Math Problem Statement

Question content area top Part 1 A particular intersection in a small town is equipped with a surveillance camera. The number of traffic tickets issued to drivers passing through the intersection follows the Poisson distribution and averages 5.6 per month. a. What is the probability that 5 traffic tickets will be issued at the intersection next​ month? b. What is the probability that 3 or fewer traffic tickets will be issued at the intersection next​ month? c. What is the probability that more than 6 traffic tickets will be issued at the intersection next​ month?

Solution

The number of traffic tickets issued follows a Poisson distribution, where the average number of tickets per month (λ\lambda) is 5.6. The probability mass function (PMF) for the Poisson distribution is given by:

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

where:

  • λ\lambda is the average number of events (5.6 in this case),
  • kk is the number of occurrences for which we want to find the probability, and
  • ee is Euler's number (e2.71828e \approx 2.71828).

Part (a)

We want to find the probability that exactly 5 traffic tickets will be issued. This is P(X=5)P(X = 5).

P(X=5)=5.65e5.65!P(X = 5) = \frac{5.6^5 e^{-5.6}}{5!}

Part (b)

We want to find the probability that 3 or fewer traffic tickets will be issued. This is P(X3)P(X \leq 3), which we compute as:

P(X3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Each of these probabilities will be calculated using the Poisson formula:

P(X=k)=5.6ke5.6k!P(X = k) = \frac{5.6^k e^{-5.6}}{k!}

for k=0,1,2,3k = 0, 1, 2, 3.

Part (c)

We want to find the probability that more than 6 traffic tickets will be issued. This is P(X>6)P(X > 6), which can be calculated as:

P(X>6)=1P(X6)P(X > 6) = 1 - P(X \leq 6)

Where P(X6)P(X \leq 6) can be calculated as:

P(X6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

Let me calculate each part for you.Here are the results for the given Poisson distribution problem:

Part (a)

The probability that exactly 5 traffic tickets will be issued next month is approximately:

P(X=5)0.1697P(X = 5) \approx 0.1697

Part (b)

The probability that 3 or fewer traffic tickets will be issued next month is approximately:

P(X3)0.1906P(X \leq 3) \approx 0.1906

Part (c)

The probability that more than 6 traffic tickets will be issued next month is approximately:

P(X>6)0.3297P(X > 6) \approx 0.3297

Let me know if you need more details or further clarifications!

Follow-up questions:

  1. How does the shape of the Poisson distribution change with varying λ\lambda?
  2. Can you calculate the variance and standard deviation for this Poisson distribution?
  3. What is the expected number of tickets issued in 3 months if the average remains the same?
  4. How does the Poisson distribution compare to the normal distribution for large λ\lambda?
  5. How would the probabilities change if the average number of tickets was 7 instead of 5.6?

Tip:

For small λ\lambda, the Poisson distribution is skewed to the right, but as λ\lambda increases, it begins to resemble a normal distribution.

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Math Problem Analysis

Mathematical Concepts

Probability
Poisson Distribution

Formulas

P(X = k) = (λ^k * e^(-λ)) / k!
P(X ≤ k) = Σ[P(X = i)] for i = 0 to k
P(X > k) = 1 - P(X ≤ k)

Theorems

Poisson Theorem

Suitable Grade Level

Grades 10-12