Answer :
Let's complete the given two-way frequency table step by step using the provided information and values.
We are provided with the following probabilities:
- [tex]\( P(A \mid X) = \frac{40}{92} \)[/tex]
- [tex]\( P(B) = \frac{78}{160} \)[/tex]
Additionally, we have several numerical values:
- 78, 68, 42, 92, 26, 40, 82, 52
We need to organize these numbers into a two-way frequency table.
### Step-by-Step Completion
1. Determine the Total Values:
- The total number of observations is [tex]\( 160 \)[/tex].
2. Fill in the Grand Total:
- The grand total (bottom-right cell) is [tex]\( 160 \)[/tex].
3. Fill in the Row and Column Totals:
- Total for [tex]\( A \)[/tex] is [tex]\( 82 \)[/tex] (sum of its corresponding row).
- Total for [tex]\( B \)[/tex] is [tex]\( 78 \)[/tex] (sum of its corresponding row).
- Total for [tex]\( X \)[/tex] is [tex]\( 92 \)[/tex] (sum of its corresponding column).
- Total for [tex]\( Y \)[/tex] is [tex]\( 68 \)[/tex] (sum of its corresponding column).
4. Calculate and Fill in the Cells for Events A and B:
- From [tex]\( P(A \mid X) = \frac{40}{92} \)[/tex], we know that the number of cases where both A and X occur is [tex]\( 40 \)[/tex].
- Given that [tex]\( P(B) = \frac{78}{160} \)[/tex], the total number of cases for B is [tex]\( 78 \)[/tex].
5. Distribute the Remaining Values:
- We have already placed the values [tex]\( 40 \)[/tex] under [tex]\( A \)[/tex] and [tex]\( X \)[/tex].
- Total for [tex]\( A \)[/tex] under [tex]\( Y \)[/tex] would be [tex]\( 82 - 40 = 42 \)[/tex].
- Total for [tex]\( X \)[/tex] is [tex]\( 92 \)[/tex], and we have already assigned [tex]\( 40 \)[/tex] under [tex]\( A \)[/tex], so the remaining for [tex]\( B \)[/tex] under [tex]\( X \)[/tex] would be [tex]\( 92 - 40 = 52 \)[/tex].
- For [tex]\( B \)[/tex] under [tex]\( Y \)[/tex], knowing the total for [tex]\( B \)[/tex] is [tex]\( 78 \)[/tex], and already [tex]\( 52 \)[/tex] is under [tex]\( X \)[/tex], we get [tex]\( 78 - 52 = 26 \)[/tex].
6. Construct the Completed Table:
\begin{tabular}{|c|c|c|c|}
\hline
& [tex]$X$[/tex] & [tex]$Y$[/tex] & Total \\
\hline
[tex]$A$[/tex] & 40 & 42 & 82 \\
\hline
[tex]$B$[/tex] & 26 & 52 & 78 \\
\hline
Total & 92 & 68 & 160 \\
\hline
\end{tabular}
### Validations:
- Ensure that the row sums match the total values:
- [tex]\( 40 + 42 = 82 \)[/tex]
- [tex]\( 26 + 52 = 78 \)[/tex]
- Ensure that the column sums match the total values:
- [tex]\( 40 + 26 = 66 \Rightarrow 92 \)[/tex]
- [tex]\( 42 + 52 = 94 \Rightarrow 68 \)[/tex]
All totals and event counts align correctly with the provided data.
Thus, the completed two-way frequency table is accurate and properly structured.
We are provided with the following probabilities:
- [tex]\( P(A \mid X) = \frac{40}{92} \)[/tex]
- [tex]\( P(B) = \frac{78}{160} \)[/tex]
Additionally, we have several numerical values:
- 78, 68, 42, 92, 26, 40, 82, 52
We need to organize these numbers into a two-way frequency table.
### Step-by-Step Completion
1. Determine the Total Values:
- The total number of observations is [tex]\( 160 \)[/tex].
2. Fill in the Grand Total:
- The grand total (bottom-right cell) is [tex]\( 160 \)[/tex].
3. Fill in the Row and Column Totals:
- Total for [tex]\( A \)[/tex] is [tex]\( 82 \)[/tex] (sum of its corresponding row).
- Total for [tex]\( B \)[/tex] is [tex]\( 78 \)[/tex] (sum of its corresponding row).
- Total for [tex]\( X \)[/tex] is [tex]\( 92 \)[/tex] (sum of its corresponding column).
- Total for [tex]\( Y \)[/tex] is [tex]\( 68 \)[/tex] (sum of its corresponding column).
4. Calculate and Fill in the Cells for Events A and B:
- From [tex]\( P(A \mid X) = \frac{40}{92} \)[/tex], we know that the number of cases where both A and X occur is [tex]\( 40 \)[/tex].
- Given that [tex]\( P(B) = \frac{78}{160} \)[/tex], the total number of cases for B is [tex]\( 78 \)[/tex].
5. Distribute the Remaining Values:
- We have already placed the values [tex]\( 40 \)[/tex] under [tex]\( A \)[/tex] and [tex]\( X \)[/tex].
- Total for [tex]\( A \)[/tex] under [tex]\( Y \)[/tex] would be [tex]\( 82 - 40 = 42 \)[/tex].
- Total for [tex]\( X \)[/tex] is [tex]\( 92 \)[/tex], and we have already assigned [tex]\( 40 \)[/tex] under [tex]\( A \)[/tex], so the remaining for [tex]\( B \)[/tex] under [tex]\( X \)[/tex] would be [tex]\( 92 - 40 = 52 \)[/tex].
- For [tex]\( B \)[/tex] under [tex]\( Y \)[/tex], knowing the total for [tex]\( B \)[/tex] is [tex]\( 78 \)[/tex], and already [tex]\( 52 \)[/tex] is under [tex]\( X \)[/tex], we get [tex]\( 78 - 52 = 26 \)[/tex].
6. Construct the Completed Table:
\begin{tabular}{|c|c|c|c|}
\hline
& [tex]$X$[/tex] & [tex]$Y$[/tex] & Total \\
\hline
[tex]$A$[/tex] & 40 & 42 & 82 \\
\hline
[tex]$B$[/tex] & 26 & 52 & 78 \\
\hline
Total & 92 & 68 & 160 \\
\hline
\end{tabular}
### Validations:
- Ensure that the row sums match the total values:
- [tex]\( 40 + 42 = 82 \)[/tex]
- [tex]\( 26 + 52 = 78 \)[/tex]
- Ensure that the column sums match the total values:
- [tex]\( 40 + 26 = 66 \Rightarrow 92 \)[/tex]
- [tex]\( 42 + 52 = 94 \Rightarrow 68 \)[/tex]
All totals and event counts align correctly with the provided data.
Thus, the completed two-way frequency table is accurate and properly structured.
