BMS 1110 · Common Core (Tronc commun)Section 4.3 · Organising and describing data

Section 19 of 30

Raw data, frequency and cumulative tables

Build a frequency table from a set of raw observations, extend it with relative and cumulative columns, check that the last cumulative total equals the number of observations, and group a large set into class intervals of a sensible width.

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Why this matters

The first thing anyone does with a ward full of measurements is count them. A pile of ages, blood pressures or lengths of stay tells you nothing until you have counted how often each value occurs, what share of patients each value represents, and how many patients fall at or below it. Those three counts are the frequency table, and almost every statistic later in this course — the median, the bar chart, the histogram, the grouped mean — is read from one. Learning to build the table by hand, column by column, and to check that the last cumulative total equals the number of patients, is what stops a summary from quietly describing the wrong number of people.

What you need to know first

You need only counting, adding and one division. If you can count how many times the number 22 appears in a list of ten ages, and work out 3 ÷ 10, you have every skill this section requires. A calculator helps with the percentages; the reasoning stays exactly the same.

Words to know (6)

Raw data: Values exactly as they were first recorded, in the order they arrived, with repeated values kept and nothing sorted or counted yet. The list 22, 35, 18, 41, 22, 29, 35, 18, 25, 22 is raw data.
Frequency: The number of times one value occurs in the data set. It is a count of observations, so it has no unit beyond the observation itself. In the ten ages above, the age 22 has frequency 3 because three patients were aged 22.
Relative frequency: The frequency divided by the total number of observations. It is a proportion between 0 and 1, and multiplying it by 100 turns it into a percentage. 3 patients out of 10 is 3 ÷ 10 = 0.30, which is 30 % of the group.
Cumulative frequency: The running total formed by adding the frequencies as you move down the table. Each entry counts the observations at or below that value. Frequencies 2, 3, 1, 1, 2, 1 give cumulative frequencies 2, 5, 6, 7, 9, 10.
Class interval: A range of values treated as one group when the data set is large, for example the ages 20 to 29 years. The class 30–39 contains every patient aged 30, 31, … up to 39 years.
Class width: The size of a class, found from successive limits. For the classes 10–19 and 20–29 the width is 20 − 10 = 10 years. Every class 10–19, 20–29, 30–39 has the same width, 10 years, which keeps the table comparable.

One idea at a timeIdea 1 / 7

In one sentence Medical data are observations and measurements collected for analysis, and they arrive from patients, records, laboratory tests, surveys and surveillance.

Where medical data come from

Data are observations or measurements collected for the purpose of analysis. In medical work they are gathered from several recognisable places, and knowing which one you are holding changes how far you can trust it. The patient is the first source: a nurse measures a temperature, a pulse or a blood pressure and writes the value down. Medical records and hospital databases supply values recorded earlier, often by someone else. Laboratory tests produce measurements such as a haemoglobin concentration or a blood glucose level. Questionnaires and surveys collect what patients report about themselves, such as smoking or diet. Clinical trials and public-health surveillance collect data to a written plan, which is why they can support comparisons that a ward notebook cannot. Whichever source you use, the numbers first obtained are raw: they have not yet been counted, sorted or summarised.

For example

A ward records the ages of ten patients admitted overnight, in the order they arrived: 22, 35, 18, 41, 22, 29, 35, 18, 25, 22.

Pause and check yourself

Name three different sources from which a nurse might obtain patient data, and say what kind of value each one gives.

Show the answer

Any three of these: the patient, giving a measured value such as temperature or pulse; medical records and hospital databases, holding values recorded earlier by other staff; laboratory tests, giving measured concentrations such as haemoglobin; questionnaires, giving what the patient reports; clinical trials and public-health surveillance, which collect data to a written plan.

