Question

Let \(x\) be the number of courses for which a randomly selected student at a certain university is registered. The probability distribution of \(x\) appears in the following table: $$ \begin{array}{lrrrrrrr} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & .02 & .03 & .09 & .25 & .40 & .16 & .05 \end{array} $$ a. What is \(P(x=4)\) ? b. What is \(P(x \leq 4)\) ? c. What is the probability that the selected student is taking at most five courses? d. What is the probability that the selected student is taking at least five courses? more than five courses? e. Calculate \(P(3 \leq x \leq 6)\) and \(P(3

Short answer

a) .25, b) .39, c) .79, d) .61 (at least 5 courses) and .21 (more than 5 courses), e) .90 (probability that \(x\) is between 3 and 6 inclusive) and .65 (probability that \(x\) is between 4 and 5). The two probabilities in e) are different because the first includes the outcomes 3 and 6 while the second does not.

Step-by-step solution

Understand the probability distribution

The first step is to understand the probability distribution table provided. It consists of different values of \(x\), representing the number of courses, and corresponding probabilities \(p(x)\). This is a discrete distribution since it's possible to clearly list all the outcomes and their corresponding probabilities.

Calculate \(P(x=4)\)

This wants to know the probability that \(x\), the number of courses a student is registered for, is exactly 4. Referring back to the given probability distribution table, this probability is .25.

Calculate \(P(x \leq 4)\)

This is the probability that \(x\) is less than or equal to 4. According to the probability distribution table, this would be the sum of the probabilities of \(x\) being 1, 2, 3, or 4. So we add up the corresponding probabilities from the table: .02 + .03 + .09 + .25 = .39.

Calculate the probability of student taking at most 5 courses

This is asking for the probability that \(x\) is less than or equal to 5. We sum the probabilities of \(x\) being 1, 2, 3, 4, or 5 according to the table: .02 + .03 + .09 + .25 + .40 = .79.

Calculate the probability of student taking at least 5 courses, and more than 5 courses

The probability of a student taking at least 5 courses means considering the cases where \(x\) is 5, 6 or 7. We sum these respective probabilities: .40 + .16 + .05 = .61. For the probability of a student taking more than 5 courses, we consider only the cases where \(x\) is 6 or 7, and sum these probabilities: .16 + .05 = .21.

Calculate \(P(3 \leq x \leq 6)\) and \(P(3

The probability \(P(3 \leq x \leq 6)\) includes the cases where \(x\) is 3, 4, 5, or 6. We sum these probabilities: .09 + .25 + .40 + .16 = .90. The probability \(P(3<x<6)\) excludes the cases where \(x\) is 3 or 6, so we sum the probabilities when \(x\) is 4 or 5: .25 + .40 = .65. The difference between these two probabilities arises because the first condition includes \(x\) being 3 or 6, while the second one does not.

Key concepts

Discrete Distribution

In the realm of statistics, a discrete distribution represents a set of probabilities assigned to finite or countably infinite outcomes. To grasp this concept, imagine rolling a six-sided die. The possible outcomes are discrete and finite—1 through 6—and the probability of any given outcome is predictable and calculable.

When faced with a probability distribution table for a discrete random variable like the number of courses a student enrolls in, we see this principle in action. Each possible number of courses (1, 2, 3, and so on) is a discrete outcome with a specific probability associated with it. For instance, if a table shows that the probability (denoted as p(x)) of a student taking exactly 4 courses is 0.25, it means there is a 25% chance for that particular scenario.

Understanding a discrete distribution is crucial because it enables us to calculate probabilities for various scenarios involving finite, distinct events. Whether it's rolling dice, flipping coins, or selecting students, knowing how to work with this type of distribution is a building block for many statistical analyses.

Cumulative Probability

The concept of cumulative probability comes into play when we want to understand the likelihood of an event occurring up to a certain point. It’s akin to stacking blocks — with each additional block representing an outcome, the stack (cumulative probability) grows.

To determine the cumulative probability, we add up the probabilities for all the outcomes up to and including a certain point. Referring back to our discrete distribution of students and their courses, if we wish to find the cumulative probability that a student is registered for up to four courses, we sum the probabilities of being registered for 1, 2, 3, and 4 courses.

P(x \( \leq \) 4), therefore, is the sum of the probabilities for 1, 2, 3, and 4 courses — as was done in the textbook solution, resulting in 0.39. This approach to probability calculation helps us foresee the likelihood of a range of outcomes and is fundamental in both theoretical and applied statistics.

Probability Calculation

At the heart of statistical analysis lies probability calculation, a method that enables us to quantify the chance of an event occurring. Each event or outcome will have its probability, typically between 0 (impossible) and 1 (certain).

Probability calculations can become intricate, especially when looking for more conditional probabilities. Yet, with our discrete distribution of students’ course enrollments, the process is straightforward. To determine the probability of a single outcome, such as a student taking exactly 4 courses, we simply observe the corresponding probability in the table (0.25 for this case).

However, for ranges of outcomes, we tally up the individual probabilities within that range—a vital step for understanding the cumulative probability. Additionally, calculating probabilities for ‘at least’ or ‘more than’ scenarios requires adding probabilities from a higher threshold, like considering the chances of a student taking at least 5 courses, resulting in a different cumulative assessment. Through consistent and correct probability calculations, we provide a foundation for making predictions, analyzing patterns, and drawing meaningful conclusions from statistical data.