To determine the value of \( R \) that most likely indicates an association between the conditional variables, let's perform a detailed analysis.
We are given the following table:
[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\cline { 2 - 5 }
\multicolumn{1}{c|}{} & A & B & C & Total \\
\hline D & 0.12 & 0.78 & 0.10 & 1.0 \\
\hline E & R & S & T & 1.0 \\
\hline
Total & U & X & Y & 1.0 \\
\hline
\end{tabular}
\][/tex]
From the data provided:
- The totals for each column (A, B, and C) add up to the overall total of 1.0.
- The total of row D is 1.0.
- The total of row E is also 1.0, which means:
[tex]\[ R + S + T = 1.0 \][/tex]
### Step-by-Step Solution
#### Step 1: Identify values for A, B, and C
Each entry in the D row must sum up to 1.0:
- \( A_D = 0.12 \)
- \( A_B = 0.78 \)
- \( A_C = 0.10 \)
#### Step 2: Calculate \( R \)
We need to calculate the left-over value \( R \).
[tex]\[ A_D + A_B + A_C = 0.12 + 0.78 + 0.10 = 1.0 \][/tex]
Since \( A_D, A_B, \) and \( A_C \) together already sum to 1.0 under row D, the sum of row E \( (R, S, T) \) should individually make up 1.0:
Let’s look at the possible values of \( R \):
- If \( R = 0.09 \):
[tex]\[ R + S + T = 0.09 + S + T = 1.0 \][/tex]
[tex]\[ S + T = 0.91 \][/tex]
- If \( R = 0.10 \):
[tex]\[ R + S + T = 0.10 + S + T = 1.0 \][/tex]
[tex]\[ S + T = 0.90 \][/tex]
- If \( R = 0.13 \):
[tex]\[ R + S + T = 0.13 + S + T = 1.0 \][/tex]
[tex]\[ S + T = 0.87 \][/tex]
- If \( R = 0.79 \):
[tex]\[ R + S + T = 0.79 + S + T = 1.0 \][/tex]
[tex]\[ S + T = 0.21 \][/tex]
#### Step 3: Choose the value of \( R \) that shows a likely natural distribution
Since values for \( S \) and \( T \) need to sum to a much more naturally balanced distribution, let’s analyze according to the mixing of higher interdependencies:
Considering our odd distributions for the plausible values:
- \( R = 0.79 \) leaves \( S \) and \( T \) very low combined \( 0.21 \).
Detailly, \( R = 0.09, 0.10, 0.13 \) leaves the remainder of \( S \) and \( T \):
- Hence \( R = 0.09, 0.13 \) is more equally probable.
Thus, a normalized association highly advises \( R = 0.13 \).
### Conclusion
[tex]\( R = 0.13 \)[/tex] is the valuable distributional figure that seems explainably natural, showing an association amongst the variables.
