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Consider the probability distribution of a random variable \(x\). Is the expected value of the distribution necessarily one of the possible values of \(x ?\) Explain or give an example.

Short Answer

Expert verified
No, the expected value can be a value not in the set of possible values for \(x\), like in the example where \(E(x) = 2\) but \(x\) can only be \(1\) or \(3\).

Step by step solution

01

Understanding the Concept

The expected value, denoted as \(E(x)\), of a random variable \(x\) is a measure of the center of the distribution of the random variable. It is calculated by summing the products of each possible value of \(x\) multiplied by their respective probabilities. This gives a weighted average, which is not necessarily one of the actual outcomes of the random variable.
02

Setting Up an Example

Consider a random variable \(x\) that takes values \(1\) and \(3\) with probabilities \(0.5\) and \(0.5\) respectively. The variable \(x\) does not take any other values.
03

Calculating the Expected Value

The expected value \(E(x)\) is given by the formula: \[ E(x) = \sum (x_i \cdot P(x_i)) \]For our distribution:\[ E(x) = 1 \cdot 0.5 + 3 \cdot 0.5 = 0.5 + 1.5 = 2 \]
04

Analyzing the Result

The expected value \(2\) is not one of the possible values \(1\) or \(3\) for the random variable \(x\). This example shows that the expected value of a probability distribution is not necessarily one of the possible values that \(x\) can take.

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Key Concepts

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

Understanding Random Variables
In probability and statistics, a random variable is a fundamental concept. It is a variable whose possible values are numerical outcomes of a random phenomenon. Random variables can be classified into two main types: discrete and continuous. A discrete random variable is one that has specific, separated values, like the roll of a die. Each side of the die represents a possible value: 1, 2, 3, 4, 5, or 6.
Continuous random variables, on the other hand, are those that can take any value within a given range, such as the height of students in a class or the time taken to run a race. Here, the potential outcomes are infinite and not countable.

A random variable is often denoted by symbols such as \(x\) or \(Y\). Understanding random variables is crucial when dealing with probability distributions since they model real-world random processes. They allow statisticians to use mathematical frameworks to examine how likely different outcomes are and make predictions based on this data.
Delving into Probability Distributions
Probability distributions describe how probabilities are distributed over the values of the random variable. In simple terms, they help us understand the likelihood of different outcomes when dealing with a random variable. Each outcome of the variable is assigned a probability, indicating the chance of its occurrence.
For discrete random variables, the probability distribution is often represented by a probability mass function (PMF). For each possible value of the random variable, the PMF provides a probability. An example is the roll of a fair six-sided die, where each of the six possible outcomes has a probability of \(\frac{1}{6}\).

In the case of continuous random variables, we use a probability density function (PDF). While PMFs give exact probabilities, PDFs provide a density, from which probabilities over an interval can be calculated. The area under the curve of a PDF over an interval represents the probability that the random variable falls within that interval.
Overall, probability distributions enable us to quantify uncertainty and make sense of how likely specific outcomes are.
Unraveling Weighted Averages
The concept of a weighted average plays a vital role in understanding expected values in probability distributions. A weighted average involves multiplying each value by a corresponding weight (or importance), summing up these products, and then dividing by the sum of the weights. This concept is clearly illustrated in calculating the expected value of a random variable.
In the context of random variables, each possible value of the variable is weighted by its probability of occurrence. This gives a comprehensive average that accounts for the different likelihood of each outcome.

  • Mathematically, the expected value \(E(x)\) of a discrete random variable \(x\) can be calculated as:
    \[ E(x) = \sum (x_i \cdot P(x_i)) \]
  • Here, \(x_i\) represents the possible values of the variable, and \(P(x_i)\) denotes the probabilities associated with these values.
Understanding weighted averages helps in realizing why the expected value does not have to be one of the possible values of the random variable. It is a balance point, or center, of the distribution that considers all possible outcomes and their probabilities.

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Most popular questions from this chapter

We now have the tools to solve the chapter Focus Problem. In the book \(A\) Guide to the Development and Use of the Myers-Briggs Type Indicators by Myers and McCaully, it was reported that approximately \(45 \%\) of all university professors are extroverted. Suppose you have classes with six different professors. (a) What is the probability that all six are extroverts? (b) What is the probability that none of your professors is an extrovert? (c) What is the probability that at least two of your professors are extroverts? (d) In a group of six professors selected at random, what is the expected number of extroverts? What is the standard deviation of the distribution? (e) Suppose you were assigned to write an article for the student newspaper and you were given a quota (by the editor) of interviewing at least three extroverted professors. How many professors selected at random would you need to interview to be at least \(90 \%\) sure of filling the quota?

Consider two binomial distributions, with \(n\) trials each. The first distribution has a higher probability of success on each trial than the second. How does the expected value of the first distribution compare to that of the second?

Only about \(70 \%\) of all donated human blood can be used in hospitals. The remaining \(30 \%\) cannot be used because of various infections in the blood. Suppose a blood bank has 10 newly donated pints of blood. Let \(r\) be a binomial random variable that represents the number of "good" pints that can be used. (a) Based on questionnaires completed by the donors, it is believed that at least 6 of the 10 pints are usable. What is the probability that at least 8 of the pints are usable, given this belief is true? Compute \(P(8 \leq r | 6 \leq r)\) (b) Assuming the belief that at least 6 of the pints are usable is true, what is the probability that all 10 pints can be used? Compute \(P(r=10 | 6 \leq r)\)

The Denver Post reported that a recent audit of Los Angeles 911 calls showed that \(85 \%\) were not emergencies. Suppose the 911 operators in Los Angeles have just received four calls. (a) What is the probability that all four calls are, in fact, emergencies? (b) What is the probability that three or more calls are not emergencies? (c) How many calls \(n\) would the 911 operators need to answer to be \(96 \%\) (or more) sure that at least one call is, in fact, an emergency?

Sara is a 60 -year-old Anglo female in reasonably good health. She wants to take out a 50,000 dollar term (i.e., straight death benefit) life insurance policy until she is \(65 .\) The policy will expire on her 65 th birthday. The probability of death in a given year is provided by the Vital Statistics Section of the Statistical Abstract of the United States (116th edition). $$\begin{array}{|l|lcccc|} \hline x=\text { age } & 60 & 61 & 62 & 63 & 64 \\\ \hline P( \text { death at this age) } & 0.00756 & 0.00825 & 0.00896 & 0.00965 & 0.01035 \\ \hline \end{array}$$ Sara is applying to Big Rock Insurance Company for her term insurance policy. (a) What is the probability that Sara will die in her 60 th year? Using this probability and the 50,000 dollar death benefit, what is the expected cost to Big Rock Insurance? (b) Repeat part (a) for years \(61,62,63,\) and \(64 .\) What would be the total expected cost to Big Rock Insurance over the years 60 through \(64 ?\) (c) If Big Rock Insurance wants to make a profit of 700 dollar above the expected total cost paid out for Sara's death, how much should it charge for the policy? (d) If Big Rock Insurance Company charges 5000 dollar for the policy, how much profit does the company expect to make?

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