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

(a) \(P(X<3) \approx 0.9988\). (b) \(P(2 \leq X \leq 4) \approx 0.01756\).

Step 1: Understanding Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space. The parameter \(\lambda\) represents the average number of events in the interval. Here, \(\lambda = 0.2\).

Step 2: Calculate Individual Probabilities for Part (a)

First, we calculate the probabilities for \(X = 0\), \(X = 1\), and \(X = 2\) using the Poisson probability mass function: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] For \(X = 0\): \[ P(X = 0) = \frac{e^{-0.2} (0.2)^0}{0!} = e^{-0.2} \approx 0.8187 \] For \(X = 1\): \[ P(X = 1) = \frac{e^{-0.2} (0.2)^1}{1!} = 0.2 \times e^{-0.2} \approx 0.1637 \] For \(X = 2\): \[ P(X = 2) = \frac{e^{-0.2} (0.2)^2}{2!} = 0.02 \times e^{-0.2} \approx 0.0164 \]

Step 3: Sum the Probabilities for Part (a)

To find \(P(X < 3)\), we add the probabilities from \(X = 0\) to \(X = 2\):\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.8187 + 0.1637 + 0.0164 = 0.9988 \]

Step 4: Calculate Individual Probabilities for Part (b)

Now, calculate the probabilities for \(X = 2\), \(X = 3\), and \(X = 4\) using the Poisson formula: We already have \( P(X = 2) \approx 0.0164 \).For \(X = 3\): \[ P(X = 3) = \frac{e^{-0.2} (0.2)^3}{3!} = 0.00133 \times e^{-0.2} \approx 0.0011 \] For \(X = 4\): \[ P(X = 4) = \frac{e^{-0.2} (0.2)^4}{4!} = 0.000133 \times e^{-0.2} \approx 0.00006 \]

Step 5: Sum the Probabilities for Part (b)

To find \(P(2 \leq X \leq 4)\), add the probabilities for \(X = 2\) to \(X = 4\): \[ P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) \approx 0.0164 + 0.0011 + 0.00006 = 0.01756 \]

Key Concepts

Probability Distribution

A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. In simpler terms, it tells us how the probabilities are distributed over the values that a random variable can take.
A key feature of a probability distribution is its total probability, which always sums up to 1. In a probabilistic experiment, this ensures that all potential outcomes are accounted for.
The Poisson distribution, specifically, is a type of probability distribution used for counting the number of events in a fixed interval of time or space. It requires an average rate, denoted by \( \lambda \), representing the average number of occurrences within that interval.

Discrete Probability

Discrete probability deals with scenarios where the set of possible outcomes is finite or countable. This makes it distinct from continuous probability, which deals with scenarios involving continuous outcomes.
In discrete setups, outcomes often relate to counts of events or specific values that a random variable can assume.

• For instance, the number of times a die lands on a six when rolled multiple times.
• Another example can be the number of emails you receive in an hour.

Discrete probability uses specific functions to determine the likelihood of each result. In the context of the provided exercise, the Poisson distribution serves as that function to describe the count of events in discrete intervals.

Probability Mass Function

A probability mass function (PMF) is a function that gives the probability that a discrete random variable is exactly equal to some value. For discrete distributions like the Poisson, the PMF is crucial to calculate the likelihood of different outcomes.

• The PMF of a discrete random variable \(X\) is denoted as \(P(X = k)\).
• The formula for the Poisson distribution’s PMF is \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \).

This formula is used to calculate the probability of counting a certain number of events within a fixed interval. For each non-negative integer \(k\), the PMF provides the exact probability of observing \(k\) occurrences, based on the average rate \(\lambda\). This is how we compute probabilities like those required in the exercise for different ranges of \(X\).