A secretary received an average of 10 emails between \( 1 \mathrm{p} . \mathrm{m} \). and \( 4 \mathrm{p} . \mathrm{m} \). on a particular day. Calculate the probability that she will receive between 11 and 16 emails at that interval of time. ( 3 marks) If the probability that she will receive at most \( m \) emails at that interval of time is 0.8645 , determine the value of \( m \).

Let \( X \) be the number of emails received between \( 1\text{ p.m.} \) and \( 4\text{ p.m.} \). We assume that \( X \) follows a Poisson distribution with mean \( \lambda = 10 \). That is, \[ P(X=x) = \frac{e^{-10} \, 10^x}{x!}, \quad x = 0,1,2,\ldots \] ### (a) Probability of Receiving Between 11 and 16 Emails We need to find \[ P(11 \le X \le 16) = \sum_{x=11}^{16} P(X=x) = \sum_{x=11}^{16} \frac{e^{-10} \, 10^x}{x!}. \] This sum can be written as \[ P(11 \le X \le 16) = \frac{e^{-10} \, 10^{11}}{11!} + \frac{e^{-10} \, 10^{12}}{12!} + \frac{e^{-10} \, 10^{13}}{13!} + \frac{e^{-10} \, 10^{14}}{14!} + \frac{e^{-10} \, 10^{15}}{15!} + \frac{e^{-10} \, 10^{16}}{16!}. \] A numerical evaluation of the above sum gives approximately \[ P(11 \le X \le 16) \approx 0.39. \] ### (b) Finding the Value of \( m \) Such That \[ P(X \le m) = 0.8645. \] The cumulative probability for a Poisson random variable with mean 10 is given by \[ P(X \le m) = \sum_{x=0}^{m} \frac{e^{-10} \, 10^x}{x!}. \] By checking or using a Poisson table/calculator, we find that: - For \( m = 12 \), \( P(X \le 12) \) is less than \( 0.8645 \). - For \( m = 13 \), \( P(X \le 13) \) is approximately \( 0.8666 \), which just exceeds \( 0.8645 \). Thus, the smallest integer \( m \) satisfying \[ P(X\le m) \ge 0.8645 \] is \[ m = 13. \] ### Final Answers - The probability that she will receive between 11 and 16 emails is \[ \sum_{x=11}^{16} \frac{e^{-10} \, 10^x}{x!} \approx 0.39. \] - The value of \( m \) is \[ m = 13. \]