Computation and Measurement of Risk - Mathematics and Statistics for the Environment - page 193

Date of onset of illness
Date of death
Date of recovery

Person #
1

2

3

4

5

6

7

8

9

10

|___ |___ |___ |___ |___ |___ |___|___ |___ |___ |___ |___|___|___|___|___|

Oct. 1, 2004 Apr. 1, 2005 Sept. 30, 2005

FIGURE 8.1 New cases of illness from October 1, 2004, to September 30, 2005. (From CDC, Principles of
Epidemiology in Public Health Practice , 3rd ed., U.S. Centers for Disease Control and Prevention, Atlanta,
GA, 2012.)

Solution:

Incidencerate numerator = Number of new case ssbetween October1andSeptember 30

= 4(th eeother 6all had onsetsbeforeOctober1andare not included)

IncidenceratedenominatorApril 1 population
18 (persons 2and

=
= 8 died before April1)

Incidencerate 4/1 = 8) 100

22 new cases per 100 population

×

=

■ EXAMPLE 8.13
Problem: Calculate the point prevalence on April 1, 2005. Point prevalence is the number of persons
ill on the date divided by the population on that date. On April 1, 7 persons (persons 1, 4, 5, 7, 9,
and 10) were ill.

Solution:

Point prevalence = (7/18) × 100 = 38.89%

■ EXAMPLE 8.14
Problem: Calculate the period prevalence from October 1, 2004, to September 30, 2005. The
numerator of period prevalence includes anyone who was ill any time during the period. In Figure
8.1, the first 10 persons were all ill at the same time during the period.

Solution:

Period prevalence = (10/20) × 100 = 50.0%