Solve the following defective rate problem using the concepts learned about the geometric distribution. A machine that produces a special type of transistor (a component of computers) has a 2% defective rate. The production is considered a random process where each transistor is independent of the others.(1) What is the probability that the 10th transistor produced is the first with a defect?(2) What is the probability that the machine produces no defective transistors in a batch of 100?(3) On average, how many transistors would you expect to be produced until the first with a defect? What is the standard deviation?(4) Another machine that also produces transistors has a 5% defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the first with a defect? What is the standard deviation?(5) Based on your answers to parts (c) and (d), how does increasing the probability of an event affect the mean and standard deviation of the wait time until success?

(1) The geometric distribution is the probability distribution of the number of Bernoulli trials needed to get one success. The probability mass function of the geometric distribution is given by:

P(X = k) = (1-p)^(k-1) * p

where p is the probability of success on each trial, and k is the number of trials. In this case, p = 0.02 (the defective rate), and k = 10 (the 10th transistor is the first with a defect). So,

P(X = 10) = (1-0.02)^(10-1) * 0.02 = 0.0176

(2) The probability that the machine produces no defective transistors in a batch of 100 is simply (1 - p)^n, where n is the number of trials. So,

P(no defects in 100) = (1 - 0.02)^100 = 0.1326

(3) The expected value (mean) of a geometric distribution is 1/p, and the standard deviation is sqrt((1 - p) / p^2). So,

E(X) = 1/0.02 = 50 SD(X) = sqrt((1 - 0.02) / 0.02^2) = 49.5

(4) If the defective rate is 5%, then p = 0.05. So,

E(X) = 1/0.05 = 20 SD(X) = sqrt((1 - 0.05) / 0.05^2) = 19.8

(5) Increasing the probability of an event (the defective rate) decreases both the mean and standard deviation of the wait time until success. This makes sense intuitively: if defects are more likely, you would expect to see one sooner, and there is less variability in the wait time.