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.)
Any help would be greatly apriciated. Thanks
To find the probability using the Poisson approximation, we need to calculate the average number of defective car batteries in a sample of 200 batteries.
Given that the wholesaler claims that 0.5 percent of the car batteries are defective, we can calculate the average using the formula:
Average = (Percentage of Defective Batteries) * (Total Number of Batteries)
Average = 0.5/100 * 200 = 1
Now, we can use the Poisson distribution to find the probability of observing four or more defective batteries in a sample of 200 batteries.
P(X ≥ 4) = 1 - P(X < 4)
Here, X represents the number of defective batteries, and we can use the Poisson distribution formula to calculate the probability for each value of X less than 4.
P(X < 4) = Σ[(e^(-λ) * λ^x) / x!]
where λ is the average number of defective batteries (1 in this case), and x ranges from 0 to 3.
Let's calculate the probabilities:
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 batteries would be found to be defective is approximately 0.0191.
To solve this problem, we will use the Poisson approximation. The Poisson distribution is commonly used to model the number of events occurring within a fixed interval of time or space when these events happen independently and at a constant average rate.
In this case, we are given that the wholesaler claims that 0.5 percent of the car batteries are defective. This implies that the average or expected number of defective batteries in a sample of 200 would be:
0.5% of 200 = 0.005 * 200 = 1
The parameter λ (lambda) of the Poisson distribution is equal to the average number of events in the given interval. So, in this case, λ = 1.
Now, we need to find the probability that four or more batteries in a random sample of 200 would be found to be defective. We can use the cumulative distribution function (CDF) of the Poisson distribution to calculate this probability.
P(X ≥ 4) = 1 - P(X < 4)
To calculate P(X < 4), we can sum the individual probabilities for X = 0, 1, 2, and 3.
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
The probability mass function (PMF) of the Poisson distribution is given by:
P(X = k) = (e^(-λ) * λ^k) / k!
Using this formula, we can calculate the probabilities for each value of X and sum them up to find P(X < 4).
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
P(X < 4) = 0.3679 + 0.3679 + 0.1839 + 0.0613 = 0.9809
Finally, we can calculate P(X ≥ 4) by subtracting P(X < 4) from 1.
P(X ≥ 4) = 1 - 0.9809 = 0.0191
Therefore, the probability that four or more car batteries would be found to be defective in a random sample of 200 batteries is approximately 0.0191 or 1.91%.