/*! 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 20 The random variable \(X\) has th... [FREE SOLUTION] | 91Ó°ÊÓ

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The random variable \(X\) has the probability distribution table shown below: \begin{tabular}{|c|c|c|c|c|c|} \hline \(\boldsymbol{x}\) & \(-2\) & \(-\mathrm{I}\) & 0 & 1 & 2 \\ \hline \(\boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x})\) & \(-\) & \(-\) & \(.4\) & \(.1\) & \(.1\) \\ \hline \end{tabular} a. Calculate \(P(X \geq 0\) ) and \(P(X<0)\). HINT [See Quick Example on page 551.] b. Assuming \(P(X=-2)=P(X=-1)\), find each of the missing values.

Short Answer

Expert verified
The short answer is: For part (a), we have \(P(X \geq 0) = 0.4 + 0.1 + 0.1 = 0.6\), and \(P(X < 0) = P(X=-2) + P(X=-1) = 1 - 0.6 = 0.4\). For part (b), after substituting and solving the equation, we get \(P(X=-2) = P(X=-1) = 0.2\).

Step by step solution

01

Part (a) – Calculate \(P(X \geq 0)\) and \(P(X < 0)\)

To calculate the probabilities, we can use the information from the given table. The probability of \(P(X \geq 0)\) can be found by summing the probabilities where \(X\) is greater than or equal to 0. This is because \(P(A \cup B) = P(A) + P(B)\) when the events are exclusive. 1. Find the values of \(X\) that are greater or equal to 0: \(X = 0, 1, 2\) 2. Sum their probabilities: \(P(X \geq 0) = P(X=0) + P(X=1) + P(X=2)\) Similarly, we need to find the probability of \(P(X < 0)\) by summing the probabilities where \(X\) is less than 0. 3. Find the values where \(X\) is less than 0: \(X = -2, -1\) 4. Sum their probabilities: \(P(X < 0) = P(X=-2) + P(X=-1)\)
02

Part (b) – Assuming \(P(X=-2)=P(X=-1)\), find each of the missing values

First, we know that the sum of probabilities for all values of the random variable must equal 1. That is, \[P(X=-2) + P(X=-1) + P(X=0) + P(X=1) + P(X=2) = 1\] We also know \(P(X=-2) = P(X=-1)\). Using this information, we can write an equation to find the missing values. 1. Substituting the given probabilities into the equation: \[P(X=-2) + P(X=-1) + 0.4 + 0.1 + 0.1 = 1\] 2. Expressing \(P(X=-1)\) in terms of \(P(X=-2)\) \[P(X=-2) + P(X=-2) + 0.4 + 0.1 + 0.1 = 1\] 3. Solving the equation for \(P(X=-2)\) and \(P(X=-1)\)

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Key Concepts

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

Probability Calculation
Probability calculation is a fundamental concept in statistics that deals with determining the likelihood of a specific outcome occurring.

For instance, in the exercise given, we are interested in calculating probabilities related to the random variable \(X\), such as \(P(X \geq 0)\) and \(P(X < 0)\). To do this, we utilize the information provided in the probability distribution table. We add together the probabilities for the events that satisfy the inequality condition, using the principal that the probability of the union of two mutually exclusive events is the sum of their respective probabilities.

To ensure accuracy in probability calculation, always remember that the probabilities of all possible outcomes must add up to 1. If you find that the sum is not equal to 1, it's time to double-check your computations or reassess the probability values given in the problem statement.
Random Variables
A random variable, often represented by \(X\), is a variable that takes on numerical values based on the outcome of a random phenomenon. In the exercise, \(X\) is discrete, which means it takes on specific values. These values and their associated probabilities are depicted in a probability distribution table.

It's vital that students recognize the type of random variable they are dealing with: if it's discrete, like in our textbook example, it will have specific values with associated probabilities. On the other hand, a continuous random variable would require different approaches such as using probability density functions.

Understanding random variables allows you to correctly interpret and calculate various probabilities, like the probability of the variable falling within a certain range, which is a common task in statistical analysis.
Sum of Probabilities
The sum of probabilities principle states that the total probability of all possible outcomes of a random variable must equal 1. This is a core tenet of probability theory and ensures that all possible outcomes are accounted for.

In the context of the exercise, this principle helps us find the missing probabilities of the random variable \(X\). Given that the sum of all probabilities equals 1, we can use algebra to solve for the unknown probabilities when a certain number of them are given. The process involves setting up an equation where the sum of the known and unknown probabilities equals 1 and solving for the unknowns.

It's crucial to check that the probability values obtained are between 0 and 1, inclusive, as probabilities cannot be negative or greater than 1. This basic check helps in validating the correctness of your results.

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