Introduction to Probability Models, Ross, 11th Ed. Ch 2, Problem 86 I’m not sure where to start with the following questions, I would love to see a solution to get a sense of how to do it. My hunch is

(a) We can use the Central Limit Theorem (CLT) to \approx imate the distribution of the total processing time of 40 books. The sum of independent random variables with the same mean and standard deviation can be \approx imated by a normal distribution. The mean of the total processing time is 40 x 10 = 400 minutes, and the standard deviation is sqrt(40) x 3 = 7.746 minutes. Let X be the total processing time of 40 books. We want to find P(X > 420). We can standardize X using the z-score formula: z = (420 - 400) / 7.746 = 2.58. The probability of X being greater than 420 is equal to the probability of a standard normal variable being greater than 2.58: P(X > 420) = P(Z > 2.58) = 0.0049. Therefore, the \approx imate probability that it will take more than 420 minutes to process all these books is 0.49%. (b) Again, we can use CLT to \approx imate the distribution of the number of books processed in the first 240 minutes. The mean number of books processed in 240 minutes is 240 / 10 = 24, and the standard deviation is sqrt(240) x 3 / 10 = 3.464. Let Y be the number of books processed in the first 240 minutes. We want to find P(Y ? 25). We can standardize Y using the z-score formula: z = (25 - 24) / 3.464 = 0.29. The probability of Y being greater than or equal to 25 is equal to the probability of a standard normal variable being greater than 0.29: P(Y ? 25) = P(Z > 0.29) = 0.3859. Therefore, the \approx imate probability that at least 25 books will be processed in the first 240 minutes is 38.59%.