If X is a Poisson random variate with mean 3, then P(|X- 3| < 1) will be:
Understanding the Poisson Distribution Problem
This problem involves a Poisson random variable. The Poisson distribution is often used to model the number of events occurring in a fixed interval of time or space, given the average rate of occurrence.
We are given that X is a Poisson random variate with a mean of 3. The mean of a Poisson distribution is denoted by $\lambda$. So, here, $\lambda = 3$.
The probability mass function (PMF) of a Poisson distribution is given by:
\(P(X=k) = \dfrac{e^{-\lambda} \lambda^k}{k!}\)
where:
- \(k\) is the number of occurrences (\(k = 0, 1, 2, \ldots\))
- \(e\) is the base of the natural logarithm (approximately 2.71828)
- \(\lambda\) is the average rate of occurrence (the mean)
- \(k!\) is the factorial of \(k\)
In this specific problem, $\lambda = 3$, so the PMF for X is:
\(P(X=k) = \dfrac{e^{-3} 3^k}{k!}\)
Calculating the Probability P(|X- 3| < 1)
We need to find the probability \(P(|X- 3| < 1)\). Let's first understand the inequality \(|X- 3| < 1\). This inequality means that the distance between X and 3 is less than 1.
We can rewrite the absolute value inequality as:
\(-1 < X - 3 < 1\)
To isolate X, we add 3 to all parts of the inequality:
\(-1 + 3 < X - 3 + 3 < 1 + 3\)
\(2 < X < 4\)
Since X is a Poisson random variate, it can only take non-negative integer values (0, 1, 2, 3, ...). The condition \(2 < X < 4\) means that X must be an integer strictly greater than 2 and strictly less than 4. The only integer value that satisfies this condition is X = 3.
Therefore, the probability \(P(|X- 3| < 1)\) is the same as the probability \(P(X=3)\).
Step-by-Step Calculation of P(X=3)
Now, we use the PMF with $\lambda = 3$ and \(k=3\) to find \(P(X=3)\):
\(P(X=3) = \dfrac{e^{-3} 3^3}{3!}\)
Let's calculate the values:
- \(3^3 = 3 \times 3 \times 3 = 27\)
- \(3! = 3 \times 2 \times 1 = 6\)
Substitute these values back into the PMF formula:
\(P(X=3) = \dfrac{e^{-3} \times 27}{6}\)
We can simplify the fraction \(\dfrac{27}{6}\) by dividing both the numerator and the denominator by their greatest common divisor, which is 3:
\(\dfrac{27}{6} = \dfrac{27 \div 3}{6 \div 3} = \dfrac{9}{2}\)
So, the probability is:
\(P(X=3) = \dfrac{9}{2} e^{-3}\)
This is the required probability \(P(|X- 3| < 1)\).
Revision Table: Poisson Distribution
| Concept | Description | Formula/Property |
|---|---|---|
| Definition | A discrete probability distribution that expresses 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. | |
| Parameter | Mean ($\lambda$) - the average number of events in the given interval. | $\lambda > 0$ |
| PMF | Probability of observing exactly \(k\) events. | \(P(X=k) = \dfrac{e^{-\lambda} \lambda^k}{k!}\), for \(k = 0, 1, 2, \ldots\) |
| Mean | Expected value of the random variable. | $E(X) = \lambda$ |
| Variance | Measure of spread. | $Var(X) = \lambda$ |
Additional Information: Properties of Probability
When working with probabilities, especially involving inequalities for discrete random variables, it's crucial to correctly identify the specific integer values included in the range. In this case, the inequality \(2 < X < 4\) for an integer random variable X only includes X = 3.
Key properties used or relevant here:
- The probability of a specific outcome for a discrete variable is given by its PMF.
- Absolute value inequalities like \(|Y - c| < d\) can be rewritten as \(-d < Y - c < d\).
- For discrete variables, inequalities like \(a < X < b\) correspond to the sum of probabilities for integers \(k\) such that \(a < k < b\).
Understanding these basic probability concepts and the specific properties of the Poisson distribution allows us to solve problems like this effectively.
Important Questions from Discrete Distributions
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X: 0 1 2 P(X = x): 3a 3b 4b with mean 1.1, is:
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. Let A denote the event that the even number of buffers are full. Then p(A) is :Let x ∼ N(μ, σ2) If μ2 = σ2, (μ > 0), then the value of P(X < -μ | X < μ) in terms of cumulative function N (0, 1) is:
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- Identify the generic probability density function that corresponds with discrete random variables.
