Problem 18 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

EFFECT OF AN EPIDEMIC A study of the effect of an epidemic of mononucleosis (mono) on the students at a particular small private college determines that during the 30 days of November, there were: 5 days when no students had mono 7 days when exactly one student had mono 4 days when exactly two students had mono 9 days when exactly six students had mono 3 days when exactly seven students had mono 2 days when exactly eight students had mono Let \(X\) be the random variable that measures the number of students with mononucleosis on a randomly selected day in November. a. Find the probability distribution for \(X\). Then construct a histogram for this distribution. b. How many students would you expect to have mononucleosis on a randomly selected day in November?

Step-by-Step Solution

Answer

P(X=x) = \( \frac{days}{30} \). E(X) \approx 3.53.

Step 1 - Organize the Data

First, list the number of days corresponding to each number of students infected with mono. This helps in identifying the total number of days and calculating probabilities.

Step 2 - Calculate Total Number of Days

Sum all the days to find the total number of days in November. 5 + 7 + 4 + 9 + 3 + 2 = 30 days.

Step 3 - Find the Probability Distribution

For each number of students, divide the number of days by the total number of days.P(X=0) = \( \frac{5}{30} \)P(X=1) = \( \frac{7}{30} \)P(X=2) = \( \frac{4}{30} \)P(X=6) = \{ \frac{9}{30} \}P(X=7) = \{ \frac{3}{30} \}P(X=8) = \{ \frac{2}{30} \}

Step 4 - Construct Histogram

Create a histogram where the x-axis represents the number of students with mono and the y-axis represents the probability for each number. Plot the probabilities against the number of students.

Step 5 - Calculate Expected Value

Calculate the expected number of students with mono using the formula for expected value:\[ E(X) = \sum{[x * P(X=x)]} \]E(X) = (0 * \( \frac{5}{30} \)) + (1 * \( \frac{7}{30} \)) + (2 * \( \frac{4}{30} \)) + (6 * \( \frac{9}{30} \)) + (7 * \( \frac{3}{30} \)) + (8 * \( \frac{2}{30} \))

Step 6 - Simplify and Solve for Expected Value

Multiply and sum the values:E(X) = 0 + \( \frac{7}{30} \) + \( \frac{8}{30} \) + \( \frac{54}{30} \) + \( \frac{21}{30} \) + \( \frac{16}{30} \)E(X) = \( \frac{106}{30} \)E(X) \( \approx 3.53 \)

Key Concepts

Random Variable

A random variable is a variable that can take on different values, each with an associated probability. In this exercise, we let the random variable \(X\) represent the number of students who have mononucleosis (mono) on any given day in November.

This means \(X\) can take on values like 0, 1, 2, 6, 7, or 8, with each value corresponding to a specific probability. By defining a random variable, we can study the variability and the distribution associated with the number of students affected by mono in a structured way.

Understanding random variables is crucial as they allow us to use statistical methods to draw insights and make informed decisions based on the variability and the distribution of data.

Expected Value

The expected value of a random variable gives us a measure of the central tendency or the average outcome we can expect over the long run.

In our context, the expected value tells us the average number of students who would be affected by mono on a typical day in November. Mathematically, it's computed as follows:

\[ E(X) = \sum{[x \times P(X=x)]} \] Where \( x \) is each possible value of the random variable \( X \) and \( P(X=x) \) is the probability of \( X=x \).

As determined in the step-by-step solution, the specific expected value for our problem is approximately 3.53. This means that on average, about 3.53 students were expected to have mono on any given day in November.

Histogram

A histogram is a type of bar chart that represents the distribution of a dataset. Each bar in a histogram represents the frequency or probability of occurrences of different values of a random variable.

For this exercise, we create a histogram with the number of students affected by mono on the x-axis and the probability of each number on the y-axis. This visual representation helps to quickly identify how often certain numbers of students were affected over the month.

Creating a histogram involves the following steps:

• List the values \(X\) can take along the x-axis (0, 1, 2, 6, 7, 8).
• Calculate and plot the probability for each value on the y-axis.
• Draw bars to represent these probabilities.

By examining the histogram, you can understand at a glance the distribution of students affected by mono, helping in identifying patterns and making comparisons.

Discrete Probability Distribution

A discrete probability distribution describes the likelihood of occurrences of different outcomes for a discrete random variable. A discrete random variable can take on a finite or countable number of values, each associated with a specific probability.

In our exercise, the discrete probability distribution is composed of the probabilities of different numbers of students affected by mono on any given day in November.

These probabilities sum to 1, reflecting the completeness of the possible outcomes. The probability distribution for our problem is:

• \(P(X=0) = \frac{5}{30}\)
• \(P(X=1) = \frac{7}{30}\)
• \(P(X=2) = \frac{4}{30}\)
• \(P(X=6) = \frac{9}{30}\)
• \(P(X=7) = \frac{3}{30}\)
• \(P(X=8) = \frac{2}{30}\)

Understanding discrete probability distributions is essential as they provide a complete picture of all possible outcomes and their likelihood, useful for various applications including predicting future events and assessing risks.