Final answer:
The student is asked to calculate probabilities using the binomial distribution for a defective rate of 5% in a sample of 20 circuit boards. The calculations would include the probability of finding no more than 2 or 5 defective boards, the probability of finding between 1 to 4 defective boards, and the probability of finding no defective boards at all. Additionally, the expected value and standard deviation for the number of defective boards in the sample are to be calculated.
Step-by-step explanation:
The student is dealing with a binomial probability distribution, where X represents the number of defective circuit boards in a sample of 20. Given that the long-run percentage of defectives is 5%, and using the binomial distribution formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
where:
- C(n, k) is the combination of n items taken k at a time,
- p is the probability of a single success,
- n is the sample size,
- k is the number of successes (defective boards).
We can calculate the probabilities as follows:
- (a) P(X ≤ 2): We sum the probabilities P(X = 0), P(X = 1), and P(X = 2).
- (b) P(X ≤ 5): Sum the probabilities from P(X = 0) to P(X = 5).
- (c) P(1 ≤ X ≤ 4): Sum the probabilities from P(X = 1) to P(X = 4).
- (d) Probability that none of the boards are defective: This is simply P(X = 0).
For the expected value and standard deviation calculations:
E(X) = np = 20 * 0.05 = 1 board
Standard Deviation (σ) = √(np(1-p)) = √(20 * 0.05 * 0.95) = √(0.95) boards