Problem 80 - Calculus for Biology and Medicine

Question

Suppose \(X\) is Poisson distributed with parameter \(\lambda=0.2\). (a) Find \(P(X<3)\). (b) Find \(P(2 \leq X \leq 4)\).

Step-by-Step Solution

Answer

\(P(X<3) \approx 0.9988\); \(P(2 \leq X \leq 4) \approx 0.0176\).

Step 1: Understanding the Poisson distribution

The Poisson distribution models the number of events occurring in a fixed interval of time/space where these events happen with a known constant mean rate and independently of the time since the last event. Given that \( X \) follows a Poisson distribution with parameter \( \lambda = 0.2 \), the probability of observing \( k \) events is given by the formula: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \].

Step 2: Finding \(P(X<3)\) using cumulative probabilities

To find \(P(X<3)\), which is equivalent to \(P(X=0) + P(X=1) + P(X=2)\), we compute each: \[ P(X=0) = \frac{e^{-0.2} (0.2)^0}{0!} = e^{-0.2} \approx 0.8187 \], \[ P(X=1) = \frac{e^{-0.2} (0.2)^1}{1!} = 0.2e^{-0.2} \approx 0.1637 \], \[ P(X=2) = \frac{e^{-0.2} (0.2)^2}{2!} = 0.02e^{-0.2} \approx 0.0164 \]. Sum these probabilities to find \(P(X<3)\): \[ P(X<3) \approx 0.8187 + 0.1637 + 0.0164 = 0.9988 \].

Step 3: Finding \(P(2 \leq X \leq 4)\) by individual probabilities

To compute \(P(2 \leq X \leq 4)\), specifically \(P(X=2) + P(X=3) + P(X=4)\), we use the Poisson probability formula: we have already calculated \(P(X=2)\): \[ P(X=2) = 0.0164 \]. Next, compute for \( k = 3 \) and \( k = 4 \):\[ P(X=3) = \frac{e^{-0.2} (0.2)^3}{3!} = 0.0011 \],\[ P(X=4) = \frac{e^{-0.2} (0.2)^4}{4!} = 0.000055 \]. Then, sum the probabilities: \[ P(2 \leq X \leq 4) \approx 0.0164 + 0.0011 + 0.000055 = 0.017555 \].

Step 4: Conclusion

By completing the above calculations, we find that the probability of \(X<3\) is approximately \(0.9988\), and the probability of \(2 \leq X \leq 4\) is approximately \(0.017555\).

Key Concepts

Probability Calculations in Poisson Distribution

In the realm of statistics, probability calculations are pivotal, especially in understanding distributions like the Poisson distribution.
Understanding this distribution involves determining the likelihood of a certain number of events occurring within a fixed period.
In the Poisson context, this is represented by the probability formula:

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

Here, \( X \) represents the number of events, \( \lambda \) is the average rate of occurrence, and \( k! \) is the factorial of \( k \).

To compute a probability using this formula, substitute the values of \( \lambda \) and \( k \) as shown in the exercise.
For example, to find \( P(X=2) \), replace \( k \) with 2 and \( \lambda \) with 0.2.
This yields \( P(X=2) = 0.0164 \), indicating the probability of exactly 2 events occurring is around 1.64%.
Calculating probabilities using this method helps in defining the distribution of events over time or space with precision.

Understanding Cumulative Probability

Cumulative probability is an aggregation of probabilities up to a certain point.
In the Poisson distribution, it refers to the likelihood of observing a certain number of events or fewer.
For example, to find the cumulative probability for \( X<3 \), you sum the probabilities \( P(X=0) \), \( P(X=1) \), and \( P(X=2) \).

• \[ P(X<3) = P(X=0) + P(X=1) + P(X=2) \]

As calculated in the solution, this summation results in approximately 0.9988.
This means that there's about a 99.88% chance of observing fewer than 3 events.

Cumulative probabilities are crucial in scenarios where the interest lies in the probability of a number of events not exceeding a threshold.
They are commonly used to illustrate the failure rates in reliability tests or queues in operations research.

Exploring Parameter Lambda (\(\lambda\))

The parameter \( \lambda \) is central to the Poisson distribution.
It represents the expected number of events in a given interval.
In simpler terms, \( \lambda \) is the average rate at which events occur.

For example, if \( \lambda = 0.2 \) as in the given exercise, it signifies an average of 0.2 events per unit of time or space.
The value of \( \lambda \) heavily influences the shape and spread of the Poisson distribution.

• A small \( \lambda \) results in a distribution skewed towards lower event counts.
• Higher values of \( \lambda \) would indicate distributions with more frequent occurrences.

Understanding \( \lambda \) is crucial because it defines how likely certain outcomes are.
It allows statisticians to predict the occurrence and to model real-world scenarios accurately, like traffic flow or birth rates.