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Control statistics

13 min

  • Calculate the mean of control measurements
  • Calculate standard deviation from stated control limits and state its unit
  • Calculate coefficient of variation from a mean and standard deviation
  • State the share of control results expected within 1, 2, and 3 SD of the mean

Read the full reference

Try first

Try first

A glucose control at level 1 has a mean of 100 mg/dL and an SD of 4 mg/dL. The level 2 control has a mean of 300 mg/dL and an SD of 9 mg/dL. Which level shows the smaller imprecision relative to its concentration?

The next section explains it.

The next section explains it.

Right. The next section explains why.

The next section explains it.

Get the idea

Four numbers

A control material is measured over and over, and its results spread around a central value. Four numbers describe those results, and every QC rule is built on them.

  • The mean is the arithmetic average: add the results and divide by how many there are.
  • The standard deviation (SD) says how far results usually sit from the mean. It carries the analyte's own unit, such as mmol/L or mg/dL.
  • The variance is the SD squared.1
  • The coefficient of variation (CV) removes the unit, so levels and methods can be compared:

CV (%) = SD ÷ mean × 100

Because the SD keeps its unit, a bigger SD at a higher concentration can still be the better performance.

SD from the limits

A control sheet often lists only the limits. Limits of ±2 SD run from 2 SD below the mean to 2 SD above it.

  • The mean is the midpoint.
  • The whole interval is 4 SD wide, so 1 SD is the width divided by 4.
  • Dividing by 2 gives an SD twice the true one and draws every limit too far out.2

What each band means

When the results of a stable procedure follow a Gaussian distribution, fixed shares fall inside each band:3

Band around the meanResults insideResults outside
±1 SD68.3%31.7%
±2 SD95.4%4.6%
±3 SD99.7%0.3%

Those shares explain how the rules treat each band:4

  • A control beyond 2 SD turns up about once in every 22 results from a procedure that is working. A run with two control levels gives two chances, so about 9% of good runs show one. That is why a single result beyond 2 SD serves as a warning to check the other rules.
  • A result beyond 3 SD happens by chance about 3 times in 1,000, rare enough to reject the run.

Laboratories set each control's mean and SD from their own repeated measurements of one control lot, collected over many runs. The examples here use a handful of results to show the arithmetic. A handful is far too few to show whether a spread is Gaussian.2

References
  1. National Institute of Standards and Technology. Measures of scale. NIST/SEMATECH e-Handbook of Statistical Methods. Accessed September 23, 2026.
  2. Clinical and Laboratory Standards Institute. Statistical Quality Control for Quantitative Measurement Procedures: Principles and Definitions. 4th ed. CLSI guideline C24Ed4E. Clinical and Laboratory Standards Institute; 2016.
  3. National Institute of Standards and Technology. What do we mean by "normal" data?. NIST/SEMATECH e-Handbook of Statistical Methods. Accessed September 23, 2026.
  4. Westgard JO. Westgard rules and multirules. Westgard QC. Accessed September 23, 2026.

Watch one

A new sodium control lot arrives with laboratory-established limits of 147.0 to 153.0 mmol/L, stated as ±2 SD. Find the mean, the SD, the CV and the ±3 SD limits.

  1. Take the midpoint: (147.0 + 153.0) ÷ 2 = 150.0 mmol/L.

    The ±2 SD limits sit the same distance on either side of the mean.

  2. Divide the width by 4: (153.0 − 147.0) ÷ 4 = 6.0 ÷ 4 = 1.5 mmol/L.

    From the lower limit to the upper limit is 2 SD down plus 2 SD up, 4 SD in all.

  3. Divide the SD by the mean: 1.5 ÷ 150.0 × 100 = 1.0%.

    The CV turns the SD into a share of the mean, so its only unit is %.

  4. Add and subtract 3 SD: 150.0 − 4.5 = 145.5 mmol/L and 150.0 + 4.5 = 154.5 mmol/L.

    Every other limit on the chart is a whole number of SDs from the mean.

Mean 150.0 mmol/L, SD 1.5 mmol/L, CV 1.0%, ±3 SD limits 145.5 to 154.5 mmol/L.

Your turn

Problem 1 of 3

A control's ±2 SD interval is 80 to 110 mg/dL. What is 1 SD?

Correct. The full interval spans 4 SD, from 2 SD below the mean to 2 SD above it. Dividing 30 mg/dL by 4 gives 7.5 mg/dL.

Incorrect. The distance from the mean of 95 mg/dL to either limit is 15 mg/dL, which represents 2 SD. Dividing that distance by 2 gives 7.5 mg/dL.

Incorrect. Subtracting 80 from 110 gives the full width of the interval. That width spans 4 SD, so 1 SD is 30 ÷ 4 = 7.5 mg/dL.

Hint
  1. The lower limit is 2 SD below the mean, and the upper limit is 2 SD above it.
  2. Count the SDs from the lower limit to the upper limit.
  3. Divide the full width of the interval by that count.

Review Statistical control of a measurement procedure

Problem 2 of 3

A stable potassium procedure runs two control levels in every run. About how often will a good run show at least one control beyond ±2 SD by chance alone?

That is the chance of one result beyond ±3 SD. The ±2 SD band leaves out far more.

About 1 result in 22 falls outside ±2 SD. With two controls in the run, the chance for the run is close to double that.

About a third of results fall outside ±1 SD. The ±2 SD band contains 95.4% of results.

Each control has a 4.6% chance of falling outside ±2 SD, and two controls make the chance about 9% for the run. That is why one control beyond 2 SD is a warning and needs the other rules before the run is rejected.

Hint
  1. About 4.6% of single results from a stable procedure fall outside ±2 SD.
  2. Each run gives the procedure two chances to produce one.

Review Statistical control of a measurement procedure

Problem 3 of 3

A magnesium control gives five results: 2.0, 2.1, 1.9, 2.0 and 2.0 mg/dL. The SD of these results is 0.07 mg/dL. What is the CV?

Show the answer

3.5 %

The mean is (2.0 + 2.1 + 1.9 + 2.0 + 2.0) ÷ 5 = 10.0 ÷ 5 = 2.00 mg/dL, and the CV is 0.07 ÷ 2.00 × 100 = 3.5%.

Review Statistical control of a measurement procedure

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