A certain kind of sheet metal has an average of 5 defects per $10 \mathrm{ft}^{2}$. Assuming a Poisson distribution, we will want to find the probability that a $13 \mathrm{ft}^{2}$ piece of sheet metal has at least 7 defects. (a) First find the mean number of defects per $13 \mathrm{ft}^{2}$ of this sheet metal. Give your answer as a fraction with no units (for example, if you find the mean to be $\frac{3}{4}\left(\frac{\text { defects }}{13 \mathrm{ft}^{2}}\right)$, then you would type "3/4" in the answer box). (b) Then find the probability that a $13 \mathrm{ft}^{2}$ piece of this sheet metal has at least 7 defects. Round this answer to 4 places after the decimal point, if necessary. $P($ at least 7 defects $)=$ Submit Question

Step 1 ：The problem involves finding probabilities related to the number of defects on a sheet metal. The mean number of defects is given as 5 per 10 square feet. This is a Poisson distribution problem because we are dealing with the number of events (defects) that occur in a fixed interval (a piece of sheet metal).

Step 2 ：First, we need to find the mean number of defects per 13 square feet. Since the average is 5 defects per 10 square feet, we can set up a proportion to find the mean for 13 square feet: \(\frac{5 \text{ defects}}{10 \text{ square feet}} = \frac{x \text{ defects}}{13 \text{ square feet}}\). Solving for x gives us x = 6.5 defects. So, the mean number of defects per 13 square feet is 6.5.

Step 3 ：The Poisson probability formula is given by: \(P(x; μ) = \frac{e^{-μ} * μ^{x}}{x!}\) where: \(P(x; μ)\) is the Poisson probability, x is the actual number of successes, e is approximately equal to 2.71828, μ is the mean number of successes, and x! is the factorial of x.

Step 4 ：For part (b), we need to find the probability of at least 7 defects. This is 1 minus the sum of the probabilities of 0, 1, 2, 3, 4, 5, and 6 defects. We can plug x=0 to 6 and μ=6.5 into the Poisson formula, sum up these probabilities, and subtract the result from 1. The calculated probability is approximately 0.3448.

Step 5 ：Final Answer: \(\boxed{0.3448}\) (Round to four decimal places as needed.)