Math Problem Statement

Data from 14 cities were combined for a 20-year period, and the total 280 city-years included a total of 73 homicides. After finding the mean number of homicides per city-year, find the probability that a randomly selected city-year has the following numbers of homicides: 0, 1, 2, or 3, using the Poisson probabilities.

Solution

The problem involves finding Poisson probabilities related to homicide data over several city-years. From the image, the key parameters are as follows:

  • There are 280 city-years in total.
  • There were 73 homicides in total.

Step 1: Find the mean number of homicides per city-year.

The mean (λ) is calculated as the total number of homicides divided by the number of city-years: λ=73280=0.2607\lambda = \frac{73}{280} = 0.2607

Step 2: Poisson probability formula

The Poisson probability formula is: P(k;λ)=eλλkk!P(k; \lambda) = \frac{e^{-\lambda} \lambda^k}{k!} Where:

  • kk is the number of homicides.
  • λ=0.2607\lambda = 0.2607
  • ee is the Euler's number (approximately 2.71828).

Step 3: Calculation for each value of kk

We already have probabilities for k=0k = 0, k=1k = 1, and k=2k = 2, so we can calculate for k=3k = 3.

a. P(0)P(0)

P(0)=e0.26070.260700!=e0.26070.7705(given)P(0) = \frac{e^{-0.2607} \cdot 0.2607^0}{0!} = e^{-0.2607} \approx 0.7705 \quad \text{(given)}

b. P(1)P(1)

P(1)=e0.26070.260711!0.2009(given)P(1) = \frac{e^{-0.2607} \cdot 0.2607^1}{1!} \approx 0.2009 \quad \text{(given)}

c. P(2)P(2)

P(2)=e0.26070.260722!0.0262(given)P(2) = \frac{e^{-0.2607} \cdot 0.2607^2}{2!} \approx 0.0262 \quad \text{(given)}

d. P(3)P(3)

P(3)=e0.26070.260733!P(3) = \frac{e^{-0.2607} \cdot 0.2607^3}{3!} Let's calculate this: P(3)=e0.2607(0.2607)36=e0.26070.0177360.77050.0177360.0136760.0023P(3) = \frac{e^{-0.2607} \cdot (0.2607)^3}{6} = \frac{e^{-0.2607} \cdot 0.01773}{6} \approx \frac{0.7705 \cdot 0.01773}{6} \approx \frac{0.01367}{6} \approx 0.0023

Thus, P(3)0.0023P(3) \approx 0.0023.

Summary of probabilities:

  • P(0)=0.7705P(0) = 0.7705
  • P(1)=0.2009P(1) = 0.2009
  • P(2)=0.0262P(2) = 0.0262
  • P(3)=0.0023P(3) = 0.0023

Would you like more details or have any further questions?


Here are five related questions to consider:

  1. How do Poisson distributions apply to real-world crime data analysis?
  2. What assumptions are made when using a Poisson distribution for modeling homicides?
  3. How does changing the mean λ\lambda affect the shape of the Poisson distribution?
  4. What would the probabilities look like for values of k4k \geq 4?
  5. How could the accuracy of these probability models be verified with further data?

Tip: When working with small means like λ=0.2607\lambda = 0.2607, the probabilities drop off quickly for higher values of kk, which is why P(3)P(3) and higher are very small.

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Math Problem Analysis

Mathematical Concepts

Probability
Poisson Distribution

Formulas

Mean (λ) = Total homicides / Total city-years
Poisson probability formula: P(k; λ) = (e^(-λ) * λ^k) / k!

Theorems

Poisson Distribution Theorem

Suitable Grade Level

College Level or Advanced High School