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 (Round to three decimal places as needed.)

To solve the problem, we will break it down into two parts: finding the mean number of radioactive atoms that decayed in a day and then finding the probability that 53 radioactive atoms decayed on a given day. ### Part a: Mean Number of Radioactive Atoms Decayed Per Day To find the mean number of radioactive atoms that decayed per day, we can use the total number of atoms that decayed over the 365 days and divide it by the number of days. Given: - Total atoms decayed = 20,723 - Total days = 365 The mean number of atoms decayed per day is calculated as follows: \[ \text{Mean} = \frac{\text{Total atoms decayed}}{\text{Total days}} = \frac{20,723}{365} \] Calculating this: \[ \text{Mean} \approx 56.75479452 \] Rounding to three decimal places: \[ \text{Mean} \approx 56.755 \] ### Part b: Probability of 53 Radioactive Atoms Decaying in a Day Assuming that the number of decayed atoms follows a Poisson distribution (which is common for counting events in a fixed interval), we can use the mean we calculated in part a as the parameter \(\lambda\) of the Poisson distribution. The probability mass function of a Poisson distribution is given by: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] Where: - \( \lambda \) is the mean (56.755), - \( k \) is the number of events (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: \[ P(X = 53) = \frac{56.755^{53} e^{-56.755}}{53!} \] Calculating \( e^{-56.755} \): \[ e^{-56.755} \approx 1.307 \times 10^{-25} \] Calculating \( 56.755^{53} \) and \( 53! \): Using a calculator or software for these calculations: 1. \( 56.755^{53} \approx 1.052 \times 10^{91} \) 2. \( 53! \approx 2.425 \times 10^{69} \) Now substituting these values into the probability formula: \[ P(X = 53) \approx \frac{1.052 \times 10^{91} \cdot 1.307 \times 10^{-25}}{2.425 \times 10^{69}} \] Calculating the numerator: \[ 1.052 \times 10^{91} \cdot 1.307 \times 10^{-25} \approx 1.376 \times 10^{66} \] Now dividing by \( 53! \): \[ P(X = 53) \approx \frac{1.376 \times 10^{66}}{2.425 \times 10^{69}} \approx 0.000567 \] Thus, the probability that exactly 53 radioactive atoms decayed on a given day is approximately: \[ P(X = 53) \approx 0.000567 \] ### Summary of Results - Mean number of radioactive atoms that decayed per day: **56.755** - Probability that exactly 53 radioactive atoms decayed on a given day: **0.000567**