Problem 83 - Calculus for Biology and Medicine

Question

Suppose \(X\) is Poisson distributed with parameter \(\lambda=2\). Find the probability that \(X\) is at least \(2 .\)

Step-by-Step Solution

Answer

The probability that \(X\) is at least 2 is approximately 0.594.

Step 1: Understanding the Poisson distribution

A Poisson random variable, like \(X\), describes the number of events occurring within a fixed interval of time or space. The probability mass function (PMF) of a Poisson distribution is given by:\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]where \(e\) is the base of natural logarithms, \(\lambda\) is the average rate (or parameter), and \(k\) is the number of occurrences.

Step 2: Define the probability of interest

We need to find \(P(X \geq 2)\). This can be found by subtracting the probabilities of \(X < 2\) from 1. So, \(P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1))\).

Step 3: Calculate \(P(X = 0)\)

Using the PMF formula, compute the probability that \(X\) is 0:\[ P(X = 0) = \frac{e^{-2} \times 2^0}{0!} = e^{-2} \times 1 = e^{-2} \approx 0.1353. \]

Step 4: Calculate \(P(X = 1)\)

Using the PMF formula, compute the probability that \(X\) is 1:\[ P(X = 1) = \frac{e^{-2} \times 2^1}{1!} = 2e^{-2} \approx 0.2707. \]

Step 5: Compute \(P(X < 2)\)

Add the probabilities found for \(X = 0\) and \(X = 1\):\[ P(X < 2) = P(X = 0) + P(X = 1) = 0.1353 + 0.2707 = 0.406. \]

Step 6: Find \(P(X \geq 2)\)

Subtract \(P(X < 2)\) from 1 to find \(P(X \geq 2)\):\[ P(X \geq 2) = 1 - P(X < 2) = 1 - 0.406 = 0.594. \]

Key Concepts

Understanding the Probability Mass Function

The probability mass function (PMF) is a fundamental concept in the world of probability distributions. It provides the probability of a discrete random variable taking on a particular value. For the Poisson distribution, the PMF is given by the formula:

• \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \)

Here, \(e\) represents the base of the natural logarithm, approximately 2.718, which is a constant. The integer \(k\) indicates the number of times an event occurs in the interval, and \(\lambda\) signifies the average number of events expected.

The PMF for the Poisson distribution helps us calculate the likelihood of observing exactly \(k\) events when the average number of occurrences is known. This function is especially useful in situations where events happen independently over a fixed space or time.

When using the PMF, remember that \(k\) must be a non-negative integer (like 0, 1, 2, etc.), because it counts occurrences, which can't be negative.

Exploring the Parameter Lambda

In a Poisson distribution, the parameter \( \lambda \) is crucial. It defines the average rate at which events occur over a given interval. In simple words, it tells us about the expected number of occurrences.

• If \( \lambda \) is small, the events happen infrequently and are more sporadic.
• Conversely, a large \( \lambda \) indicates that events are frequent.

For our exercise, \( \lambda = 2 \), meaning we expect about 2 events on average within the specified interval. The Poisson distribution becomes smoother and more spread out with larger values of \( \lambda \). But when \( \lambda \) is small, the distribution is more skewed, with higher probabilities for values close to zero.

This parameter greatly influences the shape and behavior of the Poisson distribution. Adjusting \( \lambda \) changes how likely various numbers of events are, which in turn affects the probabilities calculated using the PMF.

Calculating Probabilities with the Poisson Model

Calculating probabilities using the Poisson distribution involves using the PMF and given \( \lambda \). In our case, we need to find the probability that the random variable \( X \) is at least 2, symbolized as \( P(X \geq 2) \). This requires us to:

• First determine the probability for \( X = 0 \)
• Then calculate the probability for \( X = 1 \)
• Add these results to find \( P(X < 2) \), as it's easier to calculate \( P(X < 2) \) and subtract from 1

Applying the PMF formula:

• \( P(X = 0) = e^{-2} \times \frac{2^0}{0!} \approx 0.1353 \)
• \( P(X = 1) = e^{-2} \times \frac{2^1}{1!} \approx 0.2707 \)

After finding these, sum them up:

• \( P(X < 2) = 0.1353 + 0.2707 = 0.406 \)

Finally, use the complement rule:

• \( P(X \geq 2) = 1 - P(X < 2) = 1 - 0.406 = 0.594 \)

This step-by-step way allows for precise probability calculation by effectively using the PMF, making even complex computations manageable.