Thank you for visiting A manufacturing company employs two inspecting devices to sample a fraction of their output for quality control purposes The first inspection monitor can accurately detect. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
To solve the problem of finding the expected value and variance for the number of defective items identified by the second inspection device, given that two defective items have been identified by the first device, we need to follow these steps:
1. Understand the Problem:
- We have two independent inspection devices.
- Device 1 identifies defective items with a probability of 99.3% (or 0.993).
- Device 2 identifies defective items with a probability of 99.8% (or 0.998).
- There are 4 defective items that need to be inspected.
2. Determine the Expected Value [tex]\(E(Y \mid X=2)\)[/tex]:
- We need to find the expected number of defective items identified by the second device, given that the first device identifies 2 defective items.
- Since device 2 is independent, the expected number detected by device 2 remains the same regardless of the outcome of device 1.
Given that device 2 has a probability of 0.998 of correctly identifying a defective item out of the 4 defective items:
[tex]\[
E(Y \mid X=2) = 4 \times 0.998 = 3.992
\][/tex]
3. Determine the Variance [tex]\(V(Y \mid X=2)\)[/tex]:
- The variance of identifying defective items by device 2, given the independence of the devices, is calculated using the formula for variance of a binomial distribution:
Variance [tex]\(V(Y) = n \times p \times (1 - p)\)[/tex], where [tex]\(n\)[/tex] is the number of defective items, and [tex]\(p\)[/tex] is the probability of correctly identifying a defective item.
[tex]\[
V(Y \mid X=2) = 4 \times 0.998 \times (1 - 0.998) = 0.008
\][/tex]
In conclusion, when considering the second inspection device, given that 2 defective items are detected by the first device, we find:
- The expected number of items detected by the second device, [tex]\(E(Y \mid X=2)\)[/tex], is 3.992.
- The variance, [tex]\(V(Y \mid X=2)\)[/tex], is 0.008.
1. Understand the Problem:
- We have two independent inspection devices.
- Device 1 identifies defective items with a probability of 99.3% (or 0.993).
- Device 2 identifies defective items with a probability of 99.8% (or 0.998).
- There are 4 defective items that need to be inspected.
2. Determine the Expected Value [tex]\(E(Y \mid X=2)\)[/tex]:
- We need to find the expected number of defective items identified by the second device, given that the first device identifies 2 defective items.
- Since device 2 is independent, the expected number detected by device 2 remains the same regardless of the outcome of device 1.
Given that device 2 has a probability of 0.998 of correctly identifying a defective item out of the 4 defective items:
[tex]\[
E(Y \mid X=2) = 4 \times 0.998 = 3.992
\][/tex]
3. Determine the Variance [tex]\(V(Y \mid X=2)\)[/tex]:
- The variance of identifying defective items by device 2, given the independence of the devices, is calculated using the formula for variance of a binomial distribution:
Variance [tex]\(V(Y) = n \times p \times (1 - p)\)[/tex], where [tex]\(n\)[/tex] is the number of defective items, and [tex]\(p\)[/tex] is the probability of correctly identifying a defective item.
[tex]\[
V(Y \mid X=2) = 4 \times 0.998 \times (1 - 0.998) = 0.008
\][/tex]
In conclusion, when considering the second inspection device, given that 2 defective items are detected by the first device, we find:
- The expected number of items detected by the second device, [tex]\(E(Y \mid X=2)\)[/tex], is 3.992.
- The variance, [tex]\(V(Y \mid X=2)\)[/tex], is 0.008.
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