To complete the probability distribution for the discrete random variable [tex]\(X\)[/tex], we need to determine the probabilities [tex]\(P(X=5)\)[/tex] and [tex]\(P(X=6)\)[/tex].
Given the existing probabilities:
[tex]\[ P(X=1) = 0.21 \][/tex]
[tex]\[ P(X=2) = 0.13 \][/tex]
[tex]\[ P(X=4) = 0.11 \][/tex]
The sum of all probabilities must equal 1. Therefore, we set up the equation:
[tex]\[ P(X=1) + P(X=2) + P(X=4) + P(X=5) + P(X=6) = 1 \][/tex]
Substituting the known values:
[tex]\[ 0.21 + 0.13 + 0.11 + P(X=5) + P(X=6) = 1 \][/tex]
Now, let's calculate the sum of the known probabilities:
[tex]\[ 0.21 + 0.13 + 0.11 = 0.45 \][/tex]
Next, we find the remaining probability:
[tex]\[ 1 - 0.45 = 0.55 \][/tex]
Assuming that the probabilities [tex]\(P(X=5)\)[/tex] and [tex]\(P(X=6)\)[/tex] are equal (since no other information is given), we divide the remaining probability equally between them:
[tex]\[ P(X=5) = P(X=6) = \frac{0.55}{2} = 0.275 \][/tex]
Therefore, the values for the probability distribution are:
[tex]\[ P(X=5) = 0.275 \][/tex]
[tex]\[ P(X=6) = 0.275 \][/tex]
The completed probability distribution is:
[tex]\[
\begin{tabular}{|c|c|}
\hline Value $x$ of $X$ & $P(X=x)$ \\
\hline 1 & 0.21 \\
\hline 2 & 0.13 \\
\hline 4 & 0.11 \\
\hline 5 & 0.275 \\
\hline 6 & 0.275 \\
\hline
\end{tabular}
\][/tex]
