The quality-control inspector of a production plan will reject a batch of syringes if two or more defective syringes are found in a random sample of eight syringes taken from the batch. Suppose the batch contains 12% defective syringes.
a. What is the expected number (mean) of defective syringes the inspector will find? What is the standard deviation?
b. What is the probability that the batch will be accepted? In other words, what is the probability that less than three syringes are defective?
The easiest way to do this is to use a binomial probability table. You would need to find P(0) and P(1), add the probabilities together, then subtract that value from 1.
To answer these questions, we can use the concept of the binomial distribution. The binomial distribution is used to model the number of successes (in this case, defective syringes) in a fixed number of trials (in this case, the sample size).
First, let's calculate the expected number (mean) of defective syringes and the standard deviation.
a. Expected number (mean):
The expected number is calculated by multiplying the probability of success (defective syringe) by the number of trials (sample size).
Expected number = Probability of success * Number of trials
Expected number = 0.12 * 8
Expected number = 0.96
Therefore, the expected number of defective syringes the inspector will find is 0.96.
b. Standard deviation:
The standard deviation of a binomial distribution is calculated using the formula:
Standard deviation = √(Number of trials * Probability of success * Probability of failure)
In this case, the probability of success is the probability of finding a defective syringe, which is given as 0.12. The probability of failure is 1 - probability of success.
Standard deviation = √(8 * 0.12 * 0.88)
Standard deviation ≈ 1.02
Therefore, the standard deviation of the number of defective syringes is approximately 1.02.
To calculate the probability that the batch will be accepted (i.e., less than three syringes are defective), we need to calculate the cumulative probability of finding 0, 1, or 2 defective syringes.
Using a binomial distribution table, calculator, or software, we can find the probabilities and add them up:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
P(X = 0) = (8 choose 0) * (0.12)^0 * (1 - 0.12)^(8-0)
P(X = 1) = (8 choose 1) * (0.12)^1 * (1 - 0.12)^(8-1)
P(X = 2) = (8 choose 2) * (0.12)^2 * (1 - 0.12)^(8-2)
Calculating these probabilities using a binomial calculator or software, we find:
P(X ≤ 2) ≈ 0.990
Therefore, the probability that the batch will be accepted (i.e., less than three syringes are defective) is approximately 0.990, or 99.0%.