Question

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 1

The plastic liner thickness specification is 4.0 mm plus or minus 0.08 mm. The upper specification limit is 4.08 mm, the lower specification limit is 3.92 mm, and the process capability index (CPk) is 0.67.

Step-by-step explanation:

The thickness specification for the plastic liner is 4.0 mm plus or minus 0.08 mm. The standard deviation of the process is 0.02 mm.

To calculate the upper specification limit (USL), we add the mean thickness of 4.0 mm to the tolerance of 0.08 mm, resulting in a USL of 4.08 mm.

To calculate the lower specification limit (LSL), we subtract the tolerance of 0.08 mm from the mean thickness, resulting in an LSL of 3.92 mm.

The process capability index (CPk) can be calculated by dividing the difference between the USL and LSL by six times the standard deviation. In this case, CPk = (4.08 - 3.92) / (6 * 0.02) = 0.08 / 0.12 = 0.67.

The upper specification limit is approximately 2 standard deviations above the centerline (mean thickness) in a normal distribution.

Answer 2

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