1. Copyright  2003 Dr. John Lipp
3-133
A lot of parts contains 1000 items. 1% of the items are nonconforming, which is the same as 1 in 100, or 10 in 1000.
The question is what fraction of the parts need to be tested to have a 90% chance of at least 1 nonconforming part?
EMIS7300 Fall 2005
Copyright  2003 Dr. John Lipp

2. Copyright  2003 Dr. John Lipp
3-133 (cont.)
The problem also suggests you assume a binomial approximation can be used instead of a hypergeometric
Binomial approximation assumes that parts are selected from the population with replacement.
Hypergeometric is formulated on selecting the parts without replacement.
The two outcomes are conforming and nonconforming. The probability of nonconforming is 1%, or p = Thus the binomial PDF is
EMIS7300 Fall 2005
Copyright  2003 Dr. John Lipp

3. Copyright  2003 Dr. John Lipp
3-133 (cont.)
To find the probability of at least one nonconforming part, the PDF is summed from x = 1 to x = 1000
In this case, it is easier to solve the complimentary probability problem: how many parts have to be sampled to have less than a 10% chance of no defective parts
simplifying and solving
EMIS7300 Fall 2005
Copyright  2003 Dr. John Lipp

4. Copyright  2003 Dr. John Lipp
3-133 (cont.)
Compare this to the exact solution, that is, find the value of n where the probability of all conforming parts is less than 10% when replacement no longer occurs
The value at n = 205 is
This is the same value obtained for K = 990, N = 1000, x = n for the hypergeometric distribution
EMIS7300 Fall 2005
Copyright  2003 Dr. John Lipp

5. Copyright  2003 Dr. John Lipp
3-135
The number of surface flaws in each panel is Poisson distributed with a mean of 0.1 flaws per panel.
What is the probability that less than 5 panels have flaws when 100 panels are checked?
This is actually two problems in one!
The flawed panel distribution is a binomial distribution (a panel is either flawed, or unflawed).
The value p (probability of flawed panel) for the binomial distribution is found from examining the distribution of flaws.
EMIS7300 Fall 2005
Copyright  2003 Dr. John Lipp

6. Copyright  2003 Dr. John Lipp
3-135 (cont.)
The number of flaws per panel is Poisson distributed. The mean flaws per panel is 0.1, that is,  = 0.1.
Then the probability of a flawed panel is
The distribution of flawed panels is binomial with p = 1 - e-0.1.
EMIS7300 Fall 2005
Copyright  2003 Dr. John Lipp

7. Copyright  2003 Dr. John Lipp
3-135 (cont.)
EMIS7300 Fall 2005
Copyright  2003 Dr. John Lipp