Question 1169184: Hello, please help me with this question. Thank you in advance
In one year, there were 116 homicide deaths in Richmond,
Virginia. For a randomly selected day, find the probability that the
number of homicide deaths is:
0, 1, 2, 3, 4
Compare the calculated probabilities to these actual results:
268 days (no homicides); 79 days (1 homicides); 17 days(2
homicides); 1 day (3 homicides); no days with more than 3
homicides
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem using the Poisson distribution, which is suitable for modeling the number of events (homicide deaths in this case) occurring in a fixed interval of time.
**1. Calculate the Average Rate (λ)**
* Total homicides: 116
* Number of days in a year: 365
* Average homicide deaths per day (λ): 116 / 365 ≈ 0.3178
**2. Calculate Probabilities using the Poisson Formula**
The Poisson probability formula is:
$$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$$
Where:
* $P(X = k)$ is the probability of $k$ events occurring.
* $e$ is Euler's number (approximately 2.71828).
* $\lambda$ is the average rate (0.3178 in this case).
* $k$ is the number of events we're interested in (0, 1, 2, 3, 4).
* $k!$ is the factorial of $k$.
Let's calculate the probabilities:
* **P(X = 0):**
* $P(0) = \frac{e^{-0.3178} (0.3178)^0}{0!} \approx \frac{0.7278 \times 1}{1} \approx 0.7278$
* **P(X = 1):**
* $P(1) = \frac{e^{-0.3178} (0.3178)^1}{1!} \approx \frac{0.7278 \times 0.3178}{1} \approx 0.2313$
* **P(X = 2):**
* $P(2) = \frac{e^{-0.3178} (0.3178)^2}{2!} \approx \frac{0.7278 \times 0.10099684}{2} \approx 0.0366$
* **P(X = 3):**
* $P(3) = \frac{e^{-0.3178} (0.3178)^3}{3!} \approx \frac{0.7278 \times 0.032096}{6} \approx 0.0039$
* **P(X = 4):**
* $P(4) = \frac{e^{-0.3178} (0.3178)^4}{4!} \approx \frac{0.7278 \times 0.01020}{24} \approx 0.0003$
**3. Compare Calculated Probabilities to Actual Results**
* **Calculated Probabilities:**
* P(0) ≈ 0.7278
* P(1) ≈ 0.2313
* P(2) ≈ 0.0366
* P(3) ≈ 0.0039
* P(4) ≈ 0.0003
* **Actual Results (Converted to Probabilities):**
* 0 homicides: 268 / 365 ≈ 0.7342
* 1 homicide: 79 / 365 ≈ 0.2164
* 2 homicides: 17 / 365 ≈ 0.0466
* 3 homicides: 1 / 365 ≈ 0.0027
* 4+ homicides: 0/365 = 0
**Comparison**
The calculated Poisson probabilities are quite close to the actual results. The Poisson distribution seems to be a reasonable fit for this data. There are very slight differences, which is to be expected when using a theoretical distribution to model real-world data.
**Summary of Probabilities**
* P(0) ≈ 0.7278
* P(1) ≈ 0.2313
* P(2) ≈ 0.0366
* P(3) ≈ 0.0039
* P(4) ≈ 0.0003
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