When studying radioactive material, a nuclear engineer found that over 365 days, \( 1,000,000 \) radioactive atoms decayed to 976,041 radioactive atoms, so 23,959 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 65.641 (Round to three decimal places as needed.) b. The probability that on a given day, 51 radioactive atoms decayed, is (Round to six decimal places as needed.)

a. To find the mean number of radioactive atoms that decayed in a day, we can divide the total number of atoms that decayed over 365 days by 365. Mean number of atoms that decayed in a day = Total number of atoms decayed / Number of days Mean number of atoms that decayed in a day = 23,959 / 365 Let's calculate this. Calculate the value by following steps: - step0: Calculate: \(\frac{23959}{365}\) The mean number of radioactive atoms that decayed in a day is approximately 65.641. b. To find the probability that on a given day, 51 radioactive atoms decayed, we can use the Poisson distribution formula: \[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \] where: - \( P(X = k) \) is the probability of \( k \) events occurring in a given time period, - \( \lambda \) is the average number of events that occur in a given time period, - \( k \) is the number of events we are interested in, - \( e \) is the base of the natural logarithm. In this case, \( \lambda = 65.641 \) (mean number of atoms that decayed in a day) and \( k = 51 \) (number of atoms we are interested in). Let's calculate the probability. Simplify the expression by following steps: - step0: Solution: \(\frac{e^{-65.641}\times 65.641^{51}}{51!}\) - step1: Convert the expressions: \(\frac{e^{-\frac{65641}{1000}}\times 65.641^{51}}{51!}\) - step2: Convert the expressions: \(\frac{e^{-\frac{65641}{1000}}\left(\frac{65641}{1000}\right)^{51}}{51!}\) - step3: Multiply the numbers: \(\frac{\frac{\left(\frac{65641}{1000}\right)^{51}}{e^{\frac{65641}{1000}}}}{51!}\) - step4: Multiply by the reciprocal: \(\frac{\left(\frac{65641}{1000}\right)^{51}}{e^{\frac{65641}{1000}}}\times \frac{1}{51!}\) - step5: Multiply the terms: \(\frac{\left(\frac{65641}{1000}\right)^{51}}{e^{\frac{65641}{1000}}\times 51!}\) - step6: Simplify: \(\frac{65641^{51}}{e^{\frac{65641}{1000}}\times 1000^{51}\times 51!}\) - step7: Simplify: \(\frac{65641^{51}e^{\frac{359}{1000}}}{1000^{51}e^{66}\times 51!}\) The probability that on a given day, 51 radioactive atoms decayed is approximately \( \frac{65641^{51}e^{\frac{359}{1000}}}{1000^{51}e^{66}\times 51!} \). To round this to six decimal places, we can calculate the value of this expression. Simplify the expression by following steps: - step0: Solution: \(\frac{65641^{51}e^{\frac{359}{1000}}}{\left(1000^{51}e^{66}\times 51!\right)}\) - step1: Remove the parentheses: \(\frac{65641^{51}e^{\frac{359}{1000}}}{1000^{51}e^{66}\times 51!}\) The probability that on a given day, 51 radioactive atoms decayed is approximately 0.000000.