Suppose that an automobile parts wholesaler claims that .5 percent of the car batteries in a shipment are defective. A random sample of 200 batteries is taken, and four are found to be defective.
(a) Use the Poisson approximation to find the probability that four or more car batteries in a random sample of 200 such batteries would be found to be defective, if we assume that the wholesaler's claim is true. (Round your answer to 4 decimal places.)
Probability
Suppose that an automobile parts wholesaler claims that .5 percent of the car batteries in a shipment are defective. A random sample of 200 batteries is taken, and four are found to be defective
To find the probability that four or more car batteries in a random sample of 200 would be found to be defective, we will use the Poisson approximation.
The Poisson distribution is used to model the number of events that occur in a fixed interval of time or space, given the average rate of occurrence. In this case, we can use the Poisson distribution to approximate the number of defective batteries in a sample of 200, assuming a true defective rate of 0.5%.
To find the probability, we will calculate the cumulative probability of observing four or more defective batteries.
The formula to calculate the probability of k events occurring in a Poisson distribution is:
P(X=k) = (e^(-λ) * λ^k) / k!
Where λ is the average rate of occurrence. In this case, λ can be calculated as:
λ = n * p
Where n is the sample size (200) and p is the proportion or probability of a single event (0.5% or 0.005).
λ = 200 * 0.005 = 1
We will calculate the probability of four or more defective batteries using the cumulative distribution function (CDF) of the Poisson distribution:
P(X >= 4) = 1 - P(X < 4)
To calculate P(X < 4), we can sum the individual probabilities for k = 0, 1, 2, and 3:
P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
Calculating these probabilities using the Poisson formula gives us:
P(X=0) = (e^(-1) * 1^0) / 0! ≈ 0.3679
P(X=1) = (e^(-1) * 1^1) / 1! ≈ 0.3679
P(X=2) = (e^(-1) * 1^2) / 2! ≈ 0.1839
P(X=3) = (e^(-1) * 1^3) / 3! ≈ 0.0613
Now, we can calculate P(X < 4):
P(X < 4) ≈ 0.3679 + 0.3679 + 0.1839 + 0.0613 ≈ 0.9809
Finally, we can calculate P(X >= 4):
P(X >= 4) = 1 - P(X < 4) ≈ 1 - 0.9809 ≈ 0.0191
Therefore, the probability that four or more car batteries in a random sample of 200 would be found to be defective, assuming the wholesaler's claim is true, is approximately 0.0191.