/*! 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 17 Suppose that the distribution fu... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(P\left\\{\frac{1}{2} < X < \frac{3}{2}\right\\}\)

Short Answer

Expert verified
(a) For \(i = 1, 2, 3\), the probabilities are: $$P(X=1) = \frac{1}{4}$$ $$P(X=2) = \frac{1}{12}$$ $$P(X=3) = \frac{1}{12}$$ (b) The probability of \(X\) being in the interval \(\left(\frac{1}{2}, \frac{3}{2}\right)\) is: $$P\left(\frac{1}{2} < X < \frac{3}{2}\right) = \frac{1}{2}$$

Step by step solution

01

Find probabilities for part (a)

To find the probability \(P(X = i)\) for each \(i = 1,2,3\), we will calculate the difference between the distribution function at \(i\) and its left-neighbor. For \(i = 1\), we have: $$P(X=1) = F(1) - F(1^-)$$ where \(1^-\) denotes the left-neighbor of 1 (in this case, any value slightly smaller than 1).
02

Calculate the probabilities for i = 1, 2, 3

Using the distribution function, \(F(b)\), we can calculate the required probabilities for \(i = 1\), \(2\), and \(3\): For \(i = 1\): $$P(X=1) = F(1) - F(1^-) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$ For \(i = 2\): $$P(X=2) = F(2) - F(2^-) = \frac{11}{12} - \frac{1}{2} - \frac{1}{4} = \frac{1}{12}$$ For \(i = 3\): $$P(X=3) = F(3) - F(3^-) = 1 - \frac{11}{12} = \frac{1}{12}$$
03

Find probability for part (b)

We are asked to find the probability of \(X\) being in the interval \(\left(\frac{1}{2}, \frac{3}{2}\right)\). To find this probability, we can use the distribution function as follows: $$P\left(\frac{1}{2} < X < \frac{3}{2}\right) = F\left(\frac{3}{2}\right) - F\left(\frac{1}{2}\right)$$
04

Calculate the probability for part (b)

Using the distribution function, \(F(b)\), we can calculate the required probability: $$P\left(\frac{1}{2} < X < \frac{3}{2}\right) = F\left(\frac{3}{2}\right) - F\left(\frac{1}{2}\right) = \left(\frac{1}{2} + \frac{\frac{3}{2} - 1}{4}\right) - \frac{1}{4}$$ Simplify the expression: $$P\left(\frac{1}{2} < X < \frac{3}{2}\right) = \frac{1}{2} + \frac{1}{4} - \frac{1}{4} = \frac{1}{2}$$ So the probability of \(X\) being in the interval \(\left(\frac{1}{2}, \frac{3}{2}\right)\) is \(\frac{1}{2}\).

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Key Concepts

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

Distribution Function
A distribution function, often denoted as \( F(b) \), provides the probability that a continuous random variable \( X \) takes on a value less than or equal to \( b \). It is a fundamental tool in probability theory. It describes how probabilities are distributed over the values of the random variable. The distribution function is cumulative, meaning it adds up the probabilities as values increase.The given distribution function for \( X \) breaks down into different intervals:- From \( b < 0 \), the function is 0. This indicates that \( X \) can never be negative.- For \( 0 \leq b < 1 \), the function is \( \frac{b}{4} \), suggesting that the probability increases linearly.- When \( 1 \leq b < 2 \), it's \( \frac{1}{2} + \frac{b-1}{4} \), adjusting the probability upwards as \( b \) increases.- Next, between \( 2 \leq b < 3 \), it is constant at \( \frac{11}{12} \), which implies no further increase until \( b \geq 3 \).- Finally, for \( b \geq 3 \), the function is 1, meaning that by the time \( b \) reaches 3, the total probability is exhausted.Understanding this structure helps us calculate probabilities for specific intervals and values, as shown in the problem.
Probability Calculation
Probability calculation involves finding the likelihood that a random variable falls within a certain range. Using the distribution function, the probability \( P(X = i) \) is determined by the difference at a point and just before the point (left neighbor).To find \( P(X = i) \) for discrete points:- For \( i = 1 \), calculate: - \( P(X=1) = F(1) - F(1^-) \) - With given function values, \( \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \).- Similarly, for \( i = 2 \): - \( P(X=2) = F(2) - F(2^-) = \frac{11}{12} - \frac{3}{4} = \frac{1}{12} \).- For \( i = 3 \): - \( P(X=3) = F(3) - F(3^-) = 1 - \frac{11}{12} = \frac{1}{12} \).For continuous ranges, such as \( \( \frac{1}{2} < X < \frac{3}{2} \) \):- Calculate using: \( F(\frac{3}{2}) - F(\frac{1}{2}) \)- Plugging into the function, we find \( \frac{1}{2} \).This method allows for precise probability estimations for complex distributions.
Continuous Random Variable
A continuous random variable, like \( X \) in this scenario, can take any value in a certain interval. Unlike a discrete random variable that has distinct outcomes, a continuous variable's values are dense, meaning there could be infinitely many possibilities.To manage the infinite possibilities, a probability distribution is assigned, often described by a continuous distribution function.This function, \( F(b) \), accumulates probabilities from negative infinity up to a certain point \( b \). The smoothness of this function over intervals shows how probabilities are spread out smoothly across different possible outcomes.By understanding where the distribution is zero, linear, or constant, we know how \( X \)'s values behave over ranges: - Zero probability for \( b < 0 \) informs that negative values are impossible.- Linear growth between 0 and 1, and also between 1 and 2, reflects increasing density or likelihood of occurrences.- A constant value implies that further increases beyond a certain point do not affect the total probability.Such detail assists significantly in calculating the likelihood of events and understanding how probabilities are distributed for \( X \).

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Most popular questions from this chapter

A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted is incorrectly received with probability .2. Suppose that we want to transmit an important message consisting of one binary digit. To reduce the chance of error, we transmit 00000 instead of 0 and 11111 instead of \(1 .\) If the receiver of the message uses "majority" decoding, what is the probability that the message will be wrong when decoded? What independence assumptions are you making?

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