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Is the expected value of a discrete random variable necessarily one of its possible values?

Short Answer

Expert verified
The expected value is not necessarily one of the possible values of a discrete random variable.

Step by step solution

01

Understanding Expected Value

The expected value of a discrete random variable is a weighted average of all possible values that the variable can take on, where each value is weighted by its probability of occurrence. Mathematically, it is given by the formula \( E(X) = \sum_{i=1}^{n} x_i P(x_i) \) where \(x_i\) are the possible values and \(P(x_i)\) is the probability of \(x_i\).
02

Recognizing Characteristics of Expected Value

The expected value is essentially an average, and like other statistical averages such as the mean, it does not have to correspond to an actual value that the random variable can take. Instead, it represents the central tendency of the possible values weighted by their probabilities.
03

Examining Through an Example

Consider a simple example where a discrete random variable \(X\) can take values {0, 1, 2}, with probabilities \(P(0)=0.6\), \(P(1)=0.3\), and \(P(2)=0.1\). The expected value is \(E(X) = 0 \times 0.6 + 1 \times 0.3 + 2 \times 0.1 = 0 + 0.3 + 0.2 = 0.5\). Here, 0.5 is not one of the possible values \(X\) can take, illustrating that the expected value does not need to be one of them.
04

Drawing a Conclusion

From the analysis, it is clear that the expected value summarizes the distribution's mean but does not imply it is one of the actual outcomes. The expected value represents a sort of 'balance point' of the probability distribution of the random variable.

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Key Concepts

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

Expected Value
The expected value of a discrete random variable represents a crucial concept in probability and statistics. Think of it as the average or mean value of a random variable, but with a twist. More specifically, it is a special kind of average that accounts for the likelihood of each outcome. Mathematically, the expected value (often denoted as \( E(X) \) for a random variable \( X \)) is calculated using the formula: \[ E(X) = \sum_{i=1}^{n} x_i P(x_i) \] where \( x_i \) are the possible values that \( X \) can take, and \( P(x_i) \) represents the probability that \( X \) will take on the value \( x_i \).
Delving deeper into this concept:
  • Expected value provides a measure of the center of the distribution of the random variable.
  • It acts as a "balance point," signifying the center of mass for the probability distribution.
  • Whether the expected value is an actual attainable value for the random variable is irrelevant - it's a theoretical average.
Even though it might not always be intuitive, this measure helps in understanding the distribution as a whole by providing insights into its overall behavior.
Probability Distribution
A probability distribution takes discrete random variables and shows how likely each of those variables' possible outcomes is to occur. Think of it as a comprehensive map of possibilities and their probabilities. In our context of discrete random variables, each possible value is associated with a specific probability, all of which must sum up to one. Here are key points about probability distributions:
  • They specify the probability of each possible outcome in the sample space.
  • All probabilities must add up to 1, ensuring completeness.
  • They are used to describe, analyze, and predict behavior in random processes.
Probability distributions are visualized in charts or tables, displaying each value alongside its probability. For discrete variables, these distributions provide a fundamental framework to calculate essential metrics like the expected value.
Ultimately, probability distributions serve as the bedrock for statistical analysis, enabling more complex calculations and predictions.
Weighted Average
A weighted average is an essential concept, tightly interconnected with expected values. Unlike a simple average, which treats all values equally, a weighted average assigns different levels of importance or significance to different values. In the context of expected values of discrete random variables, this "weight" comes from the probability attached to each outcome. Let's explore this concept with these points:
  • In a weighted average, each value impacts the result according to its assigned weight (probability).
  • The weighted average can provide a more accurate representation of the dataset when certain outcomes are likelier than others.
  • This approach underlies the calculation of expected values in probability, reinforcing why expected values are "weighted" averages of possible outcomes.
Weighted averages allow us to factor in the likelihood of different outcomes, ensuring more realistic and meaningful statistical results. Whenever you hear about expected values in probability, remember - it's just like taking a weighted average, where probabilities decide the weight each possible outcome has.

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Most popular questions from this chapter

A machine requires all seven of its micro-chips to operate correctly in order to be acceptable. The probability a micro-chip is operating correctly is \(0.99\). (a) What is the probability the machine is acceptable? (b) What is the probability that six of the seven chips are operating correctly? (c) The machine is redesigned so that the original seven chips are replaced by four new chips. The probability a new chip operates correctly is \(0.98\). Is the new design more or less reliable than the original?

The probability distribution of the random variable, \(y\), is given as \begin{tabular}{llllllll} \hline\(y\) & \(-3\) & \(-2\) & \(-1\) & 0 & 1 & 2 & 3 \\ \(P(y)\) & \(0.63\) & \(0.20\) & \(0.09\) & \(0.04\) & \(0.02\) & \(0.01\) & \(0.01\) \\ \hline \end{tabular} Calculate (a) \(P(y \geqslant 0)\) (b) \(P(y \leqslant 1)\) (c) \(P(|y| \leqslant 1)\) (d) \(P\left(y^{2}>3\right)\) (e) \(P\left(y^{2}<6\right)\)

A service engineer receives on average an emergency call every 3 hours. If the time between calls follows an exponential distribution, calculate the probability that the time from one emergency call to the next is (a) greater than 3 hours (b) less than \(4.5\) hours

The probability a chip is manufactured to an acceptable standard is \(0.87 .\) A sample of six chips is picked at random from a large batch. (a) Calculate the probability all six chips are acceptable. (b) Calculate the probability none of the chips is acceptable. (c) Calculate the probability that fewer than five chips in the sample are acceptable. (d) Calculate the most likely number of acceptable chips in the sample. (e) Calculate the probability that more than two chips are unacceptable.

The probability a component is acceptable is \(0.8\). Four components are sampled. What is the probability that (a) exactly one is acceptable (b) exactly two are acceptable?

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