F is correct As part of the process for improving the quality of their cars, Toy
ID: 330402 • Letter: F
Question
F is correct
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 2.1 2.1 2.4 1.7 1.9 182.2 2.2 1.9 18 2.1 18 2.01.9 2.2 2.0 1.6 2.0 2.1 2.0 8 1.7 1.9 172.11.6 1.8 16 2.142.2 1.9 2.0 1.8 2.22. Use the appropriate Excel function to compute normal distribution probabilities in Parts a, b, e and fExplanation / Answer
Process Overall average = µ = 1.9500 mm
Process standard deviation = ? = 0.2075 mm
a.
Probability of washers is expected to be greater than 2.4 mm thickness using standard z-score is obtained as follows:
P(X > 2.4) = 1 – P(X < 2.4)
z-score for X = 2.4, z = (X – µ)/? = (2.4 – 1.95)/0.2075 = 2.1682
P(z < 2.1682) = (=NORMSDIST(2.1682)) = 0.9849
P(X > 2.4)= 1 - P(z < 2.1682) = 1 – 0.9849 = 0.0151
Percentage of output greater than 2.4 mm = 1.51%
b.
Probability of washers is expected to be greater than 2.4 mm thickness and less than 1.5 mm, using standard z-score is obtained as follows:
P(1.5 < X > 2.4) = P(X < 1.5) + [1 – P(X < 2.4)]
z-score for X = 2.4, z = (X – µ)/? = (2.4 – 1.95)/0.2075 = 2.1682
z-score for X = 1.5, z = (X – µ)/? = (1.5 – 1.95)/0.2075 = - 2.1682
P(z < 2.1682) = (=NORMSDIST(2.1682)) = 0.9849
P(X > 2.4)= 1 - P(z < 2.1682) = 1 – 0.9849 = 0.0151
P(z < -2.1682) = (=NORMSDIST(-2.1682)) = 0.0151
P(1.5 < X > 2.4) = P(X < 1.5) + [1 – P(X < 2.4)] = 0.0151 + 0.0151 = 0.302
Percentage of output greater than 2.4 mm and less than 1.5 mm = 3.02 %
c.
LSL = 1.5 mm
USL = 2.4 mm
Lower Process capability index = Cpkl = (µ – LSL)/3? = (1.95 – 1.5)/(3*0.2075) = 0.7227
Upper Process capability index = Cpku = (USL – µ)/3? = (2.4 – 1.95)/(3*0.2075) = 0.7227
Process capability index = Minimum [(USL – µ)/3?, (µ – LSL)/3?] = Min (Cpku, Cpkl) = Min (0.7227, 0.7227)
Cpk = 0.7227
D.
Target thickness = 1.5 + (2.4 – 1.5)/2 = 1.5 + 0.45 = 1.95 mm
Consider the process is centered at 1.95 mm,
µ = 1.95 mm, ? = 0.2075 mm
Lower Process capability index = Cpkl = (µ – LSL)/3? = (1.95 – 1.5)/(3*0.2075) = 0.7227
Upper Process capability index = Cpku = (USL – µ)/3? = (2.4 – 1.95)/(3*0.2075) = 0.7227
Process capability index = Minimum [(USL – µ)/3?, (µ – LSL)/3?] = Min (Cpku, Cpkl) = Min (0.7227, 0.7227)
Cpk = 0.7227
E.
percentage of output would be expected to be out of tolerance if the process were centered at 1.95 mm:
Probability of washers is expected to be greater than USL = 2.4 mm thickness and less than LSL = 1.5 mm, using standard z-score is obtained as follows:
P(LSL < X > USL) = P(X < 1.5) + [1 – P(X < 2.4)]
z-score for X = 2.4, z = (X – µ)/? = (2.4 – 1.95)/0.2075 = 2.1682
z-score for X = 1.5, z = (X – µ)/? = (1.5 – 1.95)/0.2075 = - 2.1682
P(z < 2.1682) = (=NORMSDIST(2.1682)) = 0.9849
P(X > 2.4)= 1 - P(z < 2.1682) = 1 – 0.9849 = 0.0151
P(z < -2.1682) = (=NORMSDIST(-2.1682)) = 0.0151
P(1.5 < X > 2.4) = P(X < 1.5) + [1 – P(X < 2.4)] = 0.0151 + 0.0151 = 0.302
Percentage of output expected greater than 2.4 mm and less than 1.5 mm = 3.02 %
f.
g.
Observation
Sample Range
1
2
3
4
5
6
7
8
9
10
(Max. – Min)
Sample 1
1.8
1.9
1.9
2.1
2.1
2.4
1.7
1.9
1.8
2.2
0.7
Sample 2
2.2
1.9
1.8
2.1
1.8
1.7
2
1.9
2.2
2
0.5
Sample 3
1.6
2
2.1
2
1.8
1.7
1.9
1.7
2.1
1.6
0.5
Sample 4
1.8
1.6
2.1
2.4
2.2
1.9
2
1.8
2.2
2.1
0.8
Average
0.625
Central line = CL = overall mean = Xbar = 1.95
Average range = Rbar = 0.625
For sample size of n = 10, A2 = 0.31, D3 = 0.22 and D4 = 1.78
UCLXbar = Xbar + (A2)(Rbar) = 1.95 + (0.31)(0.625) = 2.1437
LCLXbar = Xbar - (A2)(Rbar) = 1.95 - (0. 31)(0.625) = 1.7563
UCLXbar = 2.1437 and LCLXbar = 1.7563
UCLR = (D4)(Rbar) = (1.78)(0.625) = 1.1125
LCLXbar = (D3)(Rbar) = (0.22)(0.625) = 0.1375
UCLR = 1.1125 and LCLR = 0.1375
Observation
Sample Range
1
2
3
4
5
6
7
8
9
10
(Max. – Min)
Sample 1
1.8
1.9
1.9
2.1
2.1
2.4
1.7
1.9
1.8
2.2
0.7
Sample 2
2.2
1.9
1.8
2.1
1.8
1.7
2
1.9
2.2
2
0.5
Sample 3
1.6
2
2.1
2
1.8
1.7
1.9
1.7
2.1
1.6
0.5
Sample 4
1.8
1.6
2.1
2.4
2.2
1.9
2
1.8
2.2
2.1
0.8
Average
0.625
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