Use the following table to answer the question(s) below. Sample # 6 5.1 4.74.3 4

Question

Use the following table to answer the question(s) below. Sample # 6 5.1 4.74.3 4.6 4.6 4.8 4.9 4.3 4.4 4.5 4.6 4.4 4.3 4.3 5.1 4.8 4.7 4.9 4.4 5 4.7 484.5 5 4.9 | 5.1 | 4614.% | 5 | 5 | 4.6 | 4.7 4.9 5.2 4.9 4.7 4.7 4.3 4.7 5.2 33) A machine shop owner wishes to monitor the diameter of an engine boring operation where the piston bore should be 4.25". He obtains 8 samples of 5 and records the data in the table above. What should the center line be for a chart that monitors process dispersion? 33) A) 0.55 B)0.69 C)0.50 D) 0.60 E) 0.80 34) A machine shop owner wishes to monitor the diameter of an engine boring operation where the piston bore should be 425. He obtains 8 samples of 5 and records the data in the table above. What should the center line be for a chart that monitors process average? 34) A) 458 B) 4.66 C) 4.82 D)4.78 E) 4.73 35) A machine shop owner wishes to monitor the diameter of an engine boring operation where the piston bore should be 4.25". He obtains 8 samples of 5 and records the data in the table above. What should the lower and upper control limits be for an X-bar chart? 35) A) (0.00, 1.45) B) (5.00, 8.00) C) (4.33, 5.12) D) (-5.10, 5.10) E) (-1.50, 1.50)

Explanation / Answer

33. Chart to process dispersion is R chart

Calculate range of each sample by measuring difference between maximum and minimum value of diameter in each sample. Measure average of range of all 8 samples. R- bar will be the center line for R chart

e.g for sample #1 Range = 5.1-4.4 = 0.7

Answer B) 0.69

34. Chart that monitor the process avaerage is X-bar chart and central line of this chart is X-bar bar.

Measure the avergae of diameters for each sample.

Xbar bar = sum of X-bar of sample / number of sample

e.g X bar for sample 1 = (5.1+4.4+4.7+4.9+4.9)/5 = 4.8

Answer, X bar bar = 4.73 D. 4.73  

35.

For X bar chart

UCL = X-bar bar + A2 * R-bar

UCL = X-bar bar - A2 * R-bar

A2 is a factor corresponding to number of items in each sample

for this case n =5,

A2 = 0.577

UCL = 4.73 + 0.577*0.69 = 5.12

LCL = 4.73 - 0.577*0.69 = 4.33

Answer C (4.33 , 5.12)

Sample # Range 1 5.1 4.4 4.7 4.9 4.9 0.7 2 4.7 4.5 4.9 5.1 5.2 0.7 3 4.3 4.6 4.4 4.6 4.9 0.6 4 4.6 4.4 5 4.9 4.7 0.6 5 4.6 4.3 4.7 5 4.7 0.7 6 4.8 4.3 4.8 5 4.3 0.7 7 4.9 5.1 4.5 4.6 4.7 0.6 8 4.3 4.8 5 4.7 5.2 0.9 Average 0.6875

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