Problem 82 - Calculus for Biology and Medicine

Question

Suppose \(X\) is Poisson distributed with parameter \(\lambda=1.2 .\) Find the probability that \(X\) is at most 3 .

Step-by-Step Solution

Answer

The probability is approximately 0.9661.

Step 1: Understanding the problem

We are given a Poisson distribution with the parameter \( \lambda = 1.2 \). We need to find the probability that the random variable \( X \), which follows this distribution, is at most 3. This is written mathematically as \( P(X \leq 3) \).

Step 2: Applying the Poisson Probability Mass Function

The probability mass function for a Poisson distribution is given by: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where \( \lambda = 1.2 \) and \( k \) is the number of occurrences. We need to calculate this probability for \( k = 0, 1, 2, \text{ and } 3 \) and then sum them up to find \( P(X \leq 3) \).

Step 3: Calculating for k=0

Substitute \( k = 0 \) into the Poisson probability mass function: \[ P(X = 0) = \frac{e^{-1.2} \times 1.2^0}{0!} = e^{-1.2} \] Using a calculator, we find \( P(X = 0) \approx 0.3012 \).

Step 4: Calculating for k=1

Substitute \( k = 1 \) into the function: \[ P(X = 1) = \frac{e^{-1.2} \times 1.2^1}{1!} = 1.2e^{-1.2} \] Using a calculator, \( P(X = 1) \approx 0.3614 \).

Step 5: Calculating for k=2

Substitute \( k = 2 \) into the function: \[ P(X = 2) = \frac{e^{-1.2} \times 1.2^2}{2!} = \frac{1.44 e^{-1.2}}{2} \] Using a calculator, \( P(X = 2) \approx 0.2168 \).

Step 6: Calculating for k=3

Substitute \( k = 3 \) into the function: \[ P(X = 3) = \frac{e^{-1.2} \times 1.2^3}{3!} = \frac{1.728 e^{-1.2}}{6} \] Using a calculator, \( P(X = 3) \approx 0.0867 \).

Step 7: Summing the Probabilities

Now, sum the probabilities calculated for \( k = 0, 1, 2, \text{ and } 3 \): \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \approx 0.3012 + 0.3614 + 0.2168 + 0.0867 \approx 0.9661 \]

Step 8: Conclusion

The probability that \( X \) is at most 3 is approximately 0.9661. This means there is a 96.61% chance that \( X \) will be 3 or less.

Key Concepts

Probability Mass Function

The Probability Mass Function (PMF) is a fundamental concept when dealing with discrete probability distributions, such as the Poisson distribution. For a Poisson-distributed random variable, the PMF provides the probability of observing a specific number of events, say \(k\), in a fixed interval of time or space. In mathematical terms, it is expressed as: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where:

• \(e\) is the base of the natural logarithm, approximately equal to 2.71828.
• \(\lambda\) is a positive real number, known as the rate parameter.
• \(k!\) (k factorial) is the product of all positive integers up to \(k\).

The PMF tells us both the likelihood of different events occurring in your time/space interval and communicates these probabilities in a very actionable form. It is particularly powerful when predicting rare events, which are common use cases for the Poisson distribution.

Parameter Lambda

In the context of a Poisson distribution, the parameter \(\lambda\) is a pivotal term because it characterizes the distribution. It represents the average number of times an event occurs within a given timeframe or area. Understanding what \(\lambda\) stands for can help you interpret results much more effectively:

• When \(\lambda\) is small, events occur less frequently.
• When \(\lambda\) is larger, events occur more often.

The beauty of the parameter \(\lambda\) is that it fully defines the distribution. In fact, given \(\lambda\), any statistical property of the distribution can be determined. In practical terms, calculating with \(\lambda = 1.2\) implies that such a scenario involves an average of 1.2 occurrences in the defined interval. When using the PMF, you plug \(\lambda\) into the function to determine the probability of different \(k\) occurrences.

Summation of Probabilities

To find the probability of a Poisson-distributed variable being at most a certain value, such as 3 in this instance, involves summing multiple probabilities. Individual PMF values for each \(k\) are combined in such a way to yield cumulative probabilities:

• Begin by calculating probabilities for \(k = 0, 1, 2, \) and \(3\).
• Then, sum these probabilities: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]

This process results in a cumulative probability, offering an encompassing likelihood across all events up to 3. Through careful summation, this provides a well-rounded picture of how likely an array of outcomes can occur, rather than just a singular outcome. In simple language, if you sum up to \(\leq 3\), you'll see a probability of approximately 0.9661 or 96.61%. This approach answers questions about ranges of possible outcomes, rather than a stand-alone event, which is a strong technique within probability theory.