Math  /  Data & Statistics

QuestionEspañol
Fill in the P(X=x)P(X=x) values to give a legitimate probability distribution for the discrete random variable XX, whose possible values are 3,2,1,0-3,-2,-1,0, and 5 . \begin{tabular}{|c|c|} \hline Value xx of XX & P(X=x)P(X=x) \\ \hline-3 & 0.26 \\ \hline-2 & 0.26 \\ \hline-1 & 0.10 \\ \hline 0 & \square \\ \hline 5 & \square \\ \hline \end{tabular}

Studdy Solution
قم بتوزيع الاحتمال المتبقي على القيم غير المعروفة P(X=0) P(X = 0) و P(X=5) P(X = 5) . يمكننا افتراض توزيع متساوٍ كالتالي:
P(X=0)=P(X=5)=0.382=0.19 P(X = 0) = P(X = 5) = \frac{0.38}{2} = 0.19
وبالتالي، توزيع الاحتمالات الشرعي هو:
\begin{tabular}{|c|c|} \hline \text{Value } x \text{ of } X & P(X=x) \\ \hline -3 & 0.26 \\ \hline -2 & 0.26 \\ \hline -1 & 0.10 \\ \hline 0 & 0.19 \\ \hline 5 & 0.19 \\ \hline \end{tabular}

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