Question 1184195: In a study of suicides, a researcher found that the monthly distribution of adolescent
suicides follows a Poisson distribution with parameter of μ= 2.75. Find the probability that
a. a randomly selected month will be one in which three adolescent suicides occur.
b. A randomly selected month will be one in which at least three suicides occur
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem using the Poisson distribution:
**a. Probability of Exactly Three Suicides:**
The Poisson probability formula is:
P(x) = (e^-μ * μ^x) / x!
Where:
* x is the number of events (suicides in this case)
* μ is the average number of events (2.75 in this case)
* e is the base of the natural logarithm (~2.71828)
* x! is the factorial of x
We want to find P(3), the probability of exactly three suicides:
P(3) = (e^-2.75 * 2.75^3) / 3!
P(3) = (0.0639 * 20.796875) / 6
P(3) = 1.3307 / 6
P(3) ≈ 0.2218
So, the probability that a randomly selected month will have exactly three suicides is approximately 0.2218 or 22.18%.
**b. Probability of At Least Three Suicides:**
To find the probability of at least three suicides, we need to calculate P(x ≥ 3). This is equal to 1 minus the probability of zero, one, or two suicides:
P(x ≥ 3) = 1 - [P(0) + P(1) + P(2)]
Let's calculate P(0), P(1), and P(2):
* P(0) = (e^-2.75 * 2.75^0) / 0! = (0.0639 * 1) / 1 ≈ 0.0639
* P(1) = (e^-2.75 * 2.75^1) / 1! = (0.0639 * 2.75) / 1 ≈ 0.1757
* P(2) = (e^-2.75 * 2.75^2) / 2! = (0.0639 * 7.5625) / 2 ≈ 0.2415
Now, substitute these values back into the equation for P(x ≥ 3):
P(x ≥ 3) = 1 - (0.0639 + 0.1757 + 0.2415)
P(x ≥ 3) = 1 - 0.4811
P(x ≥ 3) ≈ 0.5189
Therefore, the probability that a randomly selected month will have at least three suicides is approximately 0.5189 or 51.89%.
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