The specifications for the diameter of a moulded part are 10 mm $\pm$ 0.5 mm. The actual average and standard deviation from 250 parts sampled is 10.1 mm and 0.1 mm, respectively. What is the process capability as measured by $C_p$? Multiple Choice 0.84 1.67 0.60 1.00 < Prev 7 of 32 Next >
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The specification for the diameter is $10 \text{ mm} \pm 0.5 \text{ mm}$. This means the Upper Specification Limit (USL) = $10 + 0.5 = 10.5 \text{ mm}$. And the Lower Specification Limit (LSL) = $10 - 0.5 = 9.5 \text{ mm}$. The standard deviation ($\sigma$) from Show more…
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The difference between the upper specification and the lower specification for a process is 0.60. The standard deviation is 0.05. Based on the given information, the process capability ratio, C p = ???? (round your response to two decimal places). Based on the process capability ratio (C p)n for the given information, one can say that the process is (very much capable, not at all capable, barely capable) to produce within the design specifications.
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The outside diameter of a part used in a gear assembly is known to be normally distributed with a mean of 40 mm and a short-term standard deviation of 2.5 mm. The specifications on the diameter are (36, 45), which means that part diameters between 36 mm and 45 mm are considered acceptable. Show your work. What is the capability index Cp? What is the capability index Cpk? If the process was centered, what would Cp and Cpk be? If the long-term standard deviation is 4, what would Pp be? (assume a mean of 40.) What would Ppk be?
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(a) A machine spindle has a specification (tolerance) on its diameter of 1.500 ± 0.009 inches. If the Process Capability Ratio Cp = 1.0, what is the standard deviation of the spindles being produced by the cylindrical grinder? (b) What would the standard deviation have to be to achieve a process capability index of 1.33? (c) If Cp = 1.33 and the process mean is centered within the tolerance range, how many oversized parts would be expected in grinding the spindle described in Problem 1 (b)? (d) The process mean has moved 1.5 times the center of the tolerance range. From Example 1 (b), σ = 0.00226 inches. The shift k = 1.5(0.00226) = 0.003 inches toward the USL. Now μ = 1.500 + 0.003 = 1.503. Determine Cpkl and Cpku and the effect of the shift in the process mean on the defect rate by comparing the percentage of good parts produced and defective parts produced ppm for different sigma levels of tolerance by filling the following table (Show the related calculations for the probability of parts within the tolerance range and falling outside the tolerance range for each sigma level). Process Centered | Process Mean 1.5 Sigma from Center --- | --- Tolerance Range (± σ level) | Cp | Percent good parts | Defective parts ppm | Percent good parts | Defective parts ppm ±3 sigma | 1.00 | | | | ±4 sigma | 1.33 | | | | ±6 sigma | 2.00 | | | |
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