im still having a hard time with this The task involves filling in the \( P(X = x) \) values to create a legitimate probability distribution for a discrete random variable \( X \), with possible values of \(-3\), \(-1\), \(1\), \(5\), and \(6\).### Probability Distribution Table| Value \( x \) of \( X \) | \( P(X = x) \) ||--------------------------|----------------|| \(-3\) | \(0.20\) || \(-1\) | \(0.15\) || \(1\) | \([ \, ? \, ]\)|| \(5\) | \(0.28\) || \(6\) | \([ \, ? \, ]\)|### ExplanationTo ensure a legitimate probability distribution:1. All probabilities \( P(X = x) \) must be non-negative.2. The sum of the probabilities must equal 1.Currently, the known probabilities are: \[0.20 + 0.15 + 0.28 = 0.63\]The sum of the unknown probabilities for \( X = 1 \) and \( X = 6 \) should be \(1 - 0.63 = 0.37\). You can distribute these probabilities to complete the table. Adjust the values of \( P(X = 1) \) and \( P(X = 6) \) to sum to 0.37 while ensuring all conditions for the probability distribution are met.

in the P(X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -3, -1, 1, 5, and 6. P ( X = x) Value x of X Check -3 -1 1 5 6 X 0.20 0.15 0 0.28 S M 31 Save For Later © 2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accessi Submit Assignme

in the P(X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -3, -1, 1, 5, and 6. P ( X = x) Value x of X Check -3 -1 1 5 6 X 0.20 0.15 0 0.28 S M 31 Save For Later © 2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accessi Submit Assignme

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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Hey there, Here is a proof I have been struggling to solve can someone help?
#14
P = 0 1 1 0 and D = 6 0 0 11 do not seem to be the correct answers, may there be a mistake somewhere? I went through by myself and got a similar answer, but it was also deemed incorrect. Just a bit confused.
Can you answer #12 and show work