When studying radioactive material, a nuclear engineer found that over 365 days, \( 1,000,000 \) radioactive atoms decayed to 979,277 radioactive atoms, so 20,723 atoms decayed during 365 days a. Find the mean number of radioactive atoms that decayed in a day b. Find the probability that on a given day, 53 radioactive atoms decayed a. The mean number of radioactive atoms that decay per day is 56.775 (Round to three decimal places as needed.) b. The probability that on a given day, 53 radioactive atoms decayed, is. (Round to six decimal places as needed.)

To solve the problem, we will use the Poisson distribution, which is appropriate for modeling the number of events (in this case, radioactive decays) that occur in a fixed interval of time. ### Part a: Mean number of radioactive atoms that decayed in a day We know that over 365 days, 20,723 atoms decayed. To find the mean number of radioactive atoms that decayed in a day, we divide the total number of decayed atoms by the number of days: \[ \text{Mean} = \frac{\text{Total decayed atoms}}{\text{Number of days}} = \frac{20,723}{365} \approx 56.775 \] So, the mean number of radioactive atoms that decay per day is approximately **56.775** (rounded to three decimal places). ### Part b: Probability that on a given day, 53 radioactive atoms decayed In a Poisson distribution, the probability of observing \( k \) events (in this case, decays) in a fixed interval is given by the formula: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] where: - \( \lambda \) is the mean number of events (in this case, 56.775), - \( k \) is the number of events we want to find the probability for (in this case, 53), - \( e \) is the base of the natural logarithm (approximately 2.71828). Now, we can calculate the probability that exactly 53 radioactive atoms decayed in a day: 1. Calculate \( \lambda^k \): \[ \lambda^{53} = (56.775)^{53} \] 2. Calculate \( e^{-\lambda} \): \[ e^{-56.775} \] 3. Calculate \( k! \): \[ 53! = 53 \times 52 \times 51 \times \ldots \times 1 \] 4. Combine these values into the formula: \[ P(X = 53) = \frac{(56.775)^{53} e^{-56.775}}{53!} \] Using a calculator or software to compute this: - \( \lambda = 56.775 \) - \( k = 53 \) - \( e^{-56.775} \approx 1.307 \times 10^{-25} \) - \( 53! \approx 4.789 \times 10^{70} \) Now, substituting these values into the formula: \[ P(X = 53) \approx \frac{(56.775)^{53} \cdot (1.307 \times 10^{-25})}{4.789 \times 10^{70}} \] Calculating this gives: \[ P(X = 53) \approx 0.073267 \] Thus, the probability that on a given day, 53 radioactive atoms decayed is approximately **0.073267** (rounded to six decimal places).