Probability of Defective Bulbs in a Sample Using Poisson Approximation
Probability of Defective Bulbs in a Sample Using Poisson Approximation
In this article, we will explore the probability of finding fewer than 2 defective bulbs or more than 3 defective bulbs in a sample of 200 electric bulbs produced by a certain company, where the probability of a bulb being defective is 2%. This analysis will be conducted using the Poisson approximation to the binomial distribution, as the sample size is large (n200) and the probability of defectiveness is small (p0.02).
Step 1: Calculate the Expected Number of Defective Bulbs
The expected number of defective bulbs, denoted as (lambda), can be calculated using the formula:
(lambda n times p 200 times 0.02 4)
Step 2: Use the Poisson Distribution
The number of defective bulbs in our sample can be modeled using a Poisson distribution with parameter (lambda 4). The probability mass function (pmf) of a Poisson distribution is given by:
(P(X k) frac{e^{-lambda} lambda^k}{k!})
where (k) is the number of defective bulbs.
Step 3: Calculate Probabilities for Fewer than 2 and More than 3 Defective Bulbs
Probability of Fewer than 2 Defective Bulbs
We need to calculate the probabilities for 0 and 1 defective bulbs and then sum these probabilities. The calculations are as follows:
(P(X 0) frac{e^{-4} 4^0}{0!} e^{-4} approx 0.0183)
(P(X 1) frac{e^{-4} 4^1}{1!} e^{-4} cdot 4 approx 0.0183 cdot 4 approx 0.0732)
(P(X
Probability of More than 3 Defective Bulb:
To find (P(X > 3)), we can use the complement rule:
(P(X > 3) 1 - P(X leq 3))
The probability (P(X leq 3)) is the sum of the probabilities for 0, 1, 2, and 3 defective bulbs:
(P(X leq 3) P(X 0) P(X 1) P(X 2) P(X 3))
(P(X 2) frac{e^{-4} 4^2}{2!} frac{e^{-4} cdot 16}{2} approx 0.1476)
(P(X 3) frac{e^{-4} 4^3}{3!} frac{e^{-4} cdot 64}{6} approx 0.1953)
(P(X leq 3) approx 0.0183 0.0732 0.1476 0.1953 approx 0.4344)
(P(X > 3) 1 - 0.4344 approx 0.5656)
Step 4: Final Probability
We are interested in the probability that in a sample of 200 bulbs, fewer than 2 bulbs are defective or more than 3 bulbs are defective:
(P(X 3) approx 0.0915 0.5656 0.6571)
Conclusion
The probability that in a sample of 200 bulbs fewer than 2 bulbs and more than 3 bulbs are defective is approximately 0.6571 or 65.71%.