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In Exercises \(\mathrm{P} .74\) to \(\mathrm{P} .77\), fill in the \(?\) to make \(p(x)\) a probability function. If not possible, say so. $$ \begin{array}{lccccc} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline p(x) & 0.3 & ? & 0.3 & 0.3 & 0.3 \\ \hline \end{array} $$

Short Answer

Expert verified
It is not possible to make \(p(x)\) a valid probability function in this case.

Step by step solution

01

Sum Known Probabilities

Start by adding the known probabilities. These are 0.3 for \(p(1)\), 0.3 for \(p(3)\), 0.3 for \(p(4)\), and 0.3 for \(p(5)\). This will give you a total of 1.2.
02

Determine The Remaining Probability

Since the total probability must equal 1, subtract the sum obtained in Step 1 from 1. This would be \(1 - 1.2\), which reveals a negative value ( -0.2)
03

Check if Obtained Value is Within Valid Range

Since a probability can not be a negative number, the obtained value in Step 2 ( -0.2) is not valid. Therefore, it is impossible to fill in a probability for \(p(2)\) in such a way that \(p(x)\) becomes a valid probability function.

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics concerned with analyzing random events and determining the likelihood of various outcomes. It is foundational to understanding how to evaluate risk and uncertainty and is applicable across numerous fields, from finance to physics.

At its core, probability theory deals with the concept of a 'probability', which is a numerical value ranging between 0 and 1, inclusive. This value represents the measure of how likely an event is to occur. A probability of 0 shows that an event is impossible, while a probability of 1 indicates an event is certain to happen. Problems in probability theory often require calculating the probabilities of composite events, based on the likelihood of their individual components.

In the context of the provided textbook exercise, the exercise involves determining if a set of values can form a valid probability function for a discrete random variable. This requires understanding that the sum of probabilities for all possible outcomes of a random variable must equal 1, reflecting certainty that one of the outcomes will occur.
Probability Distribution
A probability distribution is a mathematical function that describes all the possible values and likelihoods that a random variable can take within a given range. Understanding probability distributions is crucial for interpreting and predicting the outcomes of random processes.

There are two main types of probability distributions: discrete and continuous. Discrete probability distributions apply to variables that have distinct, countable outcomes, such as the roll of a die. Continuous probability distributions, on the other hand, apply to variables that can take on a continuous range of values, like heights of people.

The exercise from the textbook is dealing with a discrete random variable, as evidenced by the distinct values that the variable 'x' can take. The task given was to determine the missing probability so that the probabilities add up to 1, adhering to the fundamental property of a probability distribution.
Discrete Random Variable
A discrete random variable is a type of random variable that possesses a countable number of distinct outcomes. Each of these outcomes has an associated probability that quantifies the chance of the outcome occurring. Examples of discrete random variables include the number of heads in a series of coin tosses, or the result of rolling a six-sided die.

In practice, when developing a probability function for a discrete random variable, the probabilities assigned to each possible outcome must sum to 1, ensuring the model is complete and accounts for all possible events. If the sum of the probabilities is less than or more than 1, it indicates an error in the model. As illustrated in the solution to the exercise, the sum of the known probabilities exceeded 1, which demonstrated that it is not possible to assign a valid probability to the remaining outcome, and thereby a valid probability function cannot exist.

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Most popular questions from this chapter

Average Household Size for Renter-Occupied Units Table \(\mathrm{P} .12\) in Exercise \(\mathrm{P} .83\) gives the probability function for the random variable giving the household size for a renter-occupied housing unit in the US. (a) Find the mean household size. (b) Find the standard deviation for household size.

Use the information that, for events \(\mathrm{A}\) and \(\mathrm{B}\), we have \(P(A)=0.8, P(B)=0.4\), and \(P(A\) and \(B)=0.25.\) Find \(P(A\) or \(B)\).

The Standard and Poor 500 (S\&P 500 ) is a weighted average of the stocks for 500 large companies in the United States. It is commonly used as a measure of the overall performance of the US stock market. Between January 1,2009 and January \(1,2012,\) the S\&P 500 increased for 423 of the 756 days that the stock market was open. We will investigate whether changes to the S\&P 500 are independent from day to day. This is important, because if changes are not independent, we should be able to use the performance on the current day to help predict performance on the next day. (a) What is the probability that the S\&P 500 increased on a randomly selected market day between January 1,2009 and January \(1,2012 ?\) (b) If we assume that daily changes to the \(S \& P\) 500 are independent, what is the probability that the S\&P 500 increases for two consecutive days? What is the probability that the S\&P 500 increases on a day, given that it increased the day before? (c) Between January 1, 2009 and January 1,2012 the S\&P 500 increased on two consecutive market days 234 times out of a possible \(755 .\) Based on this information, what is the probability that the S\&P 500 increases for two consecutive days? What is the probability that the S\&P 500 increases on a day, given that it increased the day before? d) Compare your answers to part (b) and part (c). Do you think that this analysis proves that daily changes to the S\&P 500 are not independent?

Ask you to convert an area from one normal distribution to an equivalent area for a different normal distribution. Draw sketches of both normal distributions, find and label the endpoints, and shade the regions on both curves. The lower \(10 \%\) for a standard normal distribution converted to a \(N(500,80)\) distribution

Ask you to convert an area from one normal distribution to an equivalent area for a different normal distribution. Draw sketches of both normal distributions, find and label the endpoints, and shade the regions on both curves. The area above 13.4 for a \(N(10,2)\) distribution converted to a standard normal distribution

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