/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 175 Let the random variable \(X\) be... [FREE SOLUTION] | 91Ó°ÊÓ

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Let the random variable \(X\) be equally likely to assume any of the values \(1 / 8,1 / 4,\) or \(3 / 8 .\) Determine the mean and variance of \(X\).

Short Answer

Expert verified
Mean = \(\frac{1}{4}\); Variance = \(\frac{1}{96}\).

Step by step solution

01

List Possible Values and Probabilities

The random variable \(X\) can take the values \(\frac{1}{8}, \frac{1}{4},\) and \(\frac{3}{8}\). Since the problem states that \(X\) is equally likely to assume any of these values, the probability for each value is \(\frac{1}{3}\).
02

Calculate the Mean

The mean (expected value) of a random variable \(X\) is given by the formula \(E(X) = \sum_{i} x_i p(x_i)\), where \(x_i\) are the values \(X\) can take, and \(p(x_i)\) are their probabilities. Substitute to find the mean: \[ E(X) = \frac{1}{8}\cdot\frac{1}{3} + \frac{1}{4}\cdot\frac{1}{3} + \frac{3}{8}\cdot\frac{1}{3} \] Simplify the expression: \[ E(X) = \frac{1}{24} + \frac{1}{12} + \frac{1}{8}\] Convert to a common denominator (24): \[ E(X) = \frac{1}{24} + \frac{2}{24} + \frac{3}{24} = \frac{6}{24} = \frac{1}{4} \] The mean of \(X\) is \(\frac{1}{4}\).
03

Calculate the Variance

Variance is calculated using \(Var(X) = E(X^2) - (E(X))^2\). First, find \(E(X^2)\):\[E(X^2) = \sum_{i} x_i^2 p(x_i) = \left(\frac{1}{8}\right)^2\cdot\frac{1}{3} + \left(\frac{1}{4}\right)^2\cdot\frac{1}{3} + \left(\frac{3}{8}\right)^2\cdot\frac{1}{3}\] Calculate each term:\[E(X^2) = \frac{1}{192} + \frac{1}{48} + \frac{9}{192} \]Convert \(\frac{1}{48}\) to the common denominator 192: \[ \frac{1}{48} = \frac{4}{192} \]Thus,\[E(X^2) = \frac{1}{192} + \frac{4}{192} + \frac{9}{192} = \frac{14}{192} \] Simplify:\[E(X^2) = \frac{7}{96} \]Now, calculate the variance:\[Var(X) = \frac{7}{96} - \left(\frac{1}{4}\right)^2 = \frac{7}{96} - \frac{1}{16} \] Convert \(\frac{1}{16}\) to \(\frac{6}{96}\):\[ Var(X) = \frac{7}{96} - \frac{6}{96} = \frac{1}{96} \]The variance of \(X\) is \(\frac{1}{96}\).

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Key Concepts

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

Random Variable
A random variable is a fundamental concept in probability theory that assigns numerical values to each outcome of a random phenomenon. Think of it as a bridge between outcomes and numbers. These variables can be discrete, like rolling a die, or continuous, like measuring the weight of an object. In the case of the original exercise, the random variable, denoted as \( X \), can take on three possible values: \( \frac{1}{8}, \frac{1}{4}, \) and \( \frac{3}{8} \). Each of these outcomes is equally probable, meaning that every value \( X \) can assume has a probability of \( \frac{1}{3} \). This characteristic simplifies computations and aligns with the uniform distribution of outcomes.

To understand random variables better, always think about:
  • What values can the variable take?
  • What are the associated probabilities of each of these values?
In practical terms, random variables help in simplifying real-world problems into a mathematical format, making it easier to compute probabilities and make predictions based on certain models.
Expected Value
Expected value, often symbolized as \( E(X) \), represents the average or mean value you would expect from a random variable after many trials. It encompasses the weighted average of all possible values, where the weights are their respective probabilities. This concept is crucial because it provides insight into the "center" of the distribution.

In our exercise, the formula to calculate expected value is:
\[ E(X) = \sum_{i} x_i p(x_i) \]
Using this formula, we find that the expected value of \( X \) is \( \frac{1}{4} \) indicating that, on average, we'd expect the value \( X \) to be around \( \frac{1}{4} \) over a large number of observations.

When calculating expected value, always consider:
  • The possible values of the random variable.
  • The probability of each value occurring.
This mathematical expectation is pivotal in various fields like finance, insurance, and gambling to predict future events and outcomes.
Variance
Variance measures the spread or dispersion of a set of values, specifically showing how much the values of a random variable deviate from the expected value. In simpler terms, it tells us how far, on average, each value is from the mean. The formula for variance is:
\[ Var(X) = E(X^2) - (E(X))^2 \]

For our specific problem, we first find \( E(X^2) \), which gives us the average of the squares of the values of \( X \). Once calculated, we subtract the square of the expected value \( (E(X))^2 \) to get the variance, which in this case is \( \frac{1}{96} \).

The importance of understanding variance includes:
  • Knowing if the data points are close to the mean or widely scattered.
  • Identifying risks and changes in observational data.
A smaller variance implies that the data points are more consistent, while a larger variance indicates high variability, adding another layer of understanding to predictions and probabilistic models.

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