A manager of an electronics store takes a random sample of large shipments of circuit boards to determine if he will accept the delivery. he randomly samples 20 boards from a shipment and will reject the delivery if more than 1 circuit board is defective. if the manufacturer claims that only 3% of the boards he ships out are defective, what is the probability the manager will find more than 1 defective board and reject the delivery?
To calculate the probability of the manager finding more than one defective board and rejecting the delivery, we can use the binomial distribution.
The binomial distribution is used when there are two possible outcomes (in this case, defective or non-defective boards) and the probability of success (defective board) remains constant for each trial (sample).
Let's break down the problem and calculate the probability step-by-step:
1. Determine the number of trials (samples):
The manager randomly samples 20 boards from the shipment. So, the number of trials is 20.
2. Determine the probability of success (finding a defective board):
The manufacturer claims that only 3% of the boards are defective. So, the probability of success (p) is 0.03.
3. Determine the number of defective boards (desired outcomes):
In this case, the manager will reject the delivery if more than one board is defective. So, we are looking for the probability of finding 2 or more defective boards.
4. Calculate the probability using the binomial distribution formula:
P(X > 1) = 1 - P(X = 0) - P(X = 1)
P(X = k) = (n C k) * p^k * (1-p)^(n-k)
where:
- n is the number of trials (20 in this case)
- k is the number of successful outcomes (defective boards)
- (n C k) is the combination formula for choosing k elements out of n (also known as "n choose k")
- p is the probability of success on each trial
- (1-p) is the probability of failure on each trial
Let's calculate the probabilities step-by-step:
P(X = 0) = (20 C 0) * (0.03)^0 * (1-0.03)^(20-0)
= 1 * 1 * 0.97^20
≈ 0.551
P(X = 1) = (20 C 1) * (0.03)^1 * (1-0.03)^(20-1)
= 20 * 0.03 * 0.97^19
≈ 0.377
P(X > 1) = 1 - P(X = 0) - P(X = 1)
≈ 1 - 0.551 - 0.377
≈ 0.072
Therefore, the probability that the manager will find more than one defective board and reject the delivery is approximately 0.072, or 7.2%.