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# Quality Management at Toyota - Essay Example

2021-07-15
2Â pages
453 words
Categories:Â
University/College:Â
University of Richmond
Type of paper:Â
Essay
This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.

In quality management, the expected quality standards are met by standardizing the quality and setting the right size and figures to be manufactured. In the case above, no washer should be more than 2.4 millimeters and therefore if all thicknesses are equal and distributed equally, then the maximum thickness is calculated as follows:

The average thickness of the sample is 1.9625. From the table, the standard deviation is 0.209624, and the probability of the part that has a thickness which is greater than 2.4 is the remainder from 2.4 which is 2.4 1.9625 / 0.209624 = 2.0870681. Therefore, the part of the washers that will be defective is 0.018441% of the total and these will have the thickness greater than 2.4.

The upper thickness limit of 2.4 and the lower limit of 1.4, the following from the excel show the calculations;

1.9 2.2 1.7 1.7

1.8 2.4 1.8 2

2.1 1.8 2 1.7

2.1 2.2 2 1.8

2 2.1 2 1.9

2.2 1.7 2.1 2.1

2.4 2.2 1.8 1.8

2 1.9 1.6 1.6

1.9 2.1 1.9 1.9

2.1 2.2 1.6 2.2

The above table shows the data of 40 washers that are used to sample the thickness. The mean of the washers is 1.9625, and the standard deviation is 0.209. Therefore the normal dist is 0.018441 for the thickness of 2.4 and 0.003644 for the thickness of 1.4. The difference between the two is (0.018441 - 0.003644 = 0.014797.

The lower limit Z = (1.4 1.9625) / 0.209624 = -0.5625

Then, -0.5625/0.209624 = -2.683375

Norms dist = 0.00364422

Therefore, the washers expected to have a thickness less than 1.4 = 0.3644%.

Total defective fraction will be 0.018441 + 0.003644 = 0.022085

Therefore those out of tolerance will be 0.0220085 x 100 = 2.2085%.

The Cpk for the process if 0.6957

Cpk= min (UTL-x/ 3s , x-LTL/3s)

=min [0.6957, 0.89446]

= 0.6957

If the process centred within the limits, the limits would be the average of the two thickness limits.

2.4 + 1.4 = 3.8

3.8/2 = 1.90

Cpk= min [UTL-x3s, X X-LTL3sTherefore if we substitute, we get 50.6289 506289= 0.795

If the process were centred, then the percentage that will be out of tolerance will be as follows:

Z = (2.4 1.9) / 0.209624 = 2.385221

As calculated above, the Norms dist = 0.00364422

Therefore, the defective fraction will be = [1- (0.00364422 x 2.385221)] x 2

= 0.017069

The percentage is therefore 1.7%.

Assuming that the process will take samples of 10 and that the mean and range were to be calculated, the bars would be as follows:

Sample 1 2 3 4 5 6 7 8 9 10 Mean Range

1 1.9 2.0 1.9 1.8 2.2 1.7 2 1.9 1.7 1.8 1.89 0.5

2 1.8 2.2 2.1 2.2 1.9 18 2.1 1.6 1.8 1.6 1.91 0.6

3 2.1 2.4 2.2 2.1 2.1 2 1.8 1.7 1.9 1.9 2.02 0.7

4 2.1 2 2.4 1.7 2.2 2 1.6 2 2.1 2.2 2.03 0.8

The mean is therefore 0.9625. The limits for the upper limit is 2.164 and for the lower limit is 1.761 and for the average is 1.157.

Considering the control limits that have been shown in the experiment, the process seems to be in control, and it has a reliable operation level. However, since only four samples were used, it may not be reliable to use this data, and the final decision should be made based on more samples. The man, standard deviation, and other tests seem to give a positive trend which may not always be the case if a bigger sample was to be used.

If the process was improved and standard deviation became only 0.10, then the mean would be 1.9625 and a standard deviation of 0.2070 which would be the best figures. The figures can be calculated by using the table in question 6.

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