Read detailed explanations

Choosing a class width without guessing: Choosing the number of classes and their width is a judgement, not a fact hidden in the data, so the judgement should be made in two visible moves. First decide roughly how many classes you want. For a beginner, aim between five and twenty. A widely used formula, Sturges’ rule, estimates the number as 1 + 3.322 × log10(n), so for n = 50 it gives 1 + 3.322 × 1.699 ≈ 6.6, that is about seven classes. Second, divide the range by that number and round the answer up to a convenient value. For ages running from 10 to 59 years the range is 59 − 10 = 49 years, and 49 ÷ 7 = 7 years exactly, so seven classes of width 7 begin 10–16, 17–23, 24–30, 31–37, 38–44, 45–51 and 52–58. The last class ends at 58 and the value 59 has nowhere to go: the arithmetic looked perfect and the table still fails. Rounding the width up to 8 repairs it, giving classes from 10–17 through 58–65, in which the final class is partly empty. A partly empty end class is harmless; a missing observation is not. The supplied example makes the opposite choice and is equally valid: five classes of width 10, namely 10–19, 20–29, 30–39, 40–49 and 50–59, cover 10 to 59 exactly. Both tables can be defended, which is why a careful writer states how many classes were wanted and what width was chosen before showing the result. Two technical terms make the width meaningful when the table is used for further work. The midpoint, also called the class mark, is the average of the two limits: for 20–29 it is (20 + 29) ÷ 2 = 24.5 years, and it stands in for every value in the class when a mean is later estimated from grouped data. The boundary method matters when the variable is continuous rather than a whole-number count. If weight is recorded to the nearest kilogram, the class written 10–19 really stretches from 9.5 kg to 19.5 kg, a width of 10 kg, because a patient weighing 19.4 kg is rounded into it. For a discrete count such as the number of children, the class 2–3 means exactly 2 or 3 children and no adjustment is needed. Never change the width part way down a table, because unequal widths make the bars of a histogram misleading, and avoid an open-ended class such as 70 and over, whose midpoint cannot be fixed. Finish by checking that every observation falls in exactly one class and that the class frequencies add to n.
Reading a frequency table forwards and backwards: A frequency table is read forwards for one value and backwards for a group, and the two readings use different columns. Forwards is the frequency column: it answers how many patients had exactly this value. Backwards is the cumulative frequency column: it answers how many had this value or less, and the answer never shrinks as you move down the table. The cumulative relative frequency is the running total of the proportions, and it behaves in the same way. In a complete table its last value is exactly 1, or 100 %. Here is a worked reading. A ward reviews the length of stay of 40 patients, and the cumulative relative frequency after 5 days is 0.85. So 85 % of patients stayed five days or fewer. The remaining share is found by subtraction, not by a new count: 1 − 0.85 = 0.15, so 15 % stayed longer than five days, and 0.15 × 40 = 6 patients. That subtraction is called the complement, and it is valid because every patient belongs to one of the two groups and to only one. Two checks keep the table honest, and they catch different faults. First, the frequencies must add to n, and the relative frequencies must add to 1. When percentages are rounded to one decimal place they may total 99.9 % or 100.1 %; that is a rounding artefact rather than a mistake, but the unrounded proportions add to exactly 1. Second, the cumulative column must never decrease and must finish at n. If it finishes at 39 when 40 patients were measured, one observation has been lost somewhere above that row, and the fault lies not in the last addition but in the frequency that was written wrongly earlier. Reading the table backwards also gives the beginner a first, informal idea of probability: if 30 % of ten patients were aged 22, you may estimate that about 30 % of similar patients will be. A proportion built on ten patients is fragile, because adding or removing one patient moves it by ten percentage points, while the same proportion built on a thousand patients barely moves. The arithmetic looks identical; the reliability does not.
From a grouped table to a clinical summary: Grouping buys a readable pattern and pays for it with precision, and knowing the price is what makes the summary honest. Once ages are written as 20–29, the individual values are gone: the table no longer records that one patient was 22 and another 29. What survives is the shape of the distribution — how many patients are young, how many are old, and where the bulk of them sits. The standard way to recover a usable number is to treat every value in a class as if it equalled the midpoint, then compute an estimated mean as the sum of each class frequency multiplied by its midpoint, divided by n. For the supplied grouped ages: 10–19 has 8 patients and midpoint 14.5, giving 8 × 14.5 = 116; 20–29 has 15 and midpoint 24.5, giving 15 × 24.5 = 367.5; 30–39 has 12 and midpoint 34.5, giving 12 × 34.5 = 414; 40–49 has 10 and midpoint 44.5, giving 10 × 44.5 = 445; 50–59 has 5 and midpoint 54.5, giving 5 × 54.5 = 272.5. The total is 116 + 367.5 + 414 + 445 + 272.5 = 1,615, and 1,615 ÷ 50 = 32.3 years. That mean is an estimate, not the true mean, because each midpoint stands in for values that were never exactly there; with narrow classes it is close, and with very wide classes it drifts. The same table answers resource questions without any new arithmetic. How many patients were at least 40 years old? Add the last two class frequencies: 10 + 5 = 15 patients, which is 15 ÷ 50 = 0.30, or 30 % of the group. Reading upwards from the bottom of the table gives an at-or-above count, and it is the mirror image of the cumulative column read downwards. The width you chose becomes the width of each bar when the same grouped counts are drawn as a histogram, so unequal widths make a bar look important merely because it is wide, not because it holds more patients. Equal widths avoid that trap, which is why the width rule and the graph rule are really one rule. Finally, when a class is open-ended, such as 70 and over, no midpoint can be fixed and no grouped mean can be trusted; if the summary is meant for calculation, avoid open-ended classes even when they make the table look tidier.

