/*! 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 65 The range of the random variable... [FREE SOLUTION] | 91Ó°ÊÓ

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The range of the random variable \(X\) is \([0,1,2,3, x]\) where \(x\) is unknown. If each value is equally likely and the mean of \(X\) is \(6,\) determine \(x\).

Short Answer

Expert verified
The unknown value \( x \) is 24.

Step by step solution

01

Understanding the Mean Formula

The mean of a discrete random variable is calculated using the formula \( \text{Mean} = \frac{1}{n} \sum_{i=1}^{n} x_i \), where \( n \) is the number of values and \( x_i \) are the values. In this problem, the total mean is given to be 6.
02

Identifying Number of Values

Here, we have the values \( [0, 1, 2, 3, x] \). Thus, there are a total of \( n = 5 \) values.
03

Setting Up the Mean Equation

Using the formula for the mean, substitute the given values into the equation:\[ \frac{0 + 1 + 2 + 3 + x}{5} = 6 \]
04

Simplifying the Mean Equation

Calculate the sum of the known values: \( 0 + 1 + 2 + 3 = 6 \). Substitute back into the equation: \[ \frac{6 + x}{5} = 6 \].
05

Solving for \( x \)

Multiply both sides of the equation by 5 to eliminate the fraction: \( 6 + x = 30 \). Simplify to find \( x \) by subtracting 6 from both sides: \( x = 24 \).

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Key Concepts

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

Mean of Random Variables
The mean, often referred to as the expected value, of a random variable is a fundamental concept in probability theory. It provides a measure to the "center" of the variable's distribution. For discrete random variables, the mean is computed using the formula \( \text{Mean} = \frac{1}{n} \sum_{i=1}^{n} x_i \), where:
  • \( n \) represents the total number of different values the random variable can take.
  • \( x_i \) are the individual values of the random variable.
In practical terms, the mean gives us the average value we can expect from our variable if we were to observe it repeatedly. This is especially useful to summarize data sets and predict outcomes. When each value in the set is equally likely, as seen in many textbook problems, calculating the mean is straightforward by simply summing the values and dividing by the number of values.
Discrete Random Variable
A discrete random variable is one that can take on a countable number of possible values. These are typically whole numbers or integers, which makes them unique when compared to continuous random variables. In the context of this exercise, the random variable \( X \) has values \([0, 1, 2, 3, x]\). Some key characteristics of discrete random variables include:
  • They often arise from counting processes, such as the number of heads in a series of coin tosses.
  • Probabilities are assigned to each possible value differently than with continuous variables.
  • The probability distribution of a discrete random variable is often illustrated using a probability mass function, which lists the probabilities associated with each of its values.
Understanding these variables is crucial as they are commonly encountered in various fields such as statistics, computer science, and engineering.
Solving Equations
In mathematical problems, solving an equation often involves finding unknown values that satisfy given conditions. In our exercise, the task was to find the unknown \( x \) given that the mean was 6.The solution relied on setting up an equation where\[\frac{0 + 1 + 2 + 3 + x}{5} = 6\]and then solving for \( x \). The following steps highlight the solving method:
  • Simplify the equation by first calculating the sum of the known values: \(0 + 1 + 2 + 3 = 6\).
  • Substitute back into the equation to obtain \(\frac{6 + x}{5} = 6\).
  • Eliminate the fraction by multiplying both sides by 5, thus simplifying the equation to \(6 + x = 30\).
  • Finally, solve for \(x\) by subtracting 6 from both sides, resulting in \(x = 24\).
This straightforward approach ensures you can systematically find a solution to many algebraic problems, a crucial skill in both education and practical applications.

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Most popular questions from this chapter

A player of a video game is confronted with a series of opponents and has an \(80 \%\) probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to contest opponents. (a) What is the probability mass function of the number of opponents contested in a game? (b) What is the probability that a player defeats at least two opponents in a game? (c) What is the expected number of opponents contested in a game? (d) What is the probability that a player contests four or more opponents in a game? (e) What is the expected number of game plays until a player contests four or more opponents?

Consider a sequence of independent Bernoulli trials with \(p=0.2\). (a) What is the expected number of trials to obtain the first success? (b) After the eighth success occurs, what is the expected number of trials to obtain the ninth success?

A computer system uses passwords that are exactly six characters and each character is one of the 26 letters \((\mathrm{a}-\mathrm{z})\) or 10 integers \((0-9)\). Suppose that 10,000 users of the system have unique passwords. A hacker randomly selects (with replacement) one billion passwords from the potential set, and a match to a user's password is called a hit. (a) What is the distribution of the number of hits? (b) What is the probability of no hits? (c) What are the mean and variance of the number of hits?

A manufacturer of a consumer electronics product expects \(2 \%\) of units to fail during the warranty period. A sample of 500 independent units is tracked for warranty performance. (a) What is the probability that none fails during the warranty period? (b) What is the expected number of failures during the warranty period? (c) What is the probability that more than two units fail during the warranty period?

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