A particular part is specified to have a diameter of 6.5 inches with a tolerance

Question

A particular part is specified to have a diameter of 6.5 inches with a tolerance of + / - .05 inches. The process of making that part is operating in statistical control with a process mean of 6.48 inches and a standard deviation of 0.02 inches. For the purposes of this analysis, assume that the P (Z< 3.09) is approximately 0.

A. Determine the capability of the process producing this part.

B. Use a normal probability distribution table to determine what percentage of parts produced will not be within tolerance due to random variation if the process stays in control.

** Please show all, and complete work on how you arrived at the solution. Including any formulas used. Thank-you. ***

Explanation / Answer

Solution

Let X = diameter of the part.

Then, we assume X ~ N(µ, 2)

Back-up Theory

1) Process capability = 6, where is estimated by the sample standard deviation, s.

Process is considered capable (of meeting specification) if 6 < T, where T = tolerance band.

and hence the process is considered not capable if 6 > T.

2) Percentage of parts produced that are not within tolerance due to random variation

= 100 x {P(X < LSL) + P(X > USL)}, where LSL and USL are lower and upper specification limits respectively

3) 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

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

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

Standard Normal Tables or can be found using Excel Function

Now, to work out the solution,   

Part (A)

Given process mean, Xbar = 6.48 and standard deviation, s = 0.02,

Process capability = 6 x 0.02 = 0.12

Tolerance band, T = 2 x 0.05 = 0.1.

Since 0.12 > 0.1, process is not capable of meeting the specification ANSWER

Part (B)

Given the part is specified to have a diameter of 6.5 inches with a tolerance of + / - .05 inches, LSL = 6.5 – 0.05 = 6.45 and USL = 6.5 + 0.05 = 6.55 and hence

Percentage of parts produced that are not within tolerance due to random variation

= 100 x {P(X < 6.45) + P(X > 6.55)}

= 100 x {P[Z < {(6.45 – 6.48)/0.02}] + P[Z > {(6.55 – 6.48)/0.02}]}

= 100 x {P[Z < - 1.5] + P[Z > 3.5]}

= 100 x {(0.0668) + 0} [from Standard Normal Tables]

= 6.68% ANSWER

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