Explore the diagram

Look at the whole drawing first. Then select one numbered part and connect it to the idea you just read.

Raw data to a three-column frequency table

The section’s whole routine on one plate. The ten patient ages are shown raw, then ordered, and the order gives the minimum 18, the maximum 41 and the range 23. The frequency table then counts each distinct age, adds a relative frequency column that totals 1.00 and 100 %, and adds a cumulative column whose last value 10 equals n. The last two panels show the grouped table of 50 patients, whose class frequencies total 50, and the rule for choosing the class width: range divided by the number of classes, rounded up.

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Look at the whole drawing once without clicking anything. Ask yourself only: what kind of picture is this (a comparison, a route, a sequence, a table)?
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01 / 4Raw data, then ordered data

If the drawing is wider than this panel, scroll it sideways. Labels stay readable; the numbered explanations below show the same ideas.

Raw data, then ordered data

The upper panel shows the ten overnight admissions exactly as they were written down: 22, 35, 18, 41, 22, 29, 35, 18, 25 and 22 years. Ordering them from smallest to largest costs nothing and changes no value, but it makes the minimum 18 years, the maximum 41 years and the range 41 − 18 = 23 years readable at a glance. The median is the average of the fifth and sixth ordered values, (22 + 25) ÷ 2 = 23.5 years — the two middle values are not both 22.

The table values reproduce the supplied ten-patient age example at notes lines 642–688 and keep its arithmetic; the grouped row totals 50 identical observations. The plate is schematic: columns are aligned for reading, not drawn at scale.

Key points to remember

Medical data come from patients, records, laboratory tests, questionnaires, trials, surveys and surveillance; in their first, unorganised form they are called raw data.
Ordered data run from smallest to largest, which makes the minimum, maximum, range, median and quartiles readable without searching the list.
Frequency is the number of times a value occurs; the frequencies of a complete table must add to n.
Relative frequency = frequency ÷ total frequency, a proportion between 0 and 1; multiply by 100 for a percentage, and the proportions of a complete table add to 1.
Cumulative frequency is the running total of the frequencies and answers “at or below this value”; its final value must equal n.
Grouped data use class intervals with class limits, a class width and class frequencies; keep widths equal, avoid overlaps and gaps, and round the width up.

Go deeper

Worked example

Building a full frequency table from ten patient ages. A ward clerk hands the nurse a scrap of paper with the ages of the ten patients admitted overnight, in the order they arrived: 22, 35, 18, 41, 22, 29, 35, 18, 25, 22. The ward needs a table showing how often each age occurred, what share of patients each age represents, and how many patients were at or below each age.

