anywhere, disappears, or appears twice. Comparing an amount in dollars with a number of visits makes no dimensional sense anyway: saying that they “cross in year 3” is like saying that a weight crosses a length. What the eye reads as a tipping point is merely an artifact of the drawing.

Series A: $\dfrac{3.3-1.2}{1.2}=\dfrac{2.1}{1.2}=1.75$, a rise of 175%. Series B: $\dfrac{2.2-3.4}{3.4}\approx -0.353$, a fall of 35.3%. The two changes are therefore of very different size, the first being about five times stronger than the second, which the graph hides entirely by giving the two curves visually comparable slopes. This is a third effect of the dual axis, independent of the crossing.

First competing explanation: REVERSE CAUSATION, visits fall for their own reasons, and the company reacts by increasing its advertising; the correlation would then be the trace of a reaction, not of an effect. Second explanation: a CONFOUNDING VARIABLE, for example the opening of a competitor or an economic situation that at the same time makes visits drop and pushes spending up. Third explanation: the COMMON TREND, or spurious time correlation: two series, one rising and the other falling over six points, are mechanically correlated, and six annual observations almost never allow chance to be ruled out.

Two SEPARATE graphs, one for spending, the other for visits, each with its axis starting at zero and the same time span: the reader then sees the two changes without an arbitrary scale staging them. In a single graph, the correct solution is to express both series as INDEXES with base 100 at year 1, which makes them comparable in the same, dimensionless unit: we would then read 100 then 275 for spending and 100 then 65 for visits, and the difference in size would become immediately visible. In all cases, the caption must say that six annual points do not allow a causal link to be asserted.

Relative frequencies: $26.0\%$, $16.0\%$, $20.0\%$, $18.0\%$, $16.0\%$ and $4.0\%$, summing to 100%. Cumulative relative frequencies: $26.0$, $42.0$, $62.0$, $80.0$, $96.0$ and $100.0$%. Densities, frequency divided by class width: $\dfrac{39}{5}=7.8$, $\dfrac{24}{5}=4.8$, $\dfrac{30}{5}=6.0$, $\dfrac{27}{5}=5.4$, $\dfrac{24}{10}=2.4$ and $\dfrac{6}{20}=0.3$ students per hour.

The height must be the DENSITY, since the last two classes are wider than the first four. The densest class is $\left[0, 5\right)$ with 7.8, and it is also the modal class. Drawn at the height of the frequency, the class $\left[20, 30\right)$ would look as important as $\left[5, 10\right)$ although its density is half as large.

Mean from class midpoints: $\dfrac{2.5(39)+7.5(24)+12.5(30)+17.5(27)+25(24)+40(6)}{150}=\dfrac{97.5+180+375+472.5+600+240}{150}=\dfrac{1965}{150}=13.1$ hours. Median: the 75th observation falls in $\left[10, 15\right)$, since the cumulative frequency goes from 63 to 93; by interpolation, $10+\dfrac{75-63}{30}\times 5=10+2.0=12.0$ hours. The mean, 13.1, exceeds the median, 12.0: the distribution is SKEWED TO THE RIGHT, pulled by the minority who work more than 30 hours. The honest summary of such a profile is the median.

The last two classes together hold $24+6=30$ students, that is $\dfrac{30}{150}=20.0\%$. The 20-hour threshold is the one used in most research on persistence because that is around where the effects change sign: below it, paid work is neutral or even favourable, giving structure to the schedule and financing studies; above it, it competes directly with study time and with participation in school life, and the association with dropping out becomes clear. A relevant threshold is always one justified by the literature, never a round number chosen for convenience.

Title: “Distribution of the 150 students by weekly hours of paid work, winter 2026”. Note under the figure: “Density histogram; the last two classes have widths of 10 and 20 hours, the height of the bars represents the number of students per hour of work, the area of each bar represents the frequency. Source: student life survey, n equals 150, response rate of 32%.” The title says who, what, when; the note says how to read it and where the figure comes from. These are the only two things a reader in a hurry will read, and they must be enough to avoid being misled.