six sigma control 1. A process for manufacturing steel bolts continuously feeds

Question

six sigma control


1.
A process for manufacturing steel bolts continuously feeds an assembly line downstream. Historically, the thickness of these bolts follow a normal distribution with a mean of 10.0 mm and a standard deviation of 1.6 mm. The process supervisor becomes concerned about the process if the thickness begins to get larger than 10.8 mm or smaller than 9.2 mm. As the quality control person in the company, what would you tell the process supervisor if he/she asks you the following questions:
a.
If you randomly select any bolt from the production line, what is the probability that its thickness will lie between 9.2 mm and 10.8 mm?

b.
If you selected 50 bolts at random, what is the probability that the mean thickness of this sample will be smaller than 9.2 mm?



In one manufacturing process, shafts of less than 1.5-in in diameter are usable in production. If, in a large population of these shafts, this dimension is normally distributed with mean of 1.490-in and standard deviation of 0.005-in, what percent of the shafts would be usable for production?

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)

Now, to work out solution,

Let X = thickness of bolt (for Q1) and Y = shaft diameter (for Q2)

We are given, X ~ N(10.0, 1.62) and Y ~ N(1.490, 0.0052)

Q1 Part (a)

We want P(9.2 < X < 10.8)

[vide (2) under Back-up Theory], P(9.2 < X < 10.8)

= P[{(9.2 – 10.0)/1.6} < Z < {(10.8 – 10.0)/1.6 } = P(- 0.5 < Z < 0.5)

= P(Z < 0.5) - P(Z < - 0.5) = 0.6915 – 0.3085 = 0.3830 ANSWER

Q1 Part (b)

We want probability that the average of sample of size 50 is less than 9.3.

[vide (3) and (4) under Back-up Theory],

P(Xbar < 9.3) = P[Z < {(9.3 – 10.0)/(1.6/50)}] = P(Z < - 3.093) = 0.001 ANSWER

Q2

Percentage of shafts usable for production = 100 x P(Y < 1.5)

[vide (2) under Back-up Theory],

P(Y < 1.5) = P[Z < {(1.5 – 1.490)/0.005}] = P(Z < 2) = 0.9981

So, Percentage of shafts usable production = 99.81% ANSWER

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