Problem 1 Assume that the Length (L) is the most important characteristic to the

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Problem 1 Assume that the Length (L) is the most important characteristic to the quality of product X. The following Table shows the measurements of the length of 55 pieces of this product. Use these results to perform the tasks below A. Estimate the mean and the variance of the distribution of the corresponding quality characteristics. B. Use the Minitab, SPSS, or Excel to construct the Histogram. The Excel sheet of these measurements is available. on the Blackboard file name:Measurements Homewrok1) C. Assume that the specification limits of the length are uot allowances -9.5t1 and NL~(Awo, aa) (9.5 mm, 0.490 mm s the process centered? your comments D. Estimate the percentage of scrap and rework at the current situation NL-(H1,aa). No Length No Length No Length No Length No Length 3 10.20 989 25 10.50 36 9.90 47 9.93 5 9.50 16 10.02 27 10.00 38 852 11.23 6 9.50 17 9.75 28 984 39 807 50 10.42 7 9.60 18 10.03 29 10.22 40 11.82 51 904 8 10.31 19 11.34 30 9.74 41 10.02 52 10.09 9 9.38 20 10.20 31 9.80 42 10.00 53 9.43 10 10.10 21 10.03 32 10.00 43 1020 54 9.00 11 10.54 22 10.03 33 10.20 44 10.00 55 10.00 Problem 2 Assume that the Length (L) is the most important characteristic to the quality of product X. The length is normally distributed variable NL-(Ho,ot) (20 mm, 4 m The specification limits are 20 mm +3 mm. Assuming that if an item exceeds the upper specification limit it can be reworked, and if it is below the lower specification limit it must be scrapped. A. The process now producing? Show the drawings. B. The process is running now at NL-(u1, oOO (23 mm, 4 mm2). What percent of scrap and rework is the process now producing? Show the drawings. c. optional The process is running now at NL~(uo, ob (20 mm, 6 mm2). What percent of scrap and rework is the process now producing? Show the drawings. Note: this is the case when the shift occurs only in the process variance. 13 PM a E

Explanation / Answer

Mean of Length = 9.99036 = 10

Standard deviation of Length = 0.698037

Variance of Length = 0.4872

Here the specification limits are LSL = 8.5 to USL = 10.5

Pr (X > USL) = Pr (X > 10.5) = 1 - Pr (X < 10.5) = 1 - Pr (Z < 0.743) = 1 - 0.7625 = 0.2375

Percentage of rework = 23.75%

Pr (X < LSL) = Pr( X < 8.5) = Pr ( Z < -2.1428) = 0.0161

Percentage of Scrap = 1.61%

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