Question 1170116: A random variable X has a poisson distribution with parameter m. If P(X=2)=0.2565, find the value of m given that e=2.7183
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Alright, let's solve this Poisson distribution problem step-by-step.
**Understanding the Poisson Distribution**
The Poisson distribution gives the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. The probability mass function (PMF) of a Poisson distribution is:
$$P(X=k) = \frac{e^{-m} m^k}{k!}$$
where:
* X is the random variable representing the number of events.
* k is the number of events (non-negative integer).
* m is the average number of events (the parameter).
* e is Euler's number (approximately 2.7183).
**Applying the Given Information**
We are given that:
* X follows a Poisson distribution with parameter m.
* P(X=2) = 0.2565
We need to find the value of m.
Using the Poisson PMF with k=2:
$$P(X=2) = \frac{e^{-m} m^2}{2!}$$
We know P(X=2) = 0.2565, so:
$$0.2565 = \frac{e^{-m} m^2}{2}$$
Multiply both sides by 2:
$$0.513 = e^{-m} m^2$$
$$0.513 = \frac{m^2}{e^{m}}$$
Now, we need to solve for m. This equation is transcendental, meaning it can't be solved algebraically. We'll need to use numerical methods or approximations.
**Solving for m (Numerical Approximation)**
We'll use a trial-and-error approach, plugging in values of m to see which one gets us closest to 0.513.
Let's try m = 2:
$$\frac{2^2}{e^2} = \frac{4}{2.7183^2} \approx \frac{4}{7.389} \approx 0.5413$$
This is close.
Let's try m = 1.9:
$$\frac{1.9^2}{e^{1.9}} = \frac{3.61}{6.6858} \approx 0.540$$
Let's try m = 2.1:
$$\frac{2.1^2}{e^{2.1}} = \frac{4.41}{8.1662} \approx 0.540$$
Let's try m = 1.8:
$$\frac{1.8^2}{e^{1.8}} = \frac{3.24}{6.0496} \approx 0.5355$$
Let's try m = 1.6:
$$\frac{1.6^2}{e^{1.6}} = \frac{2.56}{4.9530} \approx 0.5168$$
Let's try m = 1.58:
$$\frac{1.58^2}{e^{1.58}} = \frac{2.4964}{4.8550} \approx 0.5142$$
Let's try m = 1.57:
$$\frac{1.57^2}{e^{1.57}} = \frac{2.4649}{4.8066} \approx 0.5128$$
Therefore, m is approximately 1.57.
**Final Answer**
The value of m is approximately 1.57.
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