To solve this problem, we can use the Poisson distribution formula, which is given by:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

Where:

• $\lambda$ is the mean number of occurrences,
• $k$ is the actual number of occurrences,
• $e$ is the base of the natural logarithm (approximately 2.71828).

Given Data:

• Mean number of flaws per panel $\lambda = 0.02$
• Number of panels $n = 50$

First, we calculate the expected number of flaws over 50 panels:

$\lambda_{50} = \lambda \times n = 0.02 \times 50 = 1$

So, $\lambda_{50} = 1$.

We need to find the probability of having 0 flaws (i.e., $k = 0$) in 50 panels:

$P(X = 0) = \frac{e^{-1} \times 1^0}{0!} = \frac{e^{-1}}{1} = e^{-1} \approx 0.3679$

Therefore, the probability that there are no flaws in 50 panels is approximately 0.3679.

However, the given options seem to refer to a different interpretation, possibly involving cumulative probabilities or rounding. The closest choice is not provided among the answer options, which suggests a reconsideration of the expected value might be intended by the question, but based on the clear instructions, the probability calculated is:

0.3679.

Would you like more details or have further questions?

Relative Questions:

1. What is the Poisson distribution used for in statistics?
2. How would you calculate the probability of having exactly one flaw in 50 panels?
3. What is the expected number of flaws if 100 panels are inspected?
4. How does increasing the mean $\lambda$ affect the Poisson distribution?
5. What is the significance of $e$ in the Poisson distribution formula?

Tip: Always ensure the units match when applying Poisson distribution to avoid errors in your mean calculations.