Editing Reference range (section)

====Example====
{{Hatnote|The method is described in further detail at [[differential diagnosis]].}}
Let's say, for example, that an individual takes a test that measures the [[ionized calcium]] in the blood, resulting in a value of 1.30&nbsp;mmol/L, and a reference group that appropriately represents the individual has established a reference range of 1.05 to 1.25&nbsp;mmol/L. The individual's value is higher than the upper limit of the reference range, and therefore has less than 2.5% probability of being a result of random variability, constituting a strong indication to make a [[differential diagnosis]] of possible causative conditions.

In this case, an [[Differential diagnosis#Specific methods|epidemiology-based differential diagnostic procedure]] is used, and its first step is to find candidate conditions that can explain the finding.

[[Hypercalcemia]] (usually defined as a calcium level above the reference range) is mostly caused by either [[primary hyperparathyroidism]] or malignancy,<ref name=Kumar>Table 20-4 in: {{cite book |author1=Mitchell, Richard Sheppard |author2=Kumar, Vinay |author3=Abbas, Abul K. |author4=Fausto, Nelson |title=Robbins Basic Pathology|publisher=Saunders |location=Philadelphia |year= 2007|isbn=978-1-4160-2973-1 }} 8th edition.</ref> and therefore, it is reasonable to include these in the differential diagnosis.

Using for example epidemiology and the individual's risk factors, let's say that the probability that the hypercalcemia would have been caused by primary hyperparathyroidism in the first place is estimated to be 0.00125 (or 0.125%), the equivalent probability for cancer is 0.0002, and 0.0005 for other conditions. With a probability given as less than 0.025 of no disease, this corresponds to a probability that the hypercalcemia would have occurred in the first place of up to 0.02695. However, the hypercalcemia ''has occurred'' with a probability of 100%, resulting adjusted probabilities of at least 4.6% that primary hyperparathyroidism has caused the hypercalcemia, at least 0.7% for cancer, at least 1.9% for other conditions and up to 92.8% for that there is no disease and the hypercalcemia is caused by random variability.

In this case, further processing benefits from specification of the probability of random variability:

The value is assumed to conform acceptably to a normal distribution, so the mean can be assumed to be 1.15 in the reference group. The [[standard deviation]], if not given already, can be inversely calculated by knowing that the [[absolute value]] of the difference between the mean and, for example, the upper limit of the reference range, is approximately 2 standard deviations (more accurately 1.96), and thus:
:{{math|1=Standard deviation (s.d.) ≈ {{sfrac|{{mabs | (Mean) - (Upper limit) }} |2}} = {{sfrac|{{mabs | 1.15 - 1.25 }}|2}} = {{sfrac| 0.1 |2}} = 0.05}}.

The [[standard score]] for the individual's test is subsequently calculated as:
:{{math|1= Standard score (''z'') = {{sfrac|{{mabs | (Mean) - (individual measurement) }} |s.d.}} = {{sfrac|{{mabs | 1.15 - 1.30 }}|0.05}} = {{sfrac|0.15|0.05}} = 3}}.

The probability that a value is of so much larger value than the mean as having a standard score of 3 corresponds to a probability of approximately 0.14% (given by {{math|(100% &minus; 99.7%)/2}}, with 99.7% here being given from the [[68–95–99.7 rule]]).

Using the same probabilities that the hypercalcemia would have occurred in the first place by the other candidate conditions, the probability that hypercalcemia would have occurred in the first place is 0.00335, and given the fact that hypercalcemia ''has occurred'' gives adjusted probabilities of 37.3%, 6.0%, 14.9% and 41.8%, respectively, for primary hyperparathyroidism, cancer, other conditions and no disease.