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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: $$ \begin{array}{cccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & .02 & .03 & .09 & .25 & .40 & .16 & .05 \end{array} $$ 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
The probabilities are as follows: a. \(P(x=4) = 0.25\); b. \(P(x \leq 4) = 0.39\); c. \(P(x \leq 5) = 0.79\); d. \(P(x \geq 5) = 0.61\), \(P(x > 5) = 0.21\); e. \(P(3 \leq x \leq 6) = 0.90\), \(P(3 < x < 6) = 0.65\)

Step by step solution

01

- Identify the individual probabilities

The probabilities for each value of \(x\) are given in the table. We can obtain the values as follows: \(P(x=4) = 0.25\)
02

- Calculating Cumulative Probabilities

To answer questions regarding the cumulative probability (finding the probability that \(x\) takes a value less than or equal to a particular value), we need to add up the individual probabilities up to that point. This is done as follows: \(P(x \leq 4) = P(x=1) + P(x=2) + P(x=3) + P(x=4) = 0.02 + 0.03 + 0.09 + 0.25 = 0.39\)
03

- Calculating Other Cumulative Probabilities

We can use the same method to find the probability that a student is taking at most five or at least five courses. For at most five courses, we continue from where we left: \(P(x \leq 5) = P(x \leq 4) + P(x=5) = 0.39 + 0.40 = 0.79\). For at least five courses, it’s often easier to calculate the complement (probability that a student is taking less than five courses and subtract it from one}: \(P(x \geq 5) = 1 - P(x < 5) = 1 - P(x \leq 4) = 1 - 0.39 = 0.61\).
04

- Probability for More Than Five Courses

The question about a student taking more than five courses is just as easily answered with using the complement: \(P(x > 5) = 1 - P(x \leq 5) = 1 - 0.79 = 0.21\)
05

- Probabilities In an Interval

For \(P(3 \leq x \leq 6)\), we sum up the probabilities within that range: \(P(3 \leq x \leq 6) = P(x=3) + P(x=4) + P(x=5) + P(x=6) = 0.09 + 0.25 + 0.40 + 0.16 = 0.90\). For \(P(3<x<6)\), we don’t include \(P(x=3)\) and \(P(x=6)\), so our range only contains \(x = 4\) and \(x = 5\), thus: \(P(3 < x < 6) = P(x=4) + P(x=5) = 0.25 + 0.40 = 0.65\). The two probabilities differ because the first one includes \(x = 3\) and \(x = 6\), while the second one does not. The first is therefore higher.

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Key Concepts

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

Cumulative Probability
Understanding cumulative probability involves adding up the probabilities of specific outcomes in a distribution range. This measure provides a total probability for obtaining a value less than or equal to a particular outcome. In practice, if you want to find the probability that a student registers for four or fewer courses, you would sum the individual probabilities for one to four courses. This concept is essential when determining total likelihood over a range rather than a specific point.
In essence, cumulative probabilities answer questions like "What is the likelihood of having no more than this many outcomes?" or "What is the probability of getting this result or something less?" This is helpful in numerous real-world scenarios, such as predicting maximum capacities or thresholds.
Complement Rule
The complement rule is an efficient way to calculate probabilities by considering the opposite or complementary outcome. Instead of calculating directly, you find it through subtraction from one, the total probability. For example, rather than calculating the probability of having at least five courses directly, find the complement (probability of having fewer than five courses) and subtract it from 1.
This concept is particularly useful because complementary events by definition cover all possible outcomes. Using this rule can often simplify complex calculations and save time. It's a handy trick when you're dealing with probabilities where finding the complement is straightforward. Remember, the formula is:
  • Probability of A = 1 - Probability of not A
A clear understanding of this rule helps in scenarios where direct computation would be too cumbersome.
Probability Interval
Probability intervals involve calculating the probability within a specific range of outcomes. This differs from looking at individual probabilities as we're interested in a set or interval of numbers. For instance, calculating the probability that a student is registered for between three and six courses relies on summing the probabilities of courses three, four, five, and six registered.
Important to note is the difference when the boundaries of the interval can include or exclude endpoints; for example, "between three and six" versus "greater than three and less than six." Each requires a precise understanding of which values are included.
This concept is crucial in contexts where multiple outcomes should be considered together, like collective probabilities. Ensuring you correctly establish and interpret these intervals helps in accurate probability assessments.
Random Variable
A random variable is a quantitative variable whose value depends on the outcomes of a random phenomenon. In our context, it represents the number of courses for which a student is registered at a university. Each possible number of registrations translates into an associated probability.
Random variables help simplify analysis by allowing us to treat complex situations with simple models. Consistently, they're central to probability calculations and are foundational in statistical studies. They are described by their probability distributions, which list all possible values and their corresponding probabilities.
Gaining an understanding of random variables is fundamental since every statistical experiment involves them. They provide the framework to attach meaning to seemingly random occurrences, translating real-world processes into understandable numeric outcomes.

