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Suppose that the distribution function of \(X\) is given by $$F(b)=\left\\{\begin{array}{ll} 0 & b<0 \\ \frac{b}{4} & 0 \leq b<1 \\ \frac{1}{2}+\frac{b-1}{4} & 1 \leq b<2 \\ \frac{11}{12} & 2 \leq b<3 \\ 1 & 3 \leq b \end{array}\right.$$ (a) Find \(P\\{X=i\\}, i=1,2,3\) (b) Find \(P\left\\{\frac{1}{2}

Short Answer

Expert verified
In summary, for a given random variable \(X\) with distribution function \(F(b)\), we have: a. \(P(X=1) = \frac{1}{4}\), \(P(X=2) = \frac{1}{12}\), and \(P(X=3) = \frac{1}{12}\). b. \(P\left(\frac{1}{2}<X<\frac{3}{2}\right) = \frac{5}{8}\).

Step by step solution

01

Calculate P{X=1}

To find \(P(X=1)\), we need to find the difference in \(F(b)\) at \(1\) and just before \(1\). Using the given distribution function: \(P(X=1) = F(1) - F(1^-) = \left(\frac{1}{2} + \frac{1-1}{4}\right) - \frac{1}{4} = \frac{1}{4}\)
02

Calculate P{X=2}

To find \(P(X=2)\), we need to find the difference in \(F(b)\) at \(2\) and just before \(2\). Using the given distribution function: \(P(X=2) = F(2) - F(2^-) = \frac{11}{12} - \left(\frac{1}{2} + \frac{2-1}{4}\right) = \frac{1}{12}\)
03

Calculate P{X=3}

To find \(P(X=3)\), we need to find the difference in \(F(b)\) at \(3\) and just before \(3\). Using the given distribution function: \(P(X=3) = F(3) - F(3^-) = 1 - \frac{11}{12} = \frac{1}{12}\) Therefore, \(P(X=1)=\frac{1}{4}\), \(P(X=2)=\frac{1}{12}\), and \(P(X=3)=\frac{1}{12}\). #b. Finding P{\(\frac{1}{2}<X<\frac{3}{2}\)}#
04

Calculate the distribution function values at the endpoints

To find the probability of \(X\) being between \(\frac{1}{2}\) and \(\frac{3}{2}\), we need to subtract the values of the distribution function at the endpoints. Using the given distribution function: \(F\left(\frac{1}{2}\right) = \frac{\frac{1}{2}}{4} = \frac{1}{8}\) \(F\left(\frac{3}{2}\right) = \frac{1}{2} + \frac{\frac{3}{2}-1}{4} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\)
05

Calculate P(\(\frac{1}{2}

Find the probability of \(X\) being in the given range by subtracting the distribution function values at the endpoints: \(P\left(\frac{1}{2}<X<\frac{3}{2}\right) = F\left(\frac{3}{2}\right) - F\left(\frac{1}{2}\right) = \frac{3}{4} - \frac{1}{8} = \frac{5}{8}\) Therefore, \(P\left(\frac{1}{2}<X<\frac{3}{2}\right) = \frac{5}{8}\).

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Key Concepts

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

Distribution Function
The distribution function, often called a cumulative distribution function (CDF), is a fundamental concept in probability theory. It describes the probability that a random variable takes on a value within a certain range. For a given value \( b \), the distribution function \( F(b) \) helps determine the probability that a random variable \( X \) is less than or equal to \( b \). This accumulates the probabilities from the beginning of the range up to \( b \). In the exercise, the distribution function is segmented into different intervals, indicating a piecewise function. This type of function is crucial when dealing with different behaviors of probabilities over various intervals. Understanding how to evaluate these sections individually helps in calculating probabilities for specific values or ranges of the random variable.
Random Variable
A random variable is a variable that can take on different values, each associated with a probability. It serves as a bridge between real-world phenomena and mathematical modeling, allowing for probabilistic expressions of uncertain events. There are two main types of random variables: discrete and continuous. Discrete random variables take on a countable number of values, while continuous random variables can take on values in a continuum. In the provided problem, \(X\) is a random variable that is defined by the given distribution function. This distribution function tells us how likely \(X\) is to be less than or equal to a particular value \(b\). By using the segments of the distribution function, we can determine specific probabilities for different values of \(X\).
Continuous Probability
Continuous probability deals with random variables that can take any value within a continuum. Unlike discrete probability, where probabilities are assigned to individual outcomes, continuous probability necessitates the use of a probability density function (PDF) and its integral, the distribution function (CDF), to find probabilities for ranges of outcomes.In the problem, although specific calculations are for discrete points \((X=1, X=2, X=3)\), the approach generally follows principles of continuous probability. This involves evaluating the distribution function at specific points and using the difference between these values to compute probabilities over intervals. Calculating probabilities using intervals and the distribution function is a hallmark of dealing with continuous probability scenarios.
Probability Calculation
Probability calculations allow us to quantify the likelihood of different outcomes. When working with distribution functions, calculating the probability of specific events involves understanding how to use the CDF. For discrete points, probability can be found by subtracting the CDF value just before the point from the CDF value at the point in question. This was done in steps to find \(P(X=1)\), \(P(X=2)\), and \(P(X=3)\). To calculate the probability over an interval, one subtracts the value of the CDF at the lower bound from the CDF at the upper bound. For example, to find \(P\left(\frac{1}{2} < X < \frac{3}{2}\right)\), the values found were \(F\left(\frac{3}{2}\right)\) minus \(F\left(\frac{1}{2}\right)\). Using these methods, probabilities for specific events can be systematically discovered.

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Most popular questions from this chapter

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