As part of the process for improving the quality of their cars, Toyota engineers

Question

As part of the process for improving the quality of their cars, Toyota engineers have identified a potential improvement to the process that makes a washer that is used in the accelerator assembly. The tolerances on the thickness of the washer are fairly large since the fit can be loose, but if it does happen to get too large, it can cause the accelerator to bind and create a potential problem for the driver. (Note: This part of the case has been fabricated for teaching purposes and none of these data were obtained from Toyota.) Let's assume that as a first step to improving the process, a sample of 40 washers coming from the machine that produces the washers was taken and the thickness measured in millimeters. The following table has the measurements from the sample 1.8 19 1.9 21 2 2.4 .7 1.9 1.8 2.2 2.2 1.9 1.8 2.1 1.8 1.7 2.0 9 2.2 2.0 1.6 2.0 2.1 2.0 8 1.7 19 1.7 2.1 1.6 1.8 1.6 2.1 2.4 2.2 19 2.0 18 2.2 2.1 Use the appropriate Excel function to compute normal distribution probabilities in Parts a, b, e and f. a. If the specification is such that no washer should be greater than 2.4 millimeters, assuming that the thicknesses are distributed normally, what percentage of output is expected to be greater than this thickness? (Round your answer to 2 decimal places.) Percentage of output 1.51 % b. What if there is an upper and lower specification, where the upper thickness limit is 2.4 and the lower thickness limit 1.5, what percentage of output is expected to be out of tolerance? (Round your answer to 2 decimal places.) Percentage of the output 3.01% c. What is the Cpk for the process? (Round your answer to 4 decimal places.) Cpk 7227 d. What would be the Cok for the process if it were centered (assume the process standard deviation is the same)? (Round your answer to 3 decimal places.) Cpk 7227 e. What percentage of output would be expected to be out of tolerance if the process were centered? (Round your answer to 2 decimal places.) Percentage of output f. If the process could be improved so that the standard deviation was only about 0.1 millimeters, what would be the best that could be expected with the processes relative to fraction defective? The process to be centered at 1.95 g. Setup X-bar and Range control charts for the current process. Assume the operators will take samples of 10 washers at a time. (Round your answer to 3 decimal places.) X-bar chart chart Upper control limit Lower control limit

Explanation / Answer

Answer to question e:

If the process is centered and Cpk = 0.7227, then cumulative spread on both sides of process mean = 0.7227 x 3 x Standard deviation = 2.168

Hence spread around each side of mean = 2.168 / 2 = 1.084 x standard deviation

As per standard normal distribution table:

Probability that output will be within + 1.084 Standard deviation = 0.861

Probability that output will be within – 1.084 standard deviation = 0.139

Therefore,

Probability that output will be within upper and lower specification limit ( i.e. would be non defective items ) = Difference of above two probability values = 0.861 – 0.139 = 0.722

Therefore, probability that output will be out of tolerance = 1 – 0.722 = 0.278

Answer to question g :

Given :

Sample size = n = 10

Number of samples = 4

Mean of first sample = 19.8 / 10 = 1.98   , Range for first sample = 2.4 – 1.7 = 0.7

Mean of second sample = 19.6/10 = 1.96 , Range for second sample = 2.2 – 1.7 = 0.5

Mean of third sample = 18.5/10 = 1.85, Range of 3rd sample = 2.1 – 1.6 = 0.5

Mean of fourth sample = 20.1/10 = 2.01, Range of 4th sample = 2.4 – 1.6 = 0.8

Thus ,

Mean of Mean values = Xbar-bar = ( 1.98 + 1.96 + 1.85 + 2.01)/4 = 7.8/4 = 1.95

Mean of range values = ( 0.7 + 0.5 + 0.5 + 0.8 ) /4 = 0.625

Following are values of constants as per standard table for Xbar and Range chart for N = 10:

A2 = 0.308

D4 = 1.777

D3 = 0.223

Xbar chart :

Upper Control Limit = Xbar-bar + A2 .Rbar = 1.95 + 0.308 x 0.625 = 1.95 +0.1925=2.1425

Lower Control Limit = Xbar-bar – A2.Rbar = 1.95 – 0.308 x 0.625 = 1.95 – 0.1925 = .7575

Rbar chart :

Upper Control Limit = D4.Rbar = 1.777 x 0.625 = 1.110

Lower Control Limit = D3.Rbar = 0.223 x 0.625 = 0.139

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