5.6 Poisson di - university of calgary - base content - v2015  (Page 3/18)

1. Let X = the number of airplanes arriving and departing from Hartsfield-Jackson in one hour. The average number of arrivals and departures per hour is $\frac{2,500}{24}$ ≈ 104.1667.
2. X ~ P (104.1667), so P ( X = 100) = poissonpdf(104.1667, 100) ≈ 0.0366.
3. P ( X ≤ 100) = poissoncdf(104.1667, 100) ≈ 0.3651.

The Poisson distribution can be used to approximate probabilities for a binomial distribution. This next example demonstrates the relationship between the Poisson and the binomial distributions. Let n represent the number of binomial trials and let p represent the probability of a success for each trial. If n is large enough and p is small enough then the Poisson approximates the binomial very well. In general, n is considered “large enough” if it is greater than or equal to 20. The probability p from the binomial distribution should be less than or equal to 0.05. When the Poisson is used to approximate the binomial, we use the binomial mean μ = np . The variance of X is σ ² = μ and the standard deviation is σ = $\sqrt{\mu }$ . The Poisson approximation to a binomial distribution was commonly used in the days before technology made both values very easy to calculate.

Let X = the number of days with moderate seismic activity.

Using the binomial distribution: P ( x = 5) = binompdf(100, 0.0143, 5) ≈ 0.0115

Using the Poisson distribution:

• Calculate μ = np = 100(0.0143) = 1.43
• P ( X = 5) = poissonpdf(1.43, 5) = 0.0119

We expect the approximation to be good because n is large (greater than 20) and p is small (less than 0.05). The results are close—the difference between the values is 0.0004.