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 1.9 1.9 2. 2. 2.4 1.7 1.9 1.8 2.2 2.2 1.9 1.8 2. 1.817 2.0 19 2. 2.0 1.6 2.0 2.1 2.0 1917 2.1 1.6 1.8 16 2. 242.2 1.9 2.0 18 22.1 Use the appropriate Excel function to compute normal distribution probabilities in Parts a, b, e and f

Explanation / Answer

There are 40 sets of sample data, putting this in excel we derive:

sample mean = 1.9625

sample standard deviation=  0.2096

a. If the specification is such that now washer should be greater than 2.4 millimeters, assuming that the thicknesses are distributed normally, what fraction of the output is expected to be greater than this thickness?

We need to establish the probability of thickness of washer to be 2.4 mm so we need to find value z such that

sample mean + Z*standard dev = 2.4

1.9625+Z*0.2096 = 2.4

Z = (2.4 – 1.9625)/ 0.2096 = 2.087068

1 – NORMSDIST(2.087068) =.018441 fraction defective, so 1.84 percent of the washers are expected to have a thickness greater than 2.4.

b. If there are an upper and lower specification, where the upper thickness limit is 2.4 and the lower thickness limit is 1.5, what fraction of the output is expected to be out of tolerance?

The upper limit is given in a. The lower limit is 1.5 so Z = (1.5 – 1.9625)/ .2096 = -2.2065.

NORMSDIST (- 2.2065) = 0.013674 fraction defective, so 0.13674 percent of the washers are expected to have a thickness lower than 1.4.

The total expected fraction defective would be 0.018441 + .0.013674 = .032115 or about 3.21 percent of the washers would be expected to be out of tolerance.

c. What would be the Cpk for the process ?

Cpk = min [ (UTL – X)/3?, X – LTL)/3? ] = min ( .5625/ .6289, .4375/.6289 )= min {8944, .6957} = 0.6957

Cpk = 0.696

d. if it were centered between the specification limits (assume the process standard deviation is the same)?

Cpk = min [ (UTL – X)/3?, X – LTL)/3? ] = min ( .5/ .6289, .5/.6289 ) = min (.795,.795)

Cpk = .795

e. What percentage of output would be expected to be out of tolerance if the process were centered?

Z = (2.4 – 1.9)/ .209624 = 2.385221

Fraction defective would be 2 x (1 – NORMSDIST (2.385221)) = 2 X .008534 = .017069, about 1.7 percent.

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 best that could be done is for the process to be centered at 1.9, and given a standard deviation of .10, then Z = (2.4-1.9)/.1 = 5. The fraction defective = 2 x (1 – NORMSDIST(5)) = 5.73303E-07 which would only be about 573 defects per billion washers

g. Setup X-bar and Range control charts for the current process. Assume the operators will take samples of 10 washers at a time

Sample 1         2          3          4          5          6          7          8          9          10        X-bar   R

1          1.9       2          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       1.8       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

                                                                                                                        Mean: 1.9625 0.65

            The upper control limit for X-bar chart = 1.9625 + .31 x .65 = 2.164

            The lower control limit for X-bar chart = 1.9625 - .31 x .65 = 1.761

            The lower control limit for the Range chart = 1.78 x .65 = 1.157

            The lower control limit for the Range chart = .22 x .65 = .143

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.