# 4.6 Poisson distribution  (Page 3/18)

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Text message users receive or send an average of 41.5 text messages per day.

1. How many text messages does a text message user receive or send per hour?
2. What is the probability that a text message user receives or sends two messages per hour?
3. What is the probability that a text message user receives or sends more than two messages per hour?
1. Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is $\frac{41.5}{24}$ ≈ 1.7292.
2. X ~ P (1.7292), so P ( x = 2) = poissonpdf(1.7292, 2) ≈ 0.2653
3. P ( x >2) = 1 – P ( x ≤ 2) = 1 – poissoncdf(1.7292, 2) ≈ 1 – 0.7495 = 0.2505

## Try it

Atlanta’s Hartsfield-Jackson International Airport is the busiest airport in the world. On average there are 2,500 arrivals and departures each day.

1. How many airplanes arrive and depart the airport per hour?
2. What is the probability that there are exactly 100 arrivals and departures in one hour?
3. What is the probability that there are at most 100 arrivals and departures in one hour?
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.

On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?

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

Using the binomial distribution:

• P ( x = 10) = binompdf(200, .0102, 10) ≈ 0.000039

Using the Poisson distribution:

• Calculate μ = np = 200(0.0102) ≈ 2.04
• P ( x = 10) = poissonpdf(2.04, 10) ≈ 0.000045

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—both probabilities reported are almost 0.

## Try it

On May 13, 2013, starting at 4:30 PM, the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%. Use this information for the next 100 days to find the probability that there will be low seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?

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.

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