**Title: Completing a Probability Distribution Table****Introduction:**When given a discrete random variable \( X \) with specific possible values, it's important to assign probabilities \( P(X=x) \) in such a way that they form a legitimate probability distribution. This involves ensuring the sum of probabilities equals 1, and each probability is between 0 and 1.**Problem Statement:**We need to fill in the \( P(X=x) \) values to create a valid probability distribution for the discrete random variable \( X \), whose possible values are \(-6, -4, -1, 0, \) and \( 2 \).**Probability Distribution Table:**The table below shows some probabilities already assigned to certain values of \( X \). Our task is to complete this table.| Value \( x \) of \( X \) | \( P(X = x) \) ||--------------------------|----------------|| -6 | 0.22 || -4 | 0.16 || -1 | 0.16 || 0 | [ ] || 2 | [ ] |**Explanation:**- The probabilities for \( X = -6, -4, \) and \( -1 \) are given as \( 0.22, 0.16, \) and \( 0.16 \) respectively.- To complete the table, the sum of all probabilities must equal 1, that is: \[ 0.22 + 0.16 + 0.16 + P(X=0) + P(X=2) = 1 \]- Calculate \( P(X=0) \) and \( P(X=2) \) such that the above condition holds.**Steps:**1. Add the given probabilities: \( 0.22 + 0.16 + 0.16 = 0.54 \).2. Subtract this sum from 1: \( 1 - 0.54 = 0.46 \).3. Distribute the remaining probability between \( P(X=0) \) and \( P(X=2) \) ensuring each remains between 0 and 1.**Conclusion:**The values for \( P(X=0) \) and \( P(X=2) \) can be calculated to ensure the distribution is valid

x p(x) -6 0.22 -4 0.16 -1 0.16 0 a 2 b

Answered: Fill in the P (X=x) values to give a…

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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Fill in the P (X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -6, -4,-1, 0, and 2. Value x of X -6 -4 -1 0 2 X P ( X = x) 0.22 0.16 0.16 0