Six Sigma Green Belt / Free study sprint

Cp sees spread. Cpk sees the shift.

Calculate Cp and Cpk for a stable, normally distributed process and explain why they differ.

Before you start: Interpret a mean, a standard deviation and upper/lower specification limits.

1. Try2. Understand3. Practice4. Revisit

01 / Start from memory

What would you choose?

Commit to an answer before reading the board.

One-question checkA stable normal process has Cp = 1.50 and Cpk = 0.80. What explains the difference?

Six Sigma Green Belt / Measure / Process capability

Cp sees spread. Cpk sees the shift.

Calculate Cp and Cpk for a stable, normally distributed process and explain why they differ.

01Summary board

The decision path

1

Check the conditions

Establish process stability and whether the normal model is appropriate. Specification limits express requirements; they are not control limits.

[S1, S2]
2

Compare total widths

Cp compares specification width with six standard deviations. It does not account for an off-center mean.

Cp = (USL - LSL) / (6σ)
[S1]
3

Check the nearer limit

Cpk uses the smaller standardized distance from the mean to either specification limit. Keep the smaller value, not their average.

Cpk = min[(USL - μ)/(3σ), (μ - LSL)/(3σ)]
[S1]

Worked example / Original StudyMaps scenario

The spread fits; the mean has moved

Assume a stable normal process with LSL = 10, USL = 22, mean μ = 18 and standard deviation σ = 2. All values use the same measurement unit.

LSL10Midpoint16Mean18USL22
  1. Specification width: 22 - 10 = 12. Process width: 6 × 2 = 12.
  2. Cp = 12 / 12 = 1.00.
  3. Upper side: (22 - 18) / 6 = 0.67. Lower side: (18 - 10) / 6 = 1.33.
  4. Cpk = min(0.67, 1.33) = 0.67, rounded to two decimals.
ResultCp = 1.00; Cpk = 0.67. The upper limit is closer.
Principles: [S1]. Original example: StudyMaps.
StudyMaps · Two-sided specifications, stable process, normal-model assumptions and positive process standard deviation.88f2ae22 / 1 of 3

03 / Apply it

Your turn.

Three new problems. Explain your choice on paper, then check the reasoning.

Practice 1Assume a stable normal process: LSL = 40, USL = 60, μ = 50, σ = 2. What are Cp and Cpk?
Practice 2Keep LSL = 40, USL = 60 and σ = 2, but move μ to 56. What is Cpk? Assume stability and normality.
Practice 3The process shows special-cause variation. What should precede a routine capability conclusion?

0 of 3 practice problems checked

04 / Return later

Does it still make sense tomorrow?

Bookmark this page and try the new problem without the board. This is a practice check, not an exam-readiness score.

Open the return-later problem
Unseen problemAssume stability and normality. LSL = 8, USL = 20, μ = 12, σ = 1. Find Cp and Cpk, rounded to two decimals.

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Sources and scope

Two-sided specifications, stable process, normal-model assumptions and positive process standard deviation. Focused on ASQ Green Belt capability concepts; check your certification provider's syllabus.

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