Worked example
Reading and calculating a rate with its full label. A district health office reports 45 maternal deaths during a year in which 9,000 live births took place. A colleague writes: “maternal mortality 0.5%”. Before accepting the figure, state which numbers belong in the numerator and the denominator, and write the rate in the published form.
- Name the event. The event counted is a maternal death, so 45 goes in the numerator.
- Name the population in which the events occurred. The deaths occurred among the 9,000 women who gave birth, so 9,000 is the denominator — not the district’s 30,000 women of reproductive age, many of whom did not give birth and were never at risk of a maternal death.
- Fix the time period. It is one year, and the period must appear in the written statement, because deaths accumulate with time.
- Choose K and write it down. Per 100,000 live births is the usual convention for this indicator, so K = 100,000.
- Calculate: 45 ÷ 9,000 = 0.005, then 0.005 × 100,000 = 500. State it in full: 500 maternal deaths per 100,000 live births per year. The colleague’s 0.5% is the same fraction with K = 100, since 0.005 × 100 = 0.5%. Both describe the same risk, but only the second form names the denominator and the period.
A rate is a full sentence, not a number. The numerator names the event, the denominator names the population that was actually exposed to it, the period names the time, and K says how many people the figure is reported per. The same fraction can honestly be written as 0.5% or as 500 per 100,000 per year; what must never happen is writing it without the denominator and the period.
Tutorial 1 — calculating and stating a mortality rate
A city reports 180 deaths from all causes during a calendar year. Its mid-year population is estimated at 300,000 people. Calculate the annual mortality rate with K = 1,000, and state the result in a form that a colleague could check.
Rule used: Mortality rate = deaths during the period ÷ the population during that period × K. The numerator is the number of deaths, the denominator is the whole population in which those deaths occurred, and K is the multiplier chosen by the writer and then reported; here K = 1,000, so the answer is read as deaths per 1,000 population per year.
Write the general formula in words with its units before substituting any numbers.
Reveal this step
Mortality rate = deaths during the year ÷ population during the year × K, read as deaths per 1,000 population per year when K = 1,000.
Which number is the numerator and which is the denominator? Why can nobody die twice, and what does that tell you about the size of the fraction?
Reveal the answer
180 deaths is the numerator and 300,000 people is the denominator. Each death is a different person, so the numerator counts distinct people and can never exceed the denominator. The fraction must therefore be smaller than 1, and usually very much smaller.
Substitute the numbers into the formula, keeping the words “deaths” and “people” visible for now.
Reveal this step
Mortality rate = 180 deaths ÷ 300,000 people × 1,000.
Carry out 180 ÷ 300,000 first, without the multiplier. Is the answer larger or smaller than 1, and why must it be?
Reveal the answer
180 ÷ 300,000 = 0.0006, a number far smaller than 1, because only 180 deaths occurred among 300,000 people. It must be smaller than 1 for the reason in the previous step: the numerator is a subset of the denominator.
Now multiply by K = 1,000 and attach the unit to the answer.
Reveal this step
0.0006 × 1,000 = 0.6. The rate is 0.6 deaths per 1,000 population per year.
State the rate in full, then say what would change if the writer had chosen K = 100,000 instead.
Reveal the answer
In full: 0.6 deaths per 1,000 population per year in that city. With K = 100,000 the same fraction is multiplied by 100,000 and gives 60 deaths per 100,000 population per year. The underlying risk is unchanged; only the reporting scale changes, which is why K must always be stated.
Answer: The annual mortality rate is 0.6 deaths per 1,000 population, which can equally be written 60 deaths per 100,000 population per year.
Keep this habit: Write the formula in words first, keep the numerator and the denominator in the same population, and finish the sentence with “per K per period”. A bare 0.6 is not a rate; 0.6 deaths per 1,000 people per year is a rate, because it can be checked.
Tutorial 2 — incidence and prevalence from one district
A district has 5,000 adults. On 1 January, 1,000 of them are already living with type 2 diabetes. During the year, 100 further adults are newly diagnosed, and nobody dies or leaves the district. Calculate the point prevalence on 1 January and the cumulative incidence during the year with K = 100, then explain why the two values are so different.
