Problem 15 - Applied Probability

Question

Suppose the number of Japanese visitors to Florida in a week is \(N 1 \sim\) Poisson (500) and the number of German visitors is \(N 2 \sim\) Poisson (300) . If 25 percent of the Japanese and 20 percent of the Germans visit Disney World, what is the distribution for the total number \(D\) of German and Japanese visitors to the park? Determine \(P(D \geq k)\) for \(k=150,155, \cdots, 245,250\).

Step-by-Step Solution

Answer

\(D\) is Poisson distributed with parameter 185. To find \(P(D \text{ is at least } k)\) for \(k=150,155,\cdots,245,250\), use \(1 - \text{Poisson CDF at } (k-1)\) with the mean 185.

Step 1: Understanding the distribution of visitors to Disney World

Visitors from Japan, denoted as \(N_1\), follows a Poisson distribution with a mean of 500, and visitors from Germany, denoted as \(N_2\), follows a Poisson distribution with a mean of 300. Since 25% of Japanese and 20% of German visitors go to Disney World, we find the expected number of visitors from each country: for Japanese visitors, \(E[D_1] = 0.25 \times 500 = 125\) and for German visitors, \(E[D_2] = 0.20 \times 300 = 60\). The total number of visitors \(D\) is the sum of \(D_1\) and \(D_2\), and the expected total, due to linearity of expectation, is \(E[D] = E[D_1] + E[D_2] = 125 + 60 = 185\).

Step 2: Define the distribution of the total number of visitors

Since \(N_1\) and \(N_2\) are independent, the total number \(D\) of visitors to Disney World that are either Japanese or German is also Poisson distributed, as the sum of two independent Poisson random variables is another Poisson random variable. The new mean for \(D\) is the sum of the individual means: \(\text{mean of } D = 125 + 60 = 185\). Thus, \(D \text{ has a Poisson distribution with parameter } 185\).

Step 3: Calculate the distribution function for a range of k

To find the probabilities \(P(D \text{ is at least } k)\) for \(k = 150, 155, \.\.\., 245, 250\), we use the Poisson cumulative distribution function (CDF). However, since we are interested in \(P(D \text{ is at least } k)\) and the CDF gives \(P(D \text{ is less than or equal to } k)\), we must subtract the CDF from 1: \(P(D \text{ is at least } k) = 1 - P(D \text{ is less than or equal to } k-1)\).

Step 4: Compute the probabilities

The probabilities are computed using a Poisson CDF table, software, or a calculator for each value of k in the specified range. If the value is not available directly from the table or software, the probability mass function (PMF) of the Poisson distribution can be summed instead. This sum or the use of the CDF gives the cumulative probabilities, which are then subtracted from 1 to obtain \(P(D \text{ is at least } k)\).

Key Concepts

Probability Mass Function

The probability mass function (PMF) is fundamental when discussing discrete random variables, such as those described by the Poisson distribution. The PMF outlines the probability that a random variable will take on a particular value. In the context of our exercise involving the number of visitors, the PMF for a Poisson distributed variable, given a mean rate of occurrence \(\lambda\), is expressed as

\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
For the number of visitors from Japan and Germany to Disney World, we calculate the probability of exactly \( k \) Japanese or German visitors on any given day using their respective PMFs. Since the PMF applies to exact values, it's different from the cumulative distribution function, which is explained further in the next section. It's crucial to note that PMFs are often used in conjunction with other statistical tools to understand the entirety of a distribution's characteristics.

Poisson Cumulative Distribution Function

Complementary to the PMF, the Poisson cumulative distribution function (CDF) represents the probability that the random variable will be less than or equal to a certain value. Mathematically, the CDF is defined as the sum of the probabilities for all outcomes up to and including \( k \):

\[ P(X \leq k) = \sum_{i=0}^{k} P(X=i) \]
In practice, this function is utilized to determine the probability of an event occurring a certain number of times or fewer within a fixed interval. For our example, the CDF calculates the chances that \( k \) or fewer visitors will show up at Disney World. We subtract this probability from one to determine \( P(D \geq k) \), hence for higher-than values as sought in the exercise. Understanding how to use and interpret the CDF is essential for solving problems that require cumulative probability assessments.

Independent Random Variables

The concept of independent random variables is pivotal in probability and statistics. Two random variables are independent if the outcome of one does not affect the outcome of the other. In our exercise, the number of Japanese visitors, \( N_1 \), and the number of German visitors, \( N_2 \), to Florida are treated as independent random variables. This assumption simplifies the process since it allows us to treat the events of Japanese and German visitors going to Disney World as unrelated occurrences. Consequently, the combined number of visitors, which is their sum, also follows a Poisson distribution with the rate given by the sum of their individual rates. This property stems directly from their independence and is critical for accurately modeling the scenario presented.

Linearity of Expectation

Linearity of expectation is a tremendously useful principle in probability theory that states the expected value of the sum of random variables is equal to the sum of their expected values, irrespective of their dependence or independence. To put it mathematically, if \( X \) and \( Y \) are two random variables, then:

\[ E[X + Y] = E[X] + E[Y] \]
Utilizing this property in our example, we determine the expected number of Japanese and German visitors to Disney World separately and then add those values to find the total expected number of visitors. It's essential to grasp that although independence is not required for linearity of expectation to hold, it does assume that the expected values of the individual random variables are known. This distinction is crucial because it means we can calculate expected values very efficiently, making linearity of expectation a powerful tool in solving various probabilistic problems.