16.7 Examples of Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.

The probability mass function (PMF) for the Poisson distribution is given by:

$P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$

Where:

• $P(X = x)$ is the probability of exactly $x$ events occurring.
• $\lambda$ (lambda) is the average number of events in the given interval.
• $e$ is the base of the natural logarithm (approximately 2.71828).
• $x$ is the number of occurrences.
• $x!$ is the factorial of $x$ ($x! = x \times (x-1) \times \dots \times 2 \times 1$).

Example 1: Website Visits

Problem: A blog receives an average of 12 visits per hour. What is the probability that exactly 8 people visit the blog in a given hour?

Solution:

Given:

• Average rate ($\lambda$) = 12 visits per hour.
• We want to find the probability of exactly 8 visits, so $x = 8$.

Applying the Poisson Probability Formula:

$P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$

Substitute the given values:

$P(X = 8) = \frac{e^{-12} \times 12^8}{8!}$

Calculations:

• $e^{-12} \approx 6.1442 \times 10^{-6}$
• $12^8 = 429,981,696$
• $8! = 40,320$

Now, substitute these values into the formula:

$P(X = 8) \approx \frac{(6.1442 \times 10^{-6}) \times 429,981,696}{40,320}$ $P(X = 8) \approx \frac{2.641}{40,320}$ $P(X = 8) \approx 0.0655$

Answer: The probability of exactly 8 people visiting the blog in an hour is approximately 0.0655, or 6.55%.

Example 2: Text Messages in 30 Minutes

Problem: A user receives an average of 6 text messages per hour. What is the probability that they receive no messages in the next 30 minutes?

Solution:

Given:

• Average rate = 6 messages per hour.
• The interval is 30 minutes, which is 0.5 hours.

Adjusting the Rate ($\lambda$): Since the average rate is given per hour, we need to adjust it for the 30-minute interval:

$\lambda = \text{Average rate per hour} \times \text{Time interval in hours}$ $\lambda = 6 \text{ messages/hour} \times 0.5 \text{ hours}$ $\lambda = 3 \text{ messages}$

• We want to find the probability of receiving no messages, so $x = 0$.

Applying the Poisson Probability Formula:

$P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$

Substitute the adjusted values:

$P(X = 0) = \frac{e^{-3} \times 3^0}{0!}$

Calculations:

• $e^{-3} \approx 0.0498$
• $3^0 = 1$ (any non-zero number raised to the power of 0 is 1)
• $0! = 1$ (by definition of factorial)

Now, substitute these values into the formula:

$P(X = 0) = \frac{0.0498 \times 1}{1}$ $P(X = 0) = 0.0498$

Answer: The probability of receiving no text messages in the next 30 minutes is 0.0498, or 4.98%.

Additional Example: Call Center Arrivals

Problem: A call center receives an average of 15 calls per hour. What is the probability that exactly 10 calls are received in a given hour?

Solution:

Given:

• Average rate ($\lambda$) = 15 calls per hour.
• We want to find the probability of exactly 10 calls, so $x = 10$.

Applying the Poisson Probability Formula:

$P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$

Substitute the given values:

$P(X = 10) = \frac{e^{-15} \times 15^{10}}{10!}$

Calculations:

• $e^{-15} \approx 3.059 \times 10^{-7}$
• $15^{10} \approx 5.7665 \times 10^{11}$
• $10! = 3,628,800$

Now, substitute these values into the formula:

$P(X = 10) \approx \frac{(3.059 \times 10^{-7}) \times (5.7665 \times 10^{11})}{3,628,800}$ $P(X = 10) \approx \frac{176,334.6}{3,628,800}$ $P(X = 10) \approx 0.0486$

Answer: The probability of exactly 10 calls being received in an hour is approximately 0.0486, or 4.86%.