Question 1184334: A certain kind of sheet metal has, on average, 4 defects per 18 square feet.
Assuming a Poisson distribution, find the probability that a 33 square foot metal sheet has at least 5 defects. Round your answer to three decimal places.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
1. **Calculate the average number of defects per square foot:**
The sheet metal has 4 defects per 18 square feet, so the average number of defects per square foot is $\lambda_{sqft} = \frac{4}{18} = \frac{2}{9}$.
2. **Calculate the average number of defects for a 33 square foot sheet:**
For a 33 square foot sheet, the average number of defects is $\lambda_{33} = \lambda_{sqft} \times 33 = \frac{2}{9} \times 33 = \frac{22}{3} \approx 7.333$.
3. **Use the Poisson distribution formula:**
The probability of *k* defects in a given area is given by the Poisson distribution formula:
$P(k) = \frac{e^{-\lambda} \lambda^k}{k!}$
where $\lambda$ is the average number of defects in the given area, and *k* is the number of defects.
4. **Calculate the probability of at least 5 defects:**
We want to find $P(k \ge 5)$. This is equal to $1 - P(k < 5)$, or $1 - [P(0) + P(1) + P(2) + P(3) + P(4)]$.
$P(0) = \frac{e^{-22/3} (22/3)^0}{0!} \approx 0.00065$
$P(1) = \frac{e^{-22/3} (22/3)^1}{1!} \approx 0.00160$
$P(2) = \frac{e^{-22/3} (22/3)^2}{2!} \approx 0.00587$
$P(3) = \frac{e^{-22/3} (22/3)^3}{3!} \approx 0.01438$
$P(4) = \frac{e^{-22/3} (22/3)^4}{4!} \approx 0.02646$
$P(k < 5) = P(0) + P(1) + P(2) + P(3) + P(4) \approx 0.00065 + 0.00160 + 0.00587 + 0.01438 + 0.02646 \approx 0.04896$
$P(k \ge 5) = 1 - P(k < 5) = 1 - 0.04896 \approx 0.95104$
5. **Round to three decimal places:**
The probability that a 33 square foot metal sheet has at least 5 defects is approximately 0.951.
Final Answer: The final answer is $\boxed{0.951}$
|
|
|
