To solve this problem, we need to use the Poisson distribution. The Poisson distribution is used to model the probability of a given number of events happening in a fixed interval of time or space, provided the events occur with a known constant mean rate and independently of the time since the last event.

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

The total number of homicides over 280 city-years is 108, so the mean number of homicides per city-year (denoted by $\lambda$) is:

$\lambda = \frac{108 \text{ homicides}}{280 \text{ city-years}} = 0.3857$

This $\lambda$ will be used as the rate for the Poisson distribution.

Step 2: Use the Poisson Probability Formula

The Poisson probability formula is:

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

where:

• $P(x)$ is the probability of observing exactly $x$ homicides,
• $\lambda$ is the average number of homicides (0.3857 in this case),
• $x!$ is the factorial of $x$,
• $e$ is Euler's number, approximately equal to 2.71828.

We will now calculate the probabilities for $x = 0, 1, 2, 3, 4$ homicides.

Part a: Probability of 0 homicides ($P(0)$)

$P(0) = \frac{e^{-0.3857} \cdot 0.3857^0}{0!} = e^{-0.3857} \cdot 1 = 0.6800 \, (\text{as given}).$

Part b: Probability of 1 homicide ($P(1)$)

$P(1) = \frac{e^{-0.3857} \cdot 0.3857^1}{1!} = e^{-0.3857} \cdot 0.3857$

Using $e^{-0.3857} \approx 0.6798$:

$P(1) \approx 0.6798 \cdot 0.3857 = 0.2621$

Part c: Probability of 2 homicides ($P(2)$)

$P(2) = \frac{e^{-0.3857} \cdot 0.3857^2}{2!} = \frac{e^{-0.3857} \cdot 0.1488}{2}$

$P(2) \approx \frac{0.6798 \cdot 0.1488}{2} = 0.0506$

Part d: Probability of 3 homicides ($P(3)$)

$P(3) = \frac{e^{-0.3857} \cdot 0.3857^3}{3!} = \frac{e^{-0.3857} \cdot 0.0574}{6}$

$P(3) \approx \frac{0.6798 \cdot 0.0574}{6} = 0.0065$

Part e: Probability of 4 homicides ($P(4)$)

$P(4) = \frac{e^{-0.3857} \cdot 0.3857^4}{4!} = \frac{e^{-0.3857} \cdot 0.0221}{24}$

$P(4) \approx \frac{0.6798 \cdot 0.0221}{24} = 0.0006$

Summary of Poisson Probabilities:

• $P(0) = 0.6800$
• $P(1) = 0.2621$
• $P(2) = 0.0506$
• $P(3) = 0.0065$
• $P(4) = 0.0006$

These values can be compared to the actual results provided in the problem:

• Actual results for 0 homicides: 190
• Actual results for 1 homicide: 74
• Actual results for 2 homicides: 14
• Actual results for 3 homicides: 2
• Actual results for 4 homicides: 0

Do you have any questions on how the Poisson distribution was applied, or would you like more details?

Here are some related questions to expand your understanding:

1. How do we calculate the factorial used in the Poisson formula?
2. What are the assumptions of the Poisson distribution?
3. How would the probabilities change if the mean number of homicides ($\lambda$) was higher?
4. Can we use the Poisson distribution if the homicides are not independent events?
5. How does the Poisson distribution compare to the normal distribution?

Tip: When using the Poisson distribution, ensure that the mean ($\lambda$) is consistent with the event count and interval duration, as the rate influences all subsequent probability calculations.