STDEV.P

Calculates the standard deviation for a population data set.

Syntax:

STDEV.P(numberOne, [numberTwo],…)

numberOne, [numberTwo],… are 1 to 255 arguments representing the numbers that you want to use. These arguments can be numbers, cell references or ranges.

Calculation for STDEV.P:

$\frac{\sum ( x - x ˉ ) ^{2}}{n}$

Example:

If numberOne, [numberTwo],… contains 1,1,1,1,2,2,2,2,2,2,3,3,4,4,4,4,5,5,5,5,5:

STDEV.P(1,1,1,1,2,2,2,2,2,2,3,3,4,4,4,4,5,5,5,5,5)

returns 1.48002574

If numberOne, [numberTwo],… contains 1,1,1,1,2,2,2,2,2,3,3,3,4,4,4,4,5,5,5,5,5:

STDEV.P(1,1,1,1,2,2,2,2,2,3,3,3,4,4,4,4,5,5,5,5,5)

returns 1.463075381

	A	B	C	D
1	1	1	1	1.463075381
2	1	2	2
3	2	2	2
4	3	3	3
5	4	4	4
6	4	5	5
7	5	5	5

Application:

A Production Line's Quality Control

A company manufactures 1-liter (1000 ml) bottles of soda. The quality control manager wants to assess the consistency of the filling process. To do this, they randomly select 10 bottles from a production run and measure the exact volume of soda in each. Since the company is only interested in this specific production run and considers these 10 bottles to be the entire population they are analyzing for this run, the STDEV.P function is the correct choice.

Data Set

The measured volumes (in ml) for the 10 bottles are as follows:

	Bottle	Volume (ml)
	A	B
1	1	998
2	2	1001
3	3	1005
4	4	999
5	5	1002
6	6	997
7	7	1003
8	8	1000
9	9	996
10	10	1004

Calculation

The STDEV.P function would be used to calculate the standard deviation of this population of data points.

Average (Mean): First, the average volume is calculated: (998 + 1001 + 1005 + 999 + 1002 + 997 + 1003 + 1000 + 996 + 1004) / 10 = 1000.5 ml
Variance: Next, the variance is calculated by finding the squared difference of each data point from the mean, summing them up, and then dividing by the total number of data points (n).
STDEV.P: Finally, the standard deviation is the square root of the variance.

Using the STDEV.P function on the data set, the result would be approximately 2.872 ml.

Interpretation

A standard deviation of 2.872 ml tells the quality control manager that the volumes in the sampled bottles typically vary by about 2.872 ml from the average volume of 1000.5 ml.

If the standard deviation were a small number (e.g., 0.5 ml), it would indicate a very consistent and precise filling process.
If the standard deviation were a large number (e.g., 15 ml), it would suggest a highly inconsistent process with a wide range of volumes, which could lead to customer complaints or regulatory issues.

In this case, the manager can use the 2.872 ml standard deviation to monitor the filling machine's performance and ensure it is operating within acceptable quality limits.

Result for STDEV.P:

2.872