The number of parking tickets issued daily is a Poisson random variable with the rate 10 ticket per day.
(a) Find the probability that exactly 5 tickets will be issued in one day
(b) Find the probability that less than 5 tickets will be issued in one day
(c) Find the probability that exactly 5 tickets will be issued in any two- day period
(d) Find the probability that more than 5 tickets will be issued in any two- day period

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.

---

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!

---

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!

The number of parking tickets issued daily is a Poisson random variable with the rate 10 tickets per day.
(a) Find the probability that exactly 5 tickets will be issued in one day
(b) Find the probability that less than 5 tickets will be issued in one day
(c) Find the probability that exactly 5 tickets will be issued in any two-day period
(d) Find the probability that more than 5 tickets will be issued in any two-day period

Explore the probability calculations for parking tickets issued based on a Poisson distribution with an average rate of 10 tickets per day. Includes calculations for exactly 5 tickets in one or two days and for more or less than 5 tickets. Suitable for high school students studying probability and Poisson distribution.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation