When studying radioactive material, a nuclear engineer found that over 365 days, \( 1,000,000 \) radioactive atoms decayed to 973,521 radioactive atoms, so 26,479 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, 51 radioactive atoms decayed a. The mean number of radioactive atoms that decay per day is (Round to three decimal places as needed.)

a. To find the mean number of radioactive atoms that decay in a day, we divide the total number of atoms that decayed over the 365-day period by the number of days. Given: Total atoms decayed in 365 days = 26,479 Number of days = 365 Mean decay per day = Total atoms decayed / Number of days Mean decay per day = 26,479 / 365 Now, let's calculate the mean decay per day: Mean decay per day = 26,479 / 365 ≈ 72.312 So, the mean number of radioactive atoms that decay in a day is approximately 72.312 (rounded to three decimal places). b. To find the probability that on a given day, 51 radioactive atoms decayed, we would need to know the decay distribution of the radioactive material. This could be described by a probability distribution, such as a Poisson distribution, if the decay process is assumed to be a Poisson process. The Poisson distribution is given by the formula: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where: - \( P(X = k) \) is the probability of \( k \) events occurring in a given time interval, - \( e \) is the base of the natural logarithm (approximately 2.71828), - \( \lambda \) is the average number of events (decays) per time interval, - \( k \) is the number of events we are interested in (51 in this case), - \( k! \) is the factorial of \( k \). From part a, we found that the mean decay per day is approximately 72.312. Therefore, \( \lambda \) is approximately 72.312. Now, let's calculate the probability that 51 radioactive atoms decay on a given day: \[ P(X = 51) = \frac{e^{-72.312} \cdot 72.312^{51}}{51!} \] This calculation requires the use of a calculator or computer software to handle the large numbers involved. Let's calculate it: \[ P(X = 51) = \frac{e^{-72.312} \cdot 72.312^{51}}{51!} \] Using a calculator or software, we find: \[ P(X = 51) \approx 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000