List.StandardDeviation

List

Returns the sample standard deviation of the values in the list.

Syntax

List.StandardDeviation(numberList as list) as nullable number

Parameters

Name	Type	Required	Description
`numberList`	`list`	Yes	The list of numbers to compute the standard deviation for.

Return Value

nullable number — The sample standard deviation of the numeric values in the list, or null if the list is empty.

Remarks

List.StandardDeviation computes the sample standard deviation of a numeric list, dividing by n - 1 (Bessel's correction) rather than n. This is the appropriate formula when the list represents a sample from a larger population, and it is the same calculation performed by Excel's STDEV.S function.

Standard deviation measures the average spread of values around the mean. A higher value indicates greater variability. A standard deviation of zero means all values are identical.

Returns null if the list is empty or contains only one element (sample standard deviation is undefined for fewer than two data points). Null values in the list are ignored — they are excluded from both the count and the calculation.

To compute the population standard deviation (dividing by n), there is no direct built-in function. One approach is Number.Sqrt(List.Average(List.Transform(values, each (_ - List.Average(values))^2))). For detecting outliers, a common heuristic is to flag values more than 2 or 3 standard deviations from the mean.

Examples

Example 1: Standard deviation of a simple list

List.StandardDeviation({2, 4, 4, 4, 5, 5, 7, 9})

Result

	Result
1	2

Example 2: High variability vs. low variability

let
    Consistent = List.StandardDeviation({10, 11, 10, 9, 10}),
    Volatile = List.StandardDeviation({1, 50, 3, 98, 5})
in
    #table({"Consistent StdDev", "Volatile StdDev"}, {{Consistent, Volatile}})

Applied Steps

The final output — a table comparing the standard deviations of the consistent and volatile lists side by side.

	Consistent StdDev	Volatile StdDev
1	0.71	44.38

Example 3: Detect outliers using standard deviation

let
    Values = {100, 105, 98, 102, 97, 200, 103},
    Mean = List.Average(Values),
    StdDev = List.StandardDeviation(Values),
    Outliers = List.Select(Values, each Number.Abs(_ - Mean) > 2 * StdDev)
in
    Outliers