12) Suppose that a manufactured product has 2 defects per unit of product inspected.
Using Poisson distribution, calculate the probabilities of finding a product without any
defect, 3 defects and 4 defects.
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Given:
A manufactured product has 2 defects per unit of product inspected
To find:
The probabilities of finding a product with
- out any defect
- 3 defects
- 4 defects
Answer:
Possion distribution is given by:
P(x; μ) = (e^-μ) (μ^x) / x!
where,
- x = number of successes from an experiment
- μ = mean of the distribution
- e = 2.71828
Here, the mean of number of defects, μ = 2
- Probability of getting 0 defects =
- P(0; 2) = (e^-2) (2^0) / 0!
- P(0; 2) = 0.135
The probabilities of finding a product without any defect = 0.135
- Probability of getting 3 defects =
- P(3; 2) = (e^-3) (2^3) / 3!
- P(3; 2) = 0.066
The probabilities of finding a product with 3 defects = 0.066
- Probability of getting 4 defects =
- P(4; 2) = (e^-4) (2^4) / 4!
- P(4; 2) = 0.0122
The probabilities of finding a product with 4 defects = 0.0122
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