### Understanding Random Variables and Expected Values#### Question:A random variable \( X \) could have an expected value of -1.#### Options:- \( \bigcirc \) 1) True- \( \bigcirc \) 2) False---**Explanation:**A random variable is a variable whose value is subject to variations due to chance. The expected value, also known as the mean, of a random variable is a measure of the center of its distribution. It is calculated as the sum of all possible values of the variable, each multiplied by its probability of occurring.The expected value of a random variable can indeed be any real number, including negative values. This is because the calculation of the expected value takes into account the probabilities and the values that the random variable can assume. So, it is possible for a random variable \( X \) to have an expected value of -1.To answer this question, select the option that you believe correctly evaluates the statement given.---

Given data A random variable X could have an expected value of -1

Answered: A random variable X could have an…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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In a digital communication channel, assume that the number of bits received in error can be modeled by a binomial random variable, and assume that the probability that a bit is received in error is 0.0001. If 150,000 bits are transmitted, what is the probability that errors are between 10 and 18? How to compute P(X=10)= (150000c10)(0.0001)10(1-0.0001)150000-10
An experiment can result in one of five equally likely simple events, E₁, E₂, A: E₂, E4 2' P(A) = = 0.4 B: E₁, E3, E4, E5 P(B) = 0.8 C: E₂, E3 Refer to the following probabilities. P(A n B) = 0.2 P(AIB) = 0.25 P(BIA) P(BU C) = 1 P(BIC) 0.5 P(CIB) 0.25 Use the Addition and Multiplication Rules to find the following probabilities. (a) P(AUB) (b) P(An B) = 0.5 P(A U B) = ---Select--- + ---Select--- ✓ P(A n B) = ---Select--- (c) P(B n C) Yes P(B n C) = ---Select--- No P(C) = 0.4 . P(B) = = P(C) = = ---Select--- V = Do the results agree with those obtained by listing the simple events in each? P(A U B) = P({E₁, E₂, E3, E4, E5}) = 1 P(A n B) = P({E₁}) = 0.2 P(B n C) = P({E3}) = 0.2 .., E5. Events A, B, and C are defined as follows.
56.6
The probability that a first-time tourist to the city ofChicago will visit the Art Institute is 0.4, will visit the Museumof Science and Industry is 0.3, and will visit both is 0.1. If afirst-time tourist to Chicago is randomly selected, find theprobability that the tourist will visit the Art Institute or theMuseum of Science and Industry.
Assume that random guesses are made for 3 multiple-choice questions on a test with 2 choices for each question, so that there are n=3 trials, each with probability of success (correct) given by p=0.50. Find the probability of no correct answers.
The finish of a hole depends on the edge of the bit used, only 1% of the holes with a new drill have an irregular cut, 3% have an irregular cut with average bits and 5% present irregular cut with worn bits. If 25% of the holes are made with new bits, 60% have average edges and 15% are worn out. [use 3 decimal places] a) If a piece is taken at random, what is the probability of having holes irregular? b) If a piece with irregular holes is taken, what is the probability that would have been made with an average drill bit?
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A random variable X could have an expected value of -1. 1) True 2) False