The specifications for a plastic liner for a concrete highway project calls for thickness of 5.0 mmplus or minus±0.10 mm. The standard deviation of the process is estimated to be 0.02 mm.
A) The upper specification limit for this product = ? mm (round your response to three decimalplaces).
B) The lower specification limit for this product = ? mm (round to three decimal palces)
C) The process capability index (CPk) = ? (round to three decimal places)
D) The upper specification lies about ? standard deviations from the centerline (mean thickness)

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Chrisnando Chrisnando · Answer 1 · 2020-07-05T03:57:15+02:00

Answer:

a) 5.10 mm

b) 4.90 mm

c) 1.67

d) upper specification lies at 5 standard deviations from mean.

Explanation:

Given:

Mean = 5.0mm

Standard deviation = 0.02 mm

a) The upper specification limit:

5.0 + 0.10

= 5.10 mm

b) The lower specification limit:

5.0 - 0.10

= 4.90 mm

c) The process capability index (CPk):

Use the formula below to find the process capability index

[tex] C_p_k = min (\frac{USL - mean}{3\sigma}, \frac{mean - LSL}{3\sigma}) [/tex]

Substitute figures:

[tex] C_p_k = min (\frac{5.1 - 5.0}{3*0.02}, \frac{5.0 - 4.9}{3*0.02}) [/tex]

[tex] C_p_k = min (\frac{0.1}{0.06}, \frac{0.1}{0.06}) [/tex]

[tex] C_p_k = min( 1.67, 1.67) [/tex]

[tex] C_p_k = 1.67 [/tex]

d) The upper specification lies at a distance of (5.1-5) = 0.1 mm

Standard deviation = 0.02 mm

upper specification lies at:

[tex] \frac{0.1}{0.02} = 5 [/tex]

Therefore, upper specification lies at 5 standard deviations from mean.