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) 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.
To solve the given defective rate problem using the geometric distribution, we need to understand and use the formula for the geometric distribution:
P(X=k) = (1-p)^(k-1) * p
Where:
P(X=k) represents the probability that the first success (in this case, a defective transistor) occurs on the kth trial.
p is the probability of success on a single trial.
Now, let's solve the problem step-by-step:
(1) What is the probability that the 10th transistor produced is the first with a defect?
Using the geometric distribution formula, we have:
P(X=10) = (1-0.02)^(10-1) * 0.02
= 0.98^9 * 0.02
≈ 0.0182
Therefore, the probability that the 10th transistor produced is the first with a defect is approximately 0.0182, or 1.82%.
(2) What is the probability that the machine produces no defective transistors in a batch of 100?
The probability of a single transistor being non-defective is (1 - 0.02) = 0.98.
Using the geometric distribution formula, the probability of producing no defective transistors in 100 trials is:
P(X=1) = (0.98)^100
≈ 0.1326
Therefore, the probability that the machine produces no defective transistors in a batch of 100 is approximately 0.1326, or 13.26%.
(3) On average, how many transistors would you expect to be produced until the first with a defect? What is the standard deviation?
The average (expected) number of trials until success in a geometric distribution is given by the formula: E(X) = 1/p.
In this case, p = 0.02, so:
E(X) = 1/0.02
= 50
Therefore, the average number of transistors expected to be produced until the first with a defect is 50.
The standard deviation of a geometric distribution is given by the formula: σ = sqrt((1-p)/p^2).
Using p = 0.02, we have:
σ = sqrt((1-0.02)/0.02^2)
≈ sqrt(0.98/0.0004)
≈ sqrt(2450)
≈ 49.49
Therefore, the standard deviation of the number of transistors produced until the first with a defect is approximately 49.49.
(4) Another machine that also produces transistors has a 5% defective rate. 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 the same formula as above, with p = 0.05, we have:
E(X) = 1/0.05
= 20
Therefore, on average, you would expect to produce 20 transistors before the first with a defect using this machine.
Using the standard deviation formula:
σ = sqrt((1-0.05)/0.05^2)
≈ sqrt(0.95/0.0025)
≈ sqrt(380)
≈ 19.49
Therefore, the standard deviation of the number of transistors produced until the first with a defect is approximately 19.49.
(5) Based on your answers to parts (3) and (4), 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 (defect) reduces the expected number of trials until success (mean). In this case, as the defective rate increases from 2% to 5%, the expected number of transistors produced until the first with a defect decreases from 50 to 20.
Regarding standard deviation, increasing the probability of an event narrows the spread of the distribution, resulting in a smaller standard deviation. In this case, as the defective rate increases, the standard deviation of the number of transistors produced until the first with a defect decreases from approximately 49.49 to 19.49.
To solve the given defective rate problems using the concepts of the geometric distribution, we need to understand and apply the formula for the geometric distribution:
P(X = k) = (1-p)^(k-1) * p
Where:
- P(X = k) is the probability of getting the first success (in this case, the first defective transistor) on the kth trial.
- k is the number of trials until the first success.
- p is the probability of success on each trial (in this case, the defective rate).
Let's solve the problems one by one:
(1) To find the probability that the 10th transistor produced is the first with a defect, we need to find P(X = 10). Here, we have k = 10 and p = 0.02. Substituting these values into the formula, we get:
P(X = 10) = (1 - 0.02)^(10-1) * 0.02
P(X = 10) = (0.98)^9 * 0.02
Using a calculator, you can evaluate this expression to find the probability.
(2) To find the probability that the machine produces no defective transistors in a batch of 100, we need to find P(X = 1). Here, we have k = 1 (since we want the first defective transistor to be on the first trial) and p = 0.02. Substituting these values into the formula, we get:
P(X = 1) = (1 - 0.02)^(1-1) * 0.02
P(X = 1) = 0.02
Note that in this case, the probability of success on each trial is the same as the defective rate.
(3) The average number of transistors you would expect to be produced until the first one with a defect can be found using the formula for the expected value (mean) of a geometric distribution:
E(X) = 1 / p
Substituting the value of p = 0.02, we have:
E(X) = 1 / 0.02
Using a calculator, you can find the expected value (mean).
To find the standard deviation (σ) of the geometric distribution, use the formula:
σ = sqrt((1 - p) / (p^2))
Substituting the value of p = 0.02, you can find the standard deviation (σ) using a calculator.
(4) Similarly, for the machine with a 5% defective rate, you will follow the same steps but use p = 0.05 instead of p = 0.02. Calculate E(X) and σ using the formulas mentioned in part (3) above.
(5) Based on the answers to parts (3) and (4), we can observe the following:
- Increasing the probability of an event (i.e., higher defective rate) decreases the mean (average) of the wait time until success. In other words, if the probability of an event happening is higher, we expect it to happen sooner on average.
- Increasing the probability of an event has no effect on the standard deviation of the wait time until success. The standard deviation only depends on the probability of success (defect in this case) and not on the rate.