The probability of a manufacturing defect in an aluminum beverage can is .00003. If 100,100 cans are produced,

(a) Find the approximate probability of at least two defective cans. (Round your answer to 4 decimal places.)

(b) Find the approximate probability of at least three defective cans. (Round your answer to 4 decimal places.)

(c) Why not use the binomial?
a) n is too large for the binomial application
b) the probability is too small for the poisson application
c) n is too large for the practical application

(d) Is the Poisson approximation justified?
a) yes
b) no

To find the approximate probabilities of at least two defective cans and at least three defective cans, we can use the Poisson distribution as an approximation.

(a) To find the probability of at least two defective cans, we need to calculate the complementary probability of having zero or one defective can and subtract it from 1. The Poisson distribution formula is used to calculate this probability, which is given by:

P(X ≥ k) = 1 - Σ (e^(-λ) * (λ^k) / k!)

Where X is the random variable representing the number of defects, λ is the mean number of defects, and k is the number of defective cans we want to calculate the probability for.

In this case, the mean number of defects (λ) can be calculated by multiplying the probability of a manufacturing defect (.00003) by the total number of cans produced (100,100):

λ = .00003 * 100,100 = 3.003

Using this value for λ, we can calculate the probability of at least two defective cans:

P(X ≥ 2) = 1 - (e^(-3.003) * (3.003^0) / 0!) - (e^(-3.003) * (3.003^1) / 1!)

Calculating this expression will give us the approximate probability of at least two defective cans.

(b) Similarly, to find the probability of at least three defective cans, we calculate:

P(X ≥ 3) = 1 - (e^(-3.003) * (3.003^0) / 0!) - (e^(-3.003) * (3.003^1) / 1!) - (e^(-3.003) * (3.003^2) / 2!)

Calculating this expression will give us the approximate probability of at least three defective cans.

(c) The reason we use the Poisson distribution instead of the binomial distribution in this case is that the binomial distribution assumes a fixed number of trials and a constant probability of success. However, the number of cans produced (100,100) is considered large for the binomial distribution to be accurate. The practical application of the binomial distribution is typically limited to smaller sample sizes.

(d) To determine if the Poisson approximation is justified, we can check whether the conditions for using the Poisson distribution are met. These conditions are:
1. The events occur independently. In this case, we assume that the probability of a manufacturing defect in one can does not affect the probability in another can.
2. The average rate of occurrence (λ) is constant. We have calculated λ as 3.003, assuming a constant rate of defects.
3. The probability of more than one occurrence in a short interval is negligible. This condition is subjective, and we need to assess whether the probability of more than three defective cans is sufficiently small.

If these conditions are met, then the Poisson approximation is justified.