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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

Expert verified
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

01

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.
02

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.
03

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.
04

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.
05

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.
06

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

An appliance dealer sells three different models of upright freezers having \(13.5,15.9\), and \(19.1\) cubic feet of storage space. Let \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: $$ \begin{array}{lrrr} x & 13.5 & 15.9 & 19.1 \\ p(x) & .2 & .5 & .3 \end{array} $$ a. Calculate the mean and standard deviation of \(x\). b. If the price of the freezer depends on the size of the storage space, \(x\), such that Price \(=25 x-8.5\), what is the mean value of the variable Price paid by the next customer? c. What is the standard deviation of the price paid?

An author has written a book and submitted it to a publisher. The publisher offers to print the book and gives the author the choice between a flat payment of $$\$ 10,000$$ and a royalty plan. Under the royalty plan the author would receive $$\$ 1$$ for each copy of the book sold. The author thinks that the following table gives the probability distribution of the variable \(x=\) the number of books that will be sold: $$ \begin{array}{lrrrr} x & 1000 & 5000 & 10,000 & 20,000 \\ p(x) & .05 & .30 & .40 & .25 \end{array} $$ Which payment plan should the author choose? Why?

The lifetime of a certain brand of battery is normally distributed with a mean value of \(6 \mathrm{hr}\) and a standard deviation of \(0.8 \mathrm{hr}\) when it is used in a particular cassette player. Suppose that two new batteries are independently selected and put into the player. The player ceases to function as soon as one of the batteries fails. a. What is the probability that the player functions for at least \(4 \mathrm{hr}\) ? b. What is the probability that the cassette player works for at most \(7 \mathrm{hr}\) ? c. Find a number \(z^{*}\) such that only \(5 \%\) of all cassette players will function without battery replacement for more \(\operatorname{than} z^{*} \mathrm{hr}\)

A mail-order computer software business has six telephone lines. Let \(x\) denote the number of lines in use at a specified time. The probability distribution of \(x\) is as follows: $$ \begin{array}{lrrrrrrr} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & .10 & .15 & .20 & .25 & .20 & .06 & .04 \end{array} $$ Write each of the following events in terms of \(x\), and then calculate the probability of each one: a. At most three lines are in use b. Fewer than three lines are in use c. At least three lines are in use d. Between two and five lines (inclusive) are in use e. Between two and four lines (inclusive) are not in use f. At least four lines are not in use

Because \(P(z<.44)=.67,67 \%\) of all \(z\) values are less than \(.44\), and \(.44\) is the 67 th percentile of the standard normal distribution. Determine the value of each of the following percentiles for the standard normal distribution (Hint: If the cumulative area that you must look for does not appear in the \(z\) table, use the closest entry): a. The 91 st percentile (Hint: Look for area \(.9100 .\) ) b. The 77 th percentile c. The 50 th percentile d. The 9 th percentile e. What is the relationship between the 70 th \(z\) percentile and the 30 th \(z\) percentile?

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