Count the observations first and state n. Ten ages are written on the paper, so n = 10. Writing n before building the table gives you the number that every later check must match.
Write the distinct values once each, in ascending order: 18, 22, 25, 29, 35, 41. Notice that only six rows are needed for ten patients, because repeats share a row.
Count each value and write the frequency: 18 gives 2, 22 gives 3, 25 gives 1, 29 gives 1, 35 gives 2 and 41 gives 1. Add the column: 2 + 3 + 1 + 1 + 2 + 1 = 10, which equals n.
Divide each frequency by n for the relative frequency, then multiply by 100 for the percentage: 2 ÷ 10 = 0.20 and 20 %, 3 ÷ 10 = 0.30 and 30 %, 25 and 29 give 0.10 and 10 % each, 35 gives 0.20 and 20 %, and 41 gives 0.10 and 10 %. The proportions add to 1.00 and the percentages to 100 %.
Add a cumulative column by running the frequencies together: 2, then 2 + 3 = 5, then 6, then 7, then 9, then 10. The last value is 10 = n, so the table is complete and correct.

The finished table has three useful columns. The frequency column says how many patients had each exact age, the relative frequency column says what share of the ward that is, and the cumulative column says how many patients were that age or younger. The final 10 closes the loop with n, which is the cheapest check in the section.

Tutorial 1 — building all three columns for ten patient ages

The ten overnight admissions were aged 22, 35, 18, 41, 22, 29, 35, 18, 25 and 22 years, in that order. Build a frequency table with a relative frequency column expressed as a percentage, and a cumulative frequency column, then prove that the table is complete.

Rule used: Frequency = the number of observations equal to that value, a whole number with no unit. Relative frequency = frequency ÷ total frequency, a proportion between 0 and 1; percentage = proportion × 100, carrying the symbol %. Cumulative frequency = the running total of the frequencies, and its final value must equal n, the number of observations.

Count the observations, then write the distinct ages once each in ascending order.
Reveal this step
n = 10 because ten patients were measured. The six distinct ages are 18, 22, 25, 29, 35 and 41 years.
Count how many patients take each of the six ages and write the frequency column.
Reveal this step
18 gives 2, 22 gives 3, 25 gives 1, 29 gives 1, 35 gives 2 and 41 gives 1. The sum is 2 + 3 + 1 + 1 + 2 + 1 = 10 = n.
The frequency of 22 is 3. What exactly does that number measure, and what would happen to the table if an eleventh patient aged 22 were admitted?
Reveal the answer
It measures that three separate patients were aged 22, not the value 22 itself. If an eleventh patient aged 22 arrived, n would become 11 and the frequency of 22 would become 4; every relative frequency would then have to be recalculated against the new total of 11.
Divide each frequency by 10 to get the relative frequency, then multiply by 100 to write the percentage.
Reveal this step
18: 2 ÷ 10 = 0.20, that is 20 %. 22: 3 ÷ 10 = 0.30, that is 30 %. 25 and 29: 1 ÷ 10 = 0.10, that is 10 % each. 35: 0.20, that is 20 %. 41: 0.10, that is 10 %.
Add the proportions: do they total 1? Why must they, whatever the data are?
Reveal the answer
0.20 + 0.30 + 0.10 + 0.10 + 0.20 + 0.10 = 1.00, which is 100 %. They must total 1 because every patient belongs to exactly one row: the rows share out the whole group, and a whole divided into shares always adds back to 1.
Build the cumulative frequency column by running the frequencies together, then check its last value.
Reveal this step
Start at 2, then 2 + 3 = 5, then 5 + 1 = 6, then 6 + 1 = 7, then 7 + 2 = 9, then 9 + 1 = 10. The last value is 10, which equals n, so no observation has been lost.
A colleague reads the table and says 25 patients were aged exactly 25 years. Using the two columns, explain what the colleague has mixed up.
Reveal the answer
The colleague has read the cumulative frequency as if it were the frequency. The frequency of 25 is 1: exactly one patient was 25 years old. The cumulative frequency at 25 is 6: six patients were 25 years old or younger. Cumulative frequency answers a different question from frequency.

Answer: The table is: 18 with 2, 0.20 and 20 %, cumulative 2; 22 with 3, 0.30 and 30 %, cumulative 5; 25 with 1, 0.10 and 10 %, cumulative 6; 29 with 1, 0.10 and 10 %, cumulative 7; 35 with 2, 0.20 and 20 %, cumulative 9; 41 with 1, 0.10 and 10 %, cumulative 10. Proportions total 1.00 and the last cumulative value is n = 10.

Keep this habit: Build the columns in order: list distinct values, count frequencies, check they add to n, divide by n for relative frequency, then run the frequencies together for the cumulative column. Finish by checking that the last cumulative value equals n; that single check catches almost every mistake.