Rule used: Prevalence = existing cases ÷ the whole population × 100. Incidence = new cases during the period ÷ the population at risk at the beginning × 100. The prevalence denominator includes people who already have the disease; the incidence denominator must exclude them, because a person who is already ill is no longer at risk.
Write the prevalence fraction for 1 January. Say which people go on top and which population goes underneath.
Reveal this step
The 1,000 people already living with diabetes go on top, because they have the disease on that date. The whole adult population of 5,000 goes underneath, because prevalence asks what share of everyone in the district is affected.
Calculate the point prevalence on 1 January with K = 100.
Reveal this step
1,000 ÷ 5,000 = 0.20, then 0.20 × 100 = 20%. The point prevalence on 1 January is 20%.
For the incidence, why is 5,000 the wrong denominator, and what is the right one?
Reveal the answer
The 1,000 people who already have diabetes cannot develop it again during the year, so they are not at risk and must not be counted. The population at risk at the beginning is 5,000 − 1,000 = 4,000, and that is the correct incidence denominator.
Calculate the cumulative incidence for the year with K = 100.
Reveal this step
100 ÷ 4,000 = 0.025, then 0.025 × 100 = 2.5%. The cumulative incidence during the year is 2.5%.
Prevalence is 20% and incidence is 2.5%. Why is prevalence so much larger, and what does the ratio 20 ÷ 2.5 = 8 estimate?
Reveal the answer
Prevalence counts every existing case, including the 1,000 people who have lived with diabetes for years, whereas incidence counts only the 100 diagnosed this year. The ratio 20% ÷ 2.5% = 8 estimates the average duration of the disease in years: a long-lasting disease accumulates cases, so a high prevalence can coexist with a low incidence.
If nobody died and nobody was cured, what is the point prevalence on 31 December, and what does the change show?
Reveal the answer
Existing cases rise to 1,000 + 100 = 1,100, so prevalence = 1,100 ÷ 5,000 × 100 = 22%. Prevalence grew during the year by exactly the new cases, which is why prevalence measures the burden carried while incidence measures the flow of new disease.
Answer: Point prevalence on 1 January was 20%; cumulative incidence during the year was 2.5%; if nobody died or recovered, point prevalence on 31 December was 22%.
Keep this habit: Prevalence counts who has the disease now; incidence counts who became ill during the period. Their denominators differ, and because a long duration lets cases pile up, a high prevalence does not mean a high risk of catching the disease.
Now you try
Attempt each one on paper first. Write the formula, substitute the numbers, then check the unit of your answer.
A city of 250,000 people records 1,200 deaths during one year. Calculate the annual mortality rate per 1,000 population and state it in full, naming the denominator and the period.
Hint
The denominator is the whole city population, not the number of patients in hospital. Divide first to get a fraction far below 1, then multiply by K = 1,000.
Show the worked answer
Mortality rate = 1,200 ÷ 250,000 × 1,000. First 1,200 ÷ 250,000 = 0.0048, then 0.0048 × 1,000 = 4.8. In full: 4.8 deaths per 1,000 population per year. The denominator is the 250,000 residents, all of whom could have died, and the period is one year. The same fraction with K = 100,000 would be 480 deaths per 100,000 population per year.
In a village of 2,400 people, 60 people already have a disease on 1 January and 12 new cases occur during the year. Calculate the point prevalence on 1 January and the cumulative incidence over the year, using K = 100.
Hint
Prevalence keeps all 2,400 people underneath. For incidence, subtract the 60 existing cases from the population first, because those people are no longer at risk.
Show the worked answer
Prevalence = 60 ÷ 2,400 × 100 = 2.5%. The population at risk for incidence is 2,400 − 60 = 2,340, so incidence = 12 ÷ 2,340 × 100 = 0.51% over the year. Prevalence (2.5%) is much larger than incidence (0.51%) because it counts the 60 people who have had the disease for some time as well as the 12 new cases.
A disease has an incidence of 4 new cases per 1,000 person-years and an average duration of 5 years. Using prevalence ≈ incidence × duration, estimate the prevalence and express it as a percentage.
