Poisson Distribution as an Approximation to Binomial Distribution
The Poisson distribution simplifies the Binomial distribution. The Binomial distribution deals with the number of successes in a set of trials. However, when there are many trials with a small chance of success, it becomes impractical to calculate using the Binomial probabilities. In such situations, the Poisson distribution works as a shortcut. It focuses on the frequency of events in a specific span of time or space, especially for rare occurrences.
For instance, imagine a bustling city intersection where accidents are infrequent. Computing the probability of exactly 2 accidents in a day with the Binomial distribution would be complex due to the large number of cars and low accident probability per car. In this case, employing the Poisson distribution as an estimate simplifies the process. By adjusting the parameters to match the average (λ = 0.1), we can readily determine the probability of exactly 2 accidents in a day using the Poisson distribution formula.
Example:
In a manufacturing process, there is a 2% defective rate for a certain product. If a sample of 200 products is selected, what is the probability that exactly 5 of them are defective, using the Poisson distribution as an approximation to the Binomial distribution?
Solution:
According to the given information, the probability of success (defective product) is p = 0.02 (2% defective rate), and the number of trials is n = 200. Using the formula for the average rate of occurrence in a Poisson distribution = np, we can calculate .
= np = 200 × 0.02 = 4
Now, we can use the Poisson distribution formula to find the probability of exactly 5 defective products x = 5,
Calculate the individual components:
e-4 ≈ 0.0183
45 = 1024
5! = 120
Substitute these values into the formula:
P(X=5) = 0.15616
∴ the probability of exactly 5 defective products in a sample of 200, using the Poisson distribution as an approximation to the Binomial distribution, is approximately 0.15.
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