A production line is in-control, centered and the quality characteristic of inte

Question

A production line is in-control, centered and the quality characteristic of interest is normally distributed. The Process Capability Ratio Cpk = 0.90 and the specifications limits (LSL, USL) are (75, 120).

(a) How many process standard deviations separate the process mean from the upper specification limit (USL)?

(b) What is the DPMO below the lower specification limit?

(c) What is the sigma level for the process?

(d) Calculate the Zbench and the Sigma Quality Level (SQL) for the process?

(e) By how much would the process standard deviation have to be reduced to have a true six sigma (6?) process

Explanation / Answer

(a)

This is a 'centered' process. So, the process mean will lie at the center of the [LSL, USL] range.

So, the process mean = (75+120)/2 = 97.5

Cpk = Cp = (USL - LSL)/6*sigma = 0.90
or, 6*Sigma = (120 - 75)/0.9 = 50
or, Sigma = 50/6 = 8.333

USL = process mean + Z*Sigma
or, 120 = 97.5 + Z*8.333
or, Z = 2.7

So, the USL is separated from the process mean by 2.7 standard deviation

(b)

Prob(Output < LSL) = NORMDIST(LSL, process mean, Sigma, TRUE)
= NORMDIST(75, 97.5, 8.333, TRUE) = 0.003466

So, DPMO = 3466

(c)

Sigma level (long-term) = 3*Cpk = 2.7

(d)

DPMOLSL + DPMOUSL = 3466*2 = 6932

Zbench = NORMSINV(1 - (6932/1000000)) = 2.46, SQL = (USL - mean)/sigma + 1.5 = 2.7+1.5=4.2

(e)

(USL - mean)/sigma + 1.5 = 6
or, (120 - 97.5)/Sigma = 4.5
or, Sigma = (120 - 97.5)/4.5 = 5

So, the process standard deviation have to be reduced from 8.333 to 5.0

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