The specifications for a plastic liner for a concrete highway project calls for thickness of 4.0 mmplus or minus0.08 mm. The standard deviation of the process is estimated to be 0.02 mm. a) The standard deviation of the process is estimated to be 0.02 mm.
b) The upper specification limit for this product = ? mm (round your response to three decimal places).
c) The lower specification limit for this product = ? mm (round to three decimal places)
d) The process capability index (CPk) = ? (round to three decimal places)
e) The upper specification lies about ? standard deviations from the centerline (mean thickness)

Answer and Explanation:

The computation is shown below:

b. The upper specification limit is

= 4 + 0.08

= 4.080 mm

c. The Lower specification limit is

= 4 - 0.08

= 3.920 mm

d. The process capability index is

= min ((Upper specification limit - Mean) ÷ (3 × Standard deviation)), ((Mean - Lower specification limit)÷ (3 × Standard deviation))

= min (0.08 ÷ (3 × 0.02)), (0.08 ÷ (3 × 0.02))

= min (1.333, 1.333)

So it would be 1.333

e. Upper specification = 4.08 mm

Mean line = 4.0 mm

Now,

The upper specification lies at a distance = Upper specification - Mean line

= 4.08 mm - 4.0 mm

= 0.08 mm

upper specification =Upper specification lies ÷ One standard deviation

= 0.08 mm ÷ 0.02 mm

= 4 mm which is standard deviations from the mean

answer; the above statement is (true);

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