Tutorial 2 — grouping fifty patients and choosing the class width

Fifty patients were admitted during the study month. Their ages were recorded in whole years and run from a youngest of 10 years to an oldest of 59 years. The table would need fifty separate rows, so the team decides to present the ages in about five groups. Choose a class width and build the grouped frequency table.

Rule used: Class width = (largest value − smallest value) ÷ the number of classes you want, and the result is rounded up to a convenient whole number, because the largest values must still have a class. Classes must all have the same width, must not overlap, and must not leave a gap between them; the class frequencies must add to n.

Find the smallest and largest ages and subtract to get the range.
Reveal this step
The smallest age is 10 years and the largest is 59 years, so the range is 59 − 10 = 49 years.
You want about five classes. Work out 49 ÷ 5 and decide whether to round the answer up or down.
Reveal the answer
49 ÷ 5 = 9.8 years. Round up to 10 years. Rounding down to 9 would be unsafe: five classes of width 9 would cover only 10 to 54 years, so patients aged 55 to 59 would have no class at all. Rounding up keeps every value inside the table.
Write out the five classes of width 10 years, starting at the smallest age.
Reveal this step
10–19, 20–29, 30–39, 40–49 and 50–59 years. The last class reaches 59, so the oldest patient is included.
Why are the classes 10–20 and 20–30 unsafe, and what happens to a patient aged exactly 20 years?
Reveal the answer
The value 20 sits in both classes, so the patient could be counted twice and the total would rise above n = 50. Adjacent classes such as 10–19 and 20–29 share no value, so every age falls in exactly one class and the total stays honest.
The fifty ages have already been tallied into the five classes, as in the grouped example above. Write the class frequencies into a table, add them, and check the total against n.
Reveal this step
The class frequencies are 8, 15, 12, 10 and 5. Their sum is 8 + 15 + 12 + 10 + 5 = 50, which equals n, so every patient has been placed exactly once.
What is the class width of this table, and what is the class frequency of the 20–29 class?
Reveal the answer
The width is the difference between successive limits: 20 − 10 = 10 years. The class frequency of 20–29 is 15, meaning fifteen patients were aged between 20 and 29 years inclusive.
Add a cumulative frequency column and state how many patients were younger than 30 years.
Reveal this step
The cumulative frequencies are 8, 23, 35, 45 and 50. Reading at the 20–29 row gives 8 + 15 = 23 patients younger than 30 years, and the final value 50 again equals n.

Answer: Grouped ages in five classes of width 10 years: 10–19 with 8, 20–29 with 15, 30–39 with 12, 40–49 with 10 and 50–59 with 5. Class frequencies total 50 = n and the cumulative column runs 8, 23, 35, 45, 50.

Keep this habit: Choose a class width by dividing the range by the number of classes you want and rounding up, never down. Then check three things before using the table: the widths are all equal, no value can fall into two classes, and the class frequencies add to n.

Now you try

Attempt each one on paper first. Write the formula, substitute the numbers, then check the unit of your answer.

