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Chapter 8: Problem 14

Calculate the expected value of \(X\) for the given probability distribution. $$ \begin{array}{|c|c|c|c|c|} \hline x & 2 & 4 & 6 & 8 \\ \hline P(X=x) & \frac{1}{20} & \frac{15}{20} & \frac{2}{20} & \frac{2}{20} \\ \hline \end{array} $$

Short Answer

Expert verified
The expected value of \(X\) for the given probability distribution is \(E(X) = 4.5\).

Step by step solution

01

Identify the given values and probabilities

We are given the following probability distribution: $$ \begin{array}{|c|c|c|c|c|} \hline x & 2 & 4 & 6 & 8 \\ \hline P(X=x) & \frac{1}{20} & \frac{15}{20} & \frac{2}{20} & \frac{2}{20} \\ \hline \end{array} $$ From the table, we can see that the values of \(X\) are 2, 4, 6, and 8, with corresponding probabilities of \(\frac{1}{20}\), \(\frac{15}{20}\), \(\frac{2}{20}\), and \(\frac{2}{20}\).
02

Apply the formula for the expected value

We will now apply the formula for the expected value: \[E(X) = \sum_{i} x_i P(X=x_i)\] Inserting the given values into the formula, we get: \[E(X) = 2 \times \frac{1}{20} + 4 \times \frac{15}{20} + 6 \times \frac{2}{20} + 8 \times \frac{2}{20}\]
03

Calculate the expected value

Now, we can calculate the values in the expression: \begin{align*} E(X) &= 2 \times \frac{1}{20} + 4 \times \frac{15}{20} + 6 \times \frac{2}{20} + 8 \times \frac{2}{20} \\ &= \frac{2}{20} + \frac{60}{20} + \frac{12}{20} + \frac{16}{20} \\ &= \frac{90}{20} \end{align*} Therefore, the expected value of \(X\) is: \[E(X) = \frac{90}{20} = 4.5\]

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Key Concepts

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

Probability Distribution
In mathematics and statistics, a probability distribution is a table or a function that provides the possible values of a random variable and their corresponding probabilities. When working with discrete random variables, we use discrete probability distributions. These show a list of possible outcomes along with their probabilities.

A probability distribution must satisfy the following conditions:
  • The sum of the probabilities for all possible outcomes must be equal to 1.
  • The probability for each outcome must be between 0 and 1, inclusive.
For instance, in the given exercise, the possible values for the random variable X are 2, 4, 6, and 8. Each value has its own probability assigned. The probability distribution provides this information in a concise and systematic way. Understanding these distributions helps in calculating expected values and other important statistical measures.
Random Variable
A random variable is a variable that takes on numerical values determined by the outcome of a random phenomenon. There are two types of random variables: discrete and continuous. A discrete random variable is one that has a finite or countable number of possible values. On the other hand, a continuous random variable can take on an infinite number of possible values within a given range.

In the exercise provided, the random variable X represents a set of discrete numbers. These numbers correspond to the possible outcomes of a random experiment or process along with their associated probabilities. The concept of random variables is crucial in probability and statistics as they form the basis for defining probability distributions, calculating expected values, and more.
Discrete Mathematics
Discrete mathematics is a branch of mathematics dealing with discrete elements that use distinct values. It incorporates a variety of topics such as logic, set theory, and probability. In the context of probability and expectation, discrete mathematics focuses on discrete probability distributions and discrete random variables.

It is relevant to our example as we deal with a discrete random variable and its probability distribution in the form of a table. The probability values in the table are calculated using principles from discrete mathematics to determine outcomes of processes or experiments.
  • Helps in constructing models used in real-life applications like computer science and cryptography.
  • Forms the foundation for algorithms and data structures.
Thus, discrete mathematics becomes an essential tool for analyzing statistical problems and calculating important parameters such as expected values.

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Most popular questions from this chapter

The new computer your business bought lists a mean time between failures of 1 year, with a standard deviation of 2 months. Ten months after a repair, it breaks down again. Is this surprising? (Assume that the times between failures are normally distributed.)

Your scores for the 20 surprise math quizzes last semester were (out of 10 ) $$ \begin{aligned} &4.5,9.5,10.0,3.5,8.0,9.5,7.5,6.5,7.0,8.0 \\ &8.0,8.5,7.5,7.0,8.0,9.0,10.0,8.5,7.5,8.0 \end{aligned} $$ Use these raw data to construct a frequency table with the following brackets: \(2.1-4.0,4.1-6.0,6.1-8.0,8.1-10.0\), and find the probability distribution using the (rounded) midpoint values as the values of \(X\).

The mean batting average in major league baseball is about \(0.250\). Supposing that batting averages are normally distributed, that the standard deviation in the averages is \(0.05\), and that there are 250 batters, what is the expected number of batters with an average of at least \(0.400 ?^{54}\)

Calculate the expected value, the variance, and the standard deviation of the given random variable \(X .\) You calculated the expected values in the last exercise set. Round all answers to two decimal places.) \(X\) is the number of tails that come up when a coin is tossed twice.

In a certain political poll, each person polled has a \(90 \%\) probability of telling his or her real preference. Suppose that \(55 \%\) of the population really prefer candidate Goode, and \(45 \%\) prefer candidate Slick. First find the probability that a person polled will say that he or she prefers Goode. Then find the approximate probability that, if 1,000 people are polled, more than \(52 \%\) will say they prefer Goode.
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