Poisson Distribution

The probability of exactly k events occurring in a fixed interval of time or space, given that the average rate of occurrence is lambda, is given by the Poisson distribution formula: P(X = k) = (lambda^k * e^(-lambda)) / k!

The mean of the Poisson distribution is equal to the average rate of occurrence, which is denoted by lambda.

The variance of the Poisson distribution is also equal to the average rate of occurrence, which is denoted by lambda.

The probability of no events occurring in a fixed interval of time or space, given that the average rate of occurrence is lambda, is given by P(X = 0) = e^(-lambda).

The probability of at least one event occurring in a fixed interval of time or space, given that the average rate of occurrence is lambda, is given by P(X >= 1) = 1 - e^(-lambda).

Let X be the number of calls received in the next hour. Then X follows a Poisson distribution with a mean of 10. The probability of receiving exactly 12 calls is given by P(X = 12) = (10^12 * e^(-10)) / 12! = 0.125.

Let X be the number of defective items produced in the next 100 items. Then X follows a Poisson distribution with a mean of 5 (0.05 * 100). The probability of producing exactly 2 defective items is given by P(X = 2) = (5^2 * e^(-5)) / 2! = 0.204.

Let X be the number of items sold on a particular day. Then X follows a Poisson distribution with a mean of 20. The probability of selling exactly 25 items is given by P(X = 25) = (20^25 * e^(-20)) / 25! = 0.271.

Let X be the number of emails received in the next 15 minutes. Then X follows a Poisson distribution with a mean of 12.5 (50 * 15 / 60). The probability of receiving no emails is given by P(X = 0) = e^(-12.5) = 0.203.

Let X be the number of patients seen on a particular day. Then X follows a Poisson distribution with a mean of 10. The probability of seeing at least one patient is given by P(X >= 1) = 1 - P(X = 0) = 1 - e^(-10) = 0.995.

Let X be the number of defective bolts produced in the next 100 bolts. Then X follows a Poisson distribution with a mean of 99 (0.99 * 100). The probability of producing at most 2 defective bolts is given by P(X <= 2) = P(X = 0) + P(X = 1) + P(X = 2) = e^(-99) + 99 * e^(-99) + (99^2 * e^(-99)) / 2! = 0.990.

Let X be the number of phone calls received in the next hour. Then X follows a Poisson distribution with a mean of 20. The probability of receiving between 15 and 25 phone calls is given by P(15 <= X <= 25) = P(X = 15) + P(X = 16) + ... + P(X = 25) = 0.494.

Let X be the number of items sold on a particular day. Then X follows a Poisson distribution with a mean of 50. The probability of selling more than 60 items is given by P(X > 60) = 1 - P(X <= 60) = 1 - (P(X = 0) + P(X = 1) + ... + P(X = 60)) = 0.203.

Let X be the number of patients seen on a particular day. Then X follows a Poisson distribution with a mean of 15. The probability of seeing exactly 20 patients is given by P(X = 20) = (15^20 * e^(-15)) / 20! = 0.204.