Page 118 - Six Sigma for electronics design and manufacturing

Six Sigma and Manufacturing Control Systems
Defects in 20,000 units of production
Total number of defects in a computer system per month
The Poisson distribution implies occurrences of events or defects
within a boundary of time, space, or region. It has no “memory”; that
is, the outcome or defect during one interval is in proportion to its
length, and independent of other intervals. In addition, the probabili-
ty of two or more outcomes or failures in a single time interval is zero.
3.3.4 Examples of using the Poisson distribution 87
Example 3.5
Assuming the number of defects in a part is = 5, What is the expect-
ed number of defects in a part? What is the probability of two defects?
Up to two defects?
Expected number of defects = 5
Probability of two defects = P(x = 2, = 5) = e 5 /2! = 0.0842
–5 2
Probability of up to two defects = P(0, 1, 2) = e (1 + 5 + 25/2) = 0.12
–5
Example 3.6
Assuming that the number of defects in a production line during a
single hour is = 4. What is the probability that six defects will occur
in that hour?
P(x = 6, = 4) = e 4 /6! = 0.1042
6
–4
Example 3.7
Assuming the probability of obtaining a defective product is 0.01,
what is the probability of obtaining at least three defective products
out of a lot of 100, using binomial and Poisson distributions?
For binomial distribution:
0
P(0, 1, 2, 3) = C 100,0 (0.01) (0.99) 100 + C 100,1 (0.01) (0.99) 99
1
2
3
+ C 100,2 (0.01) (0.99) 98 + C 100,3 (0.01) (0.99) 97 = 0.9816
For Poisson distribution:
= np = 1
–1
3
2
–1
1
0
–1
P(0, 1, 2, 3) = e (1 /0!) + e (1 /1!) + e (1 /2!) + e (1 /3!)
–1
1
–1
P(0, 1, 2, 3) = e (1 + 1 + – + 1/6) = 0.9810
2
The result of the Poisson distribution is in good agreement with the
value of the binomial distribution for small p and large n, but much
easier to compute.