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Use the probability distribution for the random variable \(x\) to answer the questions in Exercises 17-21. $$\begin{array}{l|lllll}x & 0 & 1 & 2 & 3 & 4 \\\\\hline p(x) & .1 & .3 & .3 & ? & .1\end{array}$$ Find \(p(3)\)

Short Answer

Expert verified
Answer: The probability p(3) is 0.2.

Step by step solution

01

Identify the known probabilities

From the given probability distribution table, we have the following probabilities: \(p(0) = 0.1\), \(p(1) = 0.3\), \(p(2) = 0.3\), and \(p(4) = 0.1\)
02

Sum up the known probabilities

Next, we sum up the probabilities of the values we know: \(0.1 + 0.3 + 0.3 + 0.1 = 0.8\)
03

Calculate the missing probability

Since the total probability must be equal to 1, we can find the missing probability \(p(3)\) by subtracting the sum of known probabilities from 1: \(p(3) = 1 - 0.8 = 0.2\) Hence, the probability \(p(3)\) is 0.2.

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Key Concepts

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

Random Variable
In understanding probability distribution, it is essential to grasp the concept of a random variable. A random variable is a numerical description of the outcome of a statistical experiment. In simpler terms, it is a variable whose possible values are numerical outcomes of a random phenomenon.

There are two types of random variables: discrete and continuous. A discrete random variable has a countable number of possible values, like the result of rolling a die (1, 2, 3, 4, 5, 6). On the other hand, a continuous random variable has an infinite number of possible values within a given range, such as the exact height of students in a school.

Each random variable has an associated probability distribution that provides the probabilities of its potential values. In our exercise, the random variable 'x' can take on the values 0, 1, 2, 3, or 4, which are discrete outcomes. The function p(x) maps each outcome with its probability, fundamentally outlining the fabric of probability theory.
Probability Theory
Probability theory is the branch of mathematics that deals with the analysis of random events. The core of the theory lies in quantifying the likelihood of various outcomes. This ranges from the probability of a simple event, like flipping a coin, to the more complex, such as forecasting weather patterns.

An understanding of probability theory aids in comprehending events that do not have a deterministic outcome. It's fundamentally based on set theory and involves elements like sample spaces, events, and the axioms of probability. The theory asserts that the sum of probabilities of all possible outcomes in a sample space is equal to 1.

In the context of the textbook exercise, probability theory guides us through the process of determining the probability of the random variable 'x' taking on the value 3. By understanding that the probabilities of all possible outcomes must add up to 1, we applied this knowledge to find that missing value.
Statistical Probability
Statistical probability refers to the use of probability theory to analyze statistical experiments, where the outcome is not deterministic but is subject to chance. This facet of mathematics is crucial in predicting the behavior of systems and making informed decisions based on likelihoods.

Statistical probability encompasses collecting data, summarizing it, and making inferences about a population based on sampled data. When we looked at the probability distribution table in our exercise, we were essentially engaging with statistical probability. We had a set of outcomes (x) with their respective probabilities which, when correctly determined, should adhere to the rules of statistical probability, such as all probabilities adding up to 1.

By following this rule, we resolved the textbook problem. However, an important practical takeaway is that understanding statistical probability can significantly enhance the analysis and interpretation of real-world phenomena, from genetics to finance and beyond.

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Most popular questions from this chapter

Let \(x\) be the number of successes observed in a sample of \(n=4\) items selected from a population of \(N=8 .\) Suppose that of the \(N=8\) items, \(M=5\) are considered "successes." Find the probabilities in Exercises \(8-10 .\) The probability of observing all successes.

In 2017 , the average of the revised SAT score (Evidence Based Reading and Writing, and Math) was 1060 out of \(1600 .^{3}\) Suppose that \(45 \%\) of all high school graduates took this test and that 100 high school graduates are randomly selected from throughout the United States. Which of the following random variables have an approximate binomial distribution? If possible, give the values of \(n\) and \(p\). a. The number of students who took the SAT. b. The scores of the 100 students on the SAT. c. The number of students who scored above average on the SAT. d. The length of time it took students to complete the SAT.

The number of visits to a website is known to have a Poisson distribution with a mean of 8 visits per minute. a. What is the probability distribution for \(x\), the number of visits per minute? b. What is the probability that the number of visits per minute is less than or equal to \(12 ?\) c. What is the probability that the number of visits per minute is greater than \(16 ?\) d. Within what limits would you expect the number of visits to this website to lie at least \(89 \%\) of the time?

Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that their offspring will develop the disease is approximately .25. Suppose a husband and wife are both carriers of the disease and the wife is pregnant on three different occasions. If the occurrence of Tay-Sachs in any one offspring is independent of the occurrence in any other, what are the probabilities of these events? a. All three children will develop Tay-Sachs disease. b. Only one child will develop Tay-Sachs disease. c. The third child will develop Tay-Sachs disease, given that the first two did not.

Talking or texting on your cell phone can be hazardous to your health! A snapshot in USA Today reports that approximately \(23 \%\) of cell phone owners have walked into someone or something while talking on their phones. A random sample of \(n=8\) cell phone owners were asked if they had ever walked into something or someone while talking on their cell phone. The following printout shows the cumulative and individual probabilities for a binomial random variable with \(n=8\) and \(p=.23 .\) Cumulative Distribution Function Binomial with \(\mathrm{n}=8\) and \(\mathrm{p}=0.23\) $$ \begin{array}{rl} \text { X } & P(X \leq X) \\ \hline 0 & 0.12357 \\ 1 & 0.41887 \\ 2 & 0.72758 \\ 3 & 0.91201 \\ 4 & 0.98087 \\ 5 & 0.99732 \\ 6 & 0.99978 \\ 7 & 0.99999 \\ 8 & 1.00000 \end{array} $$ Probability Density Function Binomial with \(n=8\) and \(p=0.23\) $$ \begin{aligned} &\begin{array}{cc} x & P(X=x) \\ \hline 0 & 0.123574 \end{array}\\\ &\begin{array}{l} 0 & 0.123574 \\ 1 & 0.295293 \\ 2 & 0.308715 \\ 3 & 0.184427 \\ 4 & 0.068861 \\ 5 & 0.016455 \\ 6 & 0.002458 \\ 7 & 0.000210 \\ 8 & 0.000008 \end{array} \end{aligned} $$ a. Use the binomial formula to find the probability that one of the eight have walked into someone or something while talking on their cell phone. b. Confirm the results of part a using the printout. c. What is the probability that at least two of the eight have walked into someone or something while talking on their cell phone.

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