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Most popular questions from this chapter

Let \(y\) denote the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose that the probability distribution of \(y\) is as follows: \(\begin{array}{cccccc}y & 0 & 1 & 2 & 3 & 4 \\ p(y) & .65 & .20 & .10 & .04 & ?\end{array}\) a. Only \(y\) values of \(0,1,2,3\), and 4 have positive probabilities. What is \(p(4)\) ? b. How would you interpret \(p(1)=.20\) ? c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2)\), the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

Sophie is a dog that loves to play catch. Unfortunately, she isn't very good, and the probability that she catches a ball is only \(.1\). Let \(x\) be the number of tosses required until Sophie catches a ball. a. Does \(x\) have a binomial or a geometric distribution? b. What is the probability that it will take exactly two tosses for Sophie to catch a ball? c. What is the probability that more than three tosses will be required?

\({ A local television station sells } 15 \text { -second, }\end{array}\) 30 -second, and 60 -second advertising spots. Let \(x\) denote the length of a randomly selected commercial appearing on this station, and suppose that the probability distribution of \(x\) is given by the following table: $$ \begin{array}{llll} x & 15 & 30 & 60 \\ p(x) & .1 & .3 & .6 \end{array} $$ a. Find the average length for commercials appearing on this station. b. If a 15 -second spot sells for \(\$ 500\), a 30 -second spot for \(\$ 800\), and a 60 -second spot for \(\$ 1000\), find the average amount paid for commercials appearing on this station. (Hint: Consider a new variable, \(y=\) cost, and then find the probability distribution and mean value of \(y .\).)

According to the paper "Commuters' Exposure to Particulate Matter and Carbon Monoxide in Hanoi. Vietnam" (Transportation Research [2008]: 206-211), the carbon monoxide exposure of someone riding a motorbike for \(5 \mathrm{~km}\) on a highway in Hanoi is approximately normally distributed with a mean of \(18.6\) ppm. Suppose that the standard deviation of carbon monoxide exposure is \(5.7\) ppm. Approximately what proportion of those who ride a motorbike for \(5 \mathrm{~km}\) on a Hanoi highway will experience a carbon monoxide exposure of more than 20 ppm? More than 25 ppm?

The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged, the risk of heart problems is increased. The paper "Left Atrial Size Increases with Body Mass Index in Children" (International Journal of Cardiology [2009]: 1-7) described a study in which the left atrial size was measured for a large number of children age 5 to 15 years. Based on this data, the authors concluded that for healthy children, left atrial diameter was approximately normally distributed with a mean of \(26.4 \mathrm{~mm}\) and a standard deviation of \(4.2 \mathrm{~mm}\). a. Approximately what proportion of healthy children has left atrial diameters less than \(24 \mathrm{~mm}\) ? b. Approximately what proportion of healthy children has left atrial diameters greater than \(32 \mathrm{~mm}\) ? c. Approximately what proportion of healthy children has left atrial diameters between 25 and \(30 \mathrm{~mm}\) ? d. For healthy children, what is the value for which only about \(20 \%\) have a larger left atrial diameter?

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