(1) To find the probability that the 10th transistor produced is the first with a defect, we can use the geometric distribution formula:
P(X = k) = (1 - p)^(k - 1) * p
where X is the number of trials until the first success (defective transistor), p is the probability of success (defective rate), and k is the specified trial number.
In this case, p = 0.02 and k = 10. Plugging these values into the formula, we get:
P(X = 10) = (1 - 0.02)^(10 - 1) * 0.02
= (0.98)^9 * 0.02
≈ 0.0226
So, the probability that the 10th transistor produced is the first with a defect is approximately 0.0226.
(2) To find the probability that the machine produces no defective transistors in a batch of 100, we can again use the geometric distribution formula:
P(X = k) = (1 - p)^(k - 1) * p
where X is the number of trials until the first success (defective transistor), p is the probability of success (defective rate), and k is the specified trial number.
In this case, p = 0.02 and k = 100. Plugging these values into the formula, we get:
P(X = 100) = (1 - 0.02)^(100 - 1) * 0.02
= (0.98)^99 * 0.02
≈ 0.366
So, the probability that the machine produces no defective transistors in a batch of 100 is approximately 0.366.
(3) The average number of transistors expected to be produced until the first with a defect can be found using the expected value formula for the geometric distribution:
E(X) = 1 / p
where X is the number of trials until the first success (defective transistor), and p is the probability of success (defective rate).
In this case, p = 0.02. Plugging this value into the formula, we get:
E(X) = 1 / 0.02
= 50
So, on average, we would expect 50 transistors to be produced until the first with a defect.
The standard deviation of the geometric distribution can be found using the formula:
σ(X) = sqrt((1 - p) / p^2)
In this case, p = 0.02. Plugging this value into the formula, we get:
σ(X) = sqrt((1 - 0.02) / 0.02^2)
≈ 49.749
So, the standard deviation of the number of transistors produced until the first with a defect is approximately 49.749.
(4) To find the average number of transistors expected to be produced with the machine with a 5% defective rate before the first with a defect, we can use the same expected value formula for the geometric distribution:
E(X) = 1 / p
where X is the number of trials until the first success (defective transistor), and p is the probability of success (defective rate).
In this case, p = 0.05. Plugging this value into the formula, we get:
E(X) = 1 / 0.05
= 20
So, on average, we would expect 20 transistors to be produced before the first with a defect with this machine.
The standard deviation of the geometric distribution can be found using the same formula as before:
σ(X) = sqrt((1 - p) / p^2)
In this case, p = 0.05. Plugging this value into the formula, we get:
σ(X) = sqrt((1 - 0.05) / 0.05^2)
≈ 19.494
So, the standard deviation of the number of transistors produced until the first with a defect with this machine is approximately 19.494.
(5) Increasing the probability of an event (defective rate) generally decreases the mean (expected value) and standard deviation of the wait time until success. This is because as the probability of success increases, it takes less time on average to achieve success. The standard deviation also decreases as the distribution becomes more concentrated around the mean.
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?
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