Hint
Put both quantities on the same scale before multiplying: 4 per 1,000 is 0.004 per person-year. The duration is already in years, so the product is a pure fraction.
Show the worked answer
Write the incidence as a fraction: 4 ÷ 1,000 = 0.004 new cases per person-year. Multiply by the average duration: 0.004 × 5 = 0.02. As a percentage, 0.02 × 100 = 2%, which is 20 existing cases per 1,000 people. The estimate holds only for a stable population in which incidence is not changing quickly.
In an outbreak, 20 people are diagnosed with a disease and 3 of them die. Elsewhere, a disease causes 500 deaths a year in a population of 2 million. Explain why the first figure is a case-fatality rate and the second a mortality rate, and calculate both.
Hint
Look at the denominator of each: the people who were diagnosed, or the whole population. For the second, choose K = 100,000 so the result reads as deaths per 100,000.
Show the worked answer
Case-fatality rate = 3 ÷ 20 × 100 = 15%, because the denominator is only the 20 patients who had the disease; it answers how lethal the disease is once contracted. Mortality rate = 500 ÷ 2,000,000 × 100,000 = 25 deaths per 100,000 population per year, because the denominator is the whole population; it answers how large a share of everyone died. The same disease can look terrifying in the first measure and small in the second when cases are rare.
Common misconception
Beginners often treat every frequency figure as a risk: if prevalence is 20%, they assume that one person in five will catch the disease this year. Prevalence counts people who already have the disease, including those diagnosed years ago, so it measures burden, not the chance of becoming ill. The chance of becoming ill is incidence, and it is calculated only among people who do not yet have the disease. A long-lasting disease can therefore show a high prevalence and a very low incidence at the same time.
Check your reasoning
A colleague reports that “the incidence of the disease fell from 5% to 2.5%, so fewer people have the disease now”. Explain what is wrong with the second half of that sentence, using the relationship between incidence, prevalence and duration.
Reveal an answer
Incidence measures only the flow of new cases; prevalence measures the stock of everyone who has the disease. A falling incidence reduces the number of new cases arriving, but the people already living with a long-lasting disease remain in the population and keep prevalence high for years. Prevalence can therefore stay level or even rise while incidence falls, for example when treatment extends survival. To claim that fewer people have the disease, the colleague must quote a prevalence, and to explain a change in prevalence both the incidence and the average duration of the disease must be examined.
Where this comes from
- CDC: Morbidity frequency measures · Optional English reading; internet required. Checked 6 October 2026. Distinguishes incidence proportion, person-time incidence rate and prevalence.
- BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (supplied) · Section 12, “Medical rates and frequencies”, lines 1159–1211 of the supplied notes (the rates block inside the extraction range 1028–1211): the list of common measures including case-fatality (1161–1166); morbidity as illness in a population, with the morbidity proportion and the note that the exact denominator and period must be specified (1167–1172); the mortality-rate formula with K = 1,000 or 100,000 (1173–1177) and the worked example 250 ÷ 100,000 × 100,000 = 250 deaths per 100,000 per year (1178–1181); incidence as new cases during a period among a population at risk (1182–1185) with the example 50 ÷ 5,000 × 100 = 1% (1186–1189); prevalence as existing cases at a point or over a period (1190–1192) with the example 160 ÷ 2,000 × 100 = 8% (1193–1196); the incidence-versus-prevalence comparison table (1197–1199).
- BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (supplied) · The notes use K = 100,000 in the mortality example (line 1180) and silently switch to K = 100 in the incidence example (line 1188). Both results — 250 deaths per 100,000 per year and 1% — are arithmetically correct and are reproduced in this section without change; the missing rule that K is chosen by the writer and must be reported is supplied here and labelled as beyond the notes.
- Beyond the supplied notes: the parts of a rate the notes leave implicit · The supplied notes give the formulas but not the rule that K is chosen and must be reported, the distinction between a proportion and a rate, person-time and the incidence rate with person-years as its denominator, the approximate relationship prevalence ≈ incidence × average duration, or the formula for the case-fatality rate that the notes list (line 1166) but never define. These are standard epidemiology taught here as beyond the notes; the worked figures from the notes are not altered.
- 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.