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In the Poisson distribution, we are interested in counting the number of events that occur within a fixed interval of time or space. This type of probability distribution is especially helpful for events that happen independently and at a constant average rate, like the number of accidents in a city per day.

To model such scenarios using the Poisson distribution, we use the formula \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] where:

• \( P(X = k) \) is the probability of \( k \) events occurring.
• \( \lambda \) is the average number of events in the given interval (0.8 accidents per day in our problem).
• \( e \) is Euler's number, approximately equal to 2.71828.
• \( k! \) is the factorial of \( k \).

For example, to find the probability of no accidents occurring on a given day, plug \( k = 0 \) and \( \lambda = 0.8 \) into the formula to find \[ P(X = 0) = \frac{0.8^0 e^{-0.8}}{0!} = 0.4493. \]
This indicates there's about 44.93% chance of zero accidents happening that day.

In a Poisson distribution, both the mean and variance have a unique property—they share the same value, denoted by \( \lambda \). This is an important feature that simplifies calculations.

So, in our scenario with accidents occurring at an average rate of 0.8 per day:

• The mean \( \mu \) is \( \lambda = 0.8 \).
• Variance, the measure of how dispersed the accident occurrences are from the mean, is also \( \lambda = 0.8 \).

The mean indicates the central or expected value of accidents per day. Since the variance is equal to the mean in a Poisson distribution, it implies a certain regularity or consistency in the occurrence of events, as they spread around this central value.

The standard deviation in the context of a Poisson distribution provides insight into the spread of the data around the mean, much like it does in other types of distributions.

For a Poisson distribution, the standard deviation \( \sigma \) is calculated as the square root of the variance. Since for Poisson distribution, the mean equals the variance, it becomes:

• \( \sigma = \sqrt{\lambda} \)

In our case, with an average of \( \lambda = 0.8 \) accidents per day:

• The standard deviation \( \sigma \) is \( \sqrt{0.8} \approx 0.8944 \).

This means the number of accidents deviates from the mean by approximately 0.8944 accidents per day. Thus, while the events occur consistently, there's still a measure of variability in their daily occurrence.