3.22 Defective rate. 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.

(a) What is the probability that the 10th transistor produced is the first with a defect?

(b) What is the probability that the machine produces no defective transistors in a batch of 100?

(c) On average, how many transistors would you expect to be produced before the first with a

defect? What is the standard deviation?

(d) 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?

(e) 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?

(a) The probability that the 10th transistor produced is the first with a defect is 0.02.

(b) The probability that the machine produces no defective transistors in a batch of 100 is 0.98^100, or 0.36.

(c) On average, you would expect to produce 50 transistors before the first with a defect. The standard deviation is 7.07.

(d) On average, you would expect to produce 20 transistors before the first with a defect. The standard deviation is 4.47.

(e) Increasing the probability of an event decreases the mean and increases the standard deviation of the wait time until success.

To answer these questions, we need to use probability concepts. Let's go through each question step by step:

(a) What is the probability that the 10th transistor produced is the first with a defect?

To calculate this probability, we can use the concept of independent events. The probability of a transistor being defective is 2% or 0.02. Since each transistor is independent of the others, the probability of the first 9 transistors being non-defective is (1 - 0.02) raised to the power of 9. The probability of the 10th transistor being defective is just 0.02.

So, the probability that the 10th transistor produced is the first with a defect is (1 - 0.02)^9 * 0.02.

(b) What is the probability that the machine produces no defective transistors in a batch of 100?

Similar to the previous question, the probability of producing a defective transistor is 0.02. The probability that a non-defective transistor is produced is (1 - 0.02). Since each transistor is independent, the probability of all 100 transistors being non-defective is (1 - 0.02) raised to the power of 100.

So, the probability that the machine produces no defective transistors in a batch of 100 is (1 - 0.02)^100.

(c) On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?

This question deals with the concept of expected value or mean. The expected value can be found by calculating the sum of the products of each possible outcome and its corresponding probability.

The probability of getting a defective transistor is 0.02, and the probability of a non-defective transistor is 0.98. The expected value for the number of transistors produced before the first defective one is calculated as follows:

Expected value = 1 / probability of defect = 1 / 0.02 = 50 transistors

The standard deviation can be found using the formula: sqrt(1 - p) / p^2, where p is the probability of success (in this case, the defect rate). Plugging in the values, we get:

Standard deviation = sqrt(1 - 0.02) / 0.02^2 ≈ 7.07 transistors

(d) 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?

Using similar steps, but now with a defect rate of 5% or 0.05:

Expected value = 1 / probability of defect = 1 / 0.05 = 20 transistors

Standard deviation = sqrt(1 - 0.05) / 0.05^2 ≈ 4.47 transistors

(e) 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?

Increasing the probability of an event (in this case, the defect rate) decreases the average number of trials needed (mean) to achieve that event. In other words, as the probability of an event increases, the average wait time until success decreases.

In terms of standard deviation, increasing the probability of an event reduces the variability or spread of the wait time until success. Therefore, increasing the probability of an event decreases the standard deviation.

Remember that these calculations assume independence between each transistor produced.

(a) To find the probability that the 10th transistor produced is the first with a defect, we can use the binomial probability formula. Let's denote a defective transistor as "D" and a non-defective transistor as "N."

P(10th transistor is the first with a defect) = P(NNNNNNNNND) = (0.98)^9 * 0.02

P(10th transistor is the first with a defect) ≈ 0.2839

So, the probability that the 10th transistor produced is the first with a defect is approximately 0.2839.

(b) To find the probability that the machine produces no defective transistors in a batch of 100, we can again use the binomial probability formula.

P(no defective transistors in a batch of 100) = P(NNN...N) = (0.98)^100

P(no defective transistors in a batch of 100) ≈ 0.1326

So, the probability that the machine produces no defective transistors in a batch of 100 is approximately 0.1326.

(c) To find the average number of transistors expected to be produced before the first with a defect, we can use the concept of geometric distribution.

The average number, or the expected value, for a geometric distribution is given by the formula:

mean = 1 / p

where p is the probability of success (in this case, the probability of a defective transistor).

mean = 1 / 0.02 = 50

So, on average, you would expect 50 transistors to be produced before the first with a defect.

The standard deviation for a geometric distribution is given by:

standard deviation = sqrt((1 - p) / p^2)

standard deviation = sqrt((1 - 0.02) / 0.02^2) = sqrt(0.98 / 0.0004) ≈ 49.7487

So, the standard deviation is approximately 49.7487.

(d) For the second machine with a 5% defective rate, we can use the same formulas as in part (c) to calculate the expected number of transistors and standard deviation.

mean = 1 / 0.05 = 20

So, on average, you would expect 20 transistors to be produced before the first with a defect.

standard deviation = sqrt((1 - p) / p^2)

standard deviation = sqrt((1 - 0.05) / 0.05^2) = sqrt(0.95 / 0.0025) ≈ 19.4936

So, the standard deviation is approximately 19.4936.

(e) Increasing the probability of an event decreases both the mean and standard deviation of the wait time until success. When the probability of an event increases, it is expected to occur sooner, thereby reducing the average number of trials needed to achieve success (mean), as well as reducing the variability around the average (standard deviation).