Eight children in a clinic had their pulse rates recorded, in beats per minute: 92, 88, 96, 88, 100, 92, 88 and 84. Build a frequency table with a cumulative frequency column, and state the total.
Hint
Write the distinct values in ascending order first, then count each one. Remember that n is the number of children, 8, not the number of different pulse rates.
Show the worked answer
The ordered values are 84, 88, 88, 88, 92, 92, 96, 100, so n = 8. The frequencies are 84 with 1, 88 with 3, 92 with 2, 96 with 1 and 100 with 1, totalling 1 + 3 + 2 + 1 + 1 = 8. The cumulative frequencies are 1, 4, 6, 7 and 8, and the last value 8 confirms that no child was missed. The relative frequency of 88 is 3 ÷ 8 = 0.375, that is 37.5 % of the children.
A hypertension clinic reviewed 40 patients. Twelve of them had a systolic pressure of 140 mmHg or more. Give the relative frequency as a proportion and as a percentage, and state how many patients were below 140 mmHg.
Hint
Relative frequency = frequency ÷ total frequency, here 12 ÷ 40. The second group is the rest of the patients, so subtract the first count from 40 rather than counting again.
Show the worked answer
The relative frequency is 12 ÷ 40 = 0.30, that is 0.30 × 100 = 30 %. So 30 % of the patients had a systolic pressure of 140 mmHg or more. The remaining patients are 40 − 12 = 28, a proportion of 28 ÷ 40 = 0.70, or 70 %. The two proportions add to 0.30 + 0.70 = 1.00, which checks that all 40 patients have been accounted for.
Ninety-six patients waited between 3 and 52 minutes to be seen. You want five classes for the waiting times. Work out a sensible class width and write out the five classes, then check that a wait of 52 minutes is included.
Hint
Class width = (largest value − smallest value) ÷ number of classes. Round the answer up, never down, or the longest waits will have no class.
Show the worked answer
The range is 52 − 3 = 49 minutes, so 49 ÷ 5 = 9.8 minutes, which rounds up to a class width of 10 minutes. The five classes are 3–12, 13–22, 23–32, 33–42 and 43–52 minutes. Successive lower limits differ by 10 (13 − 3 = 10), the classes do not overlap because one ends at 12 and the next begins at 13, and there is no gap. The last class reaches 52, so the longest wait is included.
In a survey of 25 families, the cumulative frequency after 2 children is 20. How many families had 3 or more children, and what proportion is that?
Hint
The cumulative frequency counts the families at or below that value. Everyone else is above it, so subtract from the total instead of adding frequencies.
Show the worked answer
The cumulative frequency of 20 means 20 families had 2 children or fewer. The remaining families are 25 − 20 = 5, so five families had 3 or more children. As a proportion that is 5 ÷ 25 = 0.20, which is 0.20 × 100 = 20 %. The check is that 20 + 5 = 25, the total number of families surveyed.

Common misconception

Beginners often treat the cumulative frequency column as a second copy of the frequency column, and read it as if it answered the same question. It does not. The frequency of 25 tells you how many patients were exactly 25 years old; the cumulative frequency at 25 tells you how many were 25 or younger. Because a cumulative column never decreases, reading it as a frequency overstates the count from the second row onward — here six patients instead of one. A second common error is believing that grouping costs nothing. Once ages are grouped as 20–29, the individual ages are gone, and any mean worked out from the middle of the classes is an estimate, not the true value.

Check your reasoning

Explain in your own words why the final cumulative frequency must equal n, and why that check alone can reveal a mistake that the frequency total might not.

Reveal an answer

Every observation belongs to exactly one row, so adding the frequencies one after another eventually counts every observation once and no observation twice. That is why the last cumulative value is n. The frequency total is a single sum, so two mistakes that cancel out — one value counted twice and another missed — can leave the total looking correct. The cumulative column makes such a pair harder to hide, because it must also rise by the right amount at every single row and stay consistent with every value in order. When a table is long, checking the final value against n, and each step against the row above, is the fastest way to catch a lost patient.

Where this comes from

BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (supplied) · Chapter Two “Organization of data”, lines 630–688: meaning of data and the eight sources listed; raw data and the ten-patient age example; ordered data and what ordering makes possible; frequency distribution; relative frequency as a proportion and a percentage; cumulative frequency and the requirement that the final total equals the number of observations; grouped frequency distribution and the terms class interval, class limits, class width and class frequency.
Note on the course code · The supplied cover page reads “Course Number: BMS 110”. The curriculum catalogue in this project records “BMS 1110”; the catalogue is treated as authoritative and the notes’ cover value is retained here for provenance.
Beyond the supplied notes: choosing and using class widths · The supplied notes define class width but do not say how to choose one. Sturges’ rule for the number of classes, the rule of rounding a width up, the class midpoint, the boundary convention for a continuous variable, the complement reading of a cumulative relative frequency, and estimating a mean from a grouped table are standard descriptive-statistics practice added here and labelled as beyond the notes.

Lesson progress is saved on this device for the current learner. Practice answers stay separate and are not stored.

About this course and its sources

Bilingual lessons organised from the supplied BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (Mr. Ebangan Prosper Achuanoh). Diagrams are original educational schematics; sections that go beyond the supplied notes say so in their sources. Mathematics, clinical and French-language review are pending.