99）FX food channel receives an average of six orders in every four minutes．Assuming an approximate Poisson distribution that five or more orders will be received during a period of four minutes. a） 0.2849 b） \( \mathbf{0 . 7 1 4 9} \) c） 0.44550 d） 0.5545

To solve the problem, we start with the Poisson distribution, which is used for modeling the number of events (in this case, orders) occurring within a fixed interval of time. The average rate of orders (λ) is given as six orders every four minutes. We need to find the probability of receiving five or more orders in four minutes. To do this, we calculate the probabilities for receiving zero to four orders and subtract their sum from 1. The probability mass function (PMF) of a Poisson distribution is given by the formula: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] for \( k = 0, 1, 2, ... \). 1. **Calculate individual probabilities** for \( k = 0, 1, 2, 3, 4 \): - \( P(X = 0) = \frac{e^{-6} 6^0}{0!} = e^{-6} \approx 0.002478752 \) - \( P(X = 1) = \frac{e^{-6} 6^1}{1!} = 6 \cdot e^{-6} \approx 0.014872512 \) - \( P(X = 2) = \frac{e^{-6} 6^2}{2!} = 18 \cdot e^{-6} \approx 0.044617537 \) - \( P(X = 3) = \frac{e^{-6} 6^3}{3!} = 36 \cdot e^{-6} \approx 0.089235073 \) - \( P(X = 4) = \frac{e^{-6} 6^4}{4!} = 54 \cdot e^{-6} \approx 0.133852777 \) 2. **Sum these probabilities** for \( k = 0, 1, 2, 3, 4 \): \[ P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \approx 0.002478752 + 0.014872512 + 0.044617537 + 0.089235073 + 0.133852777 \approx 0.285156651 \] 3. **Find the probability of five or more orders**: \[ P(X \geq 5) = 1 - P(X < 5) \approx 1 - 0.285156651 \approx 0.714843349 \] When rounded to four decimal places, this result matches option (b) \( \mathbf{0 . 7 1 4 9} \). Thus, the answer is option **b**.