Question

When ˚uit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. If 25 such discs are tested, what is the probability that:

a) Exactly 20 are not defective.
b) At least 20 are not defective.
c) Find the average number of defectives.
d) Exactly 20 are defective.

Answer 1

To solve this problem, we can use the binomial probability formula. For part a), the probability that exactly 20 discs are not defective is given by P(X = 20) = (0.95)^5 * (0.05)^20 * C(25, 20), where C(25, 20) represents the number of ways to choose 20 out of 25 discs.

Step-by-step explanation:

To solve this problem, we can use the binomial probability formula. For part a), the probability that exactly 20 discs are not defective is given by P(X = 20) = (0.95)^5 * (0.05)^20 * C(25, 20), where C(25, 20) represents the number of ways to choose 20 out of 25 discs. For part b), the probability that at least 20 discs are not defective is given by P(X >= 20) = P(X = 20) + P(X = 21) + ... + P(X = 25). For part c), the average number of defectives can be found using the formula E(X) = np, where n is the number of trials and p is the probability of success. For part d), the probability that exactly 20 discs are defective is given by P(X = 20) = (0.05)^20 * (0.95)^5 * C(25, 20).