We are dealing with a Poisson distribution where the number of parking tickets issued per day is described by a Poisson random variable. The Poisson distribution is characterized by its rate parameter $\lambda$, which represents the average number of occurrences within a fixed interval of time or space. In this case, $\lambda = 10$ tickets per day.

The probability mass function (PMF) of a Poisson distribution is given by:

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

Where:

• $X$ is the random variable representing the number of occurrences (in this case, parking tickets issued),
• $\lambda$ is the rate (mean) of occurrences,
• $k$ is the number of occurrences for which we want to find the probability, and
• $e$ is Euler's number (approximately 2.71828).

(a) Find the probability that exactly 5 tickets will be issued in one day.

For one day, the rate $\lambda = 10$ tickets/day, and we want the probability of exactly 5 tickets (i.e., $k = 5$):

$P(X = 5) = \frac{10^5 e^{-10}}{5!}$

Let’s calculate this.

(b) Find the probability that less than 5 tickets will be issued in one day.

The probability of less than 5 tickets being issued means we need to sum the probabilities for 0, 1, 2, 3, and 4 tickets:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Each term can be calculated using the Poisson PMF.

(c) Find the probability that exactly 5 tickets will be issued in any two-day period.

For a two-day period, the rate $\lambda$ changes to $2 \times 10 = 20$ tickets for two days. Now, we want the probability of exactly 5 tickets (i.e., $k = 5$) in this two-day period:

$P(X = 5) = \frac{20^5 e^{-20}}{5!}$

(d) Find the probability that more than 5 tickets will be issued in any two-day period.

To find the probability that more than 5 tickets will be issued, we calculate the complement of the probability of 5 or fewer tickets:

$P(X > 5) = 1 - P(X \leq 5)$

Where:

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

We can now compute each part.

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Let me calculate these probabilities for you.Here are the computed probabilities:

(a) The probability that exactly 5 tickets will be issued in one day:

$P(X = 5) = 0.0378$

(b) The probability that less than 5 tickets will be issued in one day:

$P(X < 5) = 0.0293$

(c) The probability that exactly 5 tickets will be issued in any two-day period:

$P(X = 5) = 0.00005496$

(d) The probability that more than 5 tickets will be issued in any two-day period:

$P(X > 5) = 0.9999$

Let me know if you would like any further clarification or a breakdown of the calculations!

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Here are 5 related questions to explore further:

1. What is the probability of exactly 10 tickets being issued in a single day?
2. How would the probabilities change if the rate increased to 15 tickets per day?
3. What is the probability of fewer than 3 tickets being issued in a two-day period?
4. What is the expected number of tickets issued over a three-day period?
5. How does the standard deviation relate to the Poisson distribution in this case?

Tip: In a Poisson distribution, the mean and the variance are both equal to the rate parameter $\lambda$. This can help in estimating variability!