Two different sizes of a certain type of fastener are being manufactured by a co

Question

Two different sizes of a certain type of fastener are being manufactured by a cold heading process. the nominal outside diameters of the fastener heads, with associated specifications, are Fastener A: 0.600 plusminus 0.020 centimeter, Fastener B: 0.800 plusminus 0.020 centimeter. SPC studies show the processes to be in good statistical control. the individual measurements follow the normal distribution. After collecting 50 samples of size n = 4: For each fastener dimension of interest, find the percent conforming. If the process mean for fastener A could be brought to the nominal, what would be the percent conforming? If the variation in the process for fastener B could be reduced by a factor of one-third, what would be the percent conforming?

Explanation / Answer

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or using Excel Function……………………………………..(5)

Estimate of = Rbar/d2, where d2 is a constant which can be obtained from Standard Control Chart Constants Tables…………………………………………………..(6)

If L and U are respectively the lower and upper specification limits, proportion of items conforming to specification is P(L X U)……………………………………..(7)

Now, to work out solution,

Let X = outside dimension of the fastener. Then, X is normally distributed with mean µ and variance 2. Given,

Number of samples, k = 50

Sample size. n = 4

From Tables, d2 for n = 4, is 2.059.

Lower Specification Limit, L = 0.580 for A and 0.780 for B

Upper Specification Limit,U = 0.620 for A and 0.820 for B

Part (a)

Average = sum(Xbar)k = 30.310/50 = 0.6062 for A

= 40.011/50 = 0.80022 for B

Average Range, Rbar = sum(R)/50 = 0.654/50 = 0.01308

= 1.034/50 = 0.02068

Estimate of µ = 0.6062 for A and 0.80022 for B

Estimate of [vide (7) under Back-up Theory], = 0.01308/2.059 = 0.00635 for A

= 0.02068/2.059 = 0.01004 for B

[vide (7) and (2) under Back-up Theory], proportion conforming to specification for A is:

P(0.580 X 0.620) = P(- 4.126 Z 2.173) = P(Z 2.173) - P(Z - 4.126)

= 0.9851 – 0 = 0.9851

Percent conforming to specification for A is: 98.5%

Proportion conforming to specification for B is:

P(0.780 X 0.820) = P(- 2.014 Z 1.970) = P(Z 1.970) - P(Z - 2.014)

= 0.9756 – 0,0220 = 0.9536

Percent conforming to specification for B is: 95.4%

Part (b)

When the process mean for A is shifted to nominal => µ = 0.600. Then, proportion conforming to specification is:

P(0.580 X 0.620) = P(- 3.150 Z 3.150) = P(Z 3.150) - P(Z - 3.150)

= 0.999183 – 0.000817 = 0.998366

Percent conforming to specification is 99.8%

Part (c)

Variation in process for B is reduced by a factor of one third => = 0.01004 x (2/3) = 0.00669. Then, proportion conforming to specification is:

P(0.780 X 0.820) = P(- 3.022 Z 2.957) = P(Z 2.957) - P(Z - 3.002)

= 0.9985 – 0.00135 = 0.99715

Percent conforming to specification is 99.7%

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