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Sagot :
To complete the two-way frequency table using the given probabilities and tabulated values, we have to appropriately place the given numbers [tex]\(42, 68, 82, 52, 78, 26, 92, 40\)[/tex].
Let's place these numbers step-by-step.
First, we have:
[tex]\[ P(A \mid X) = \frac{40}{92} \][/tex]
This information tells us that [tex]\( A \cap X \)[/tex] (the number of outcomes that are both A and X) is 40 and the total number of outcomes in X is 92. Thus:
- [tex]\( A_X = 40 \)[/tex]
- [tex]\( X_{total} = 92 \)[/tex]
Next, we know:
[tex]\[ P(B) = \frac{78}{160} \][/tex]
We are given the total number of outcomes is 160.
So:
- [tex]\( B_{total} = 78 \)[/tex]
- [tex]\( A_{total} = 160 - 78 = 82 \)[/tex]
Similarly, because the total is divided into X and Y:
- [tex]\( Y_{total} = 160 - 92 = 68 \)[/tex]
Now, let's find other missing values:
1. Since [tex]\( A_X = 40 \)[/tex] and [tex]\( A_{total} = 82 \)[/tex]:
- [tex]\( A_Y = A_{total} - A_X = 82 - 40 = 42 \)[/tex]
2. We know [tex]\( B_{total} = 78 \)[/tex] and [tex]\( X_{total} = 92 \)[/tex]:
- [tex]\( B_X = X_{total} - A_X = 92 - 40 = 52 \)[/tex]
3. Finally, knowing [tex]\( B_{total} = 78 \)[/tex] and [tex]\( B_X = 52 \)[/tex]:
- [tex]\( B_Y = B_{total} - B_X = 78 - 52 = 26 \)[/tex]
Now, let's place all these values into the table.
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & X & Y & Total \\ \hline A & 40 & 42 & 82 \\ \hline B & 52 & 26 & 78 \\ \hline Total & 92 & 68 & 160 \\ \hline \end{tabular} \][/tex]
Here's how the numbers are distributed into the table:
- [tex]\( 40 \)[/tex] is placed at [tex]\( A \cap X \)[/tex]
- [tex]\( 42 \)[/tex] is placed at [tex]\( A \cap Y \)[/tex]
- [tex]\( 52 \)[/tex] is placed at [tex]\( B \cap X \)[/tex]
- [tex]\( 26 \)[/tex] is placed at [tex]\( B \cap Y \)[/tex]
- [tex]\( 92 \)[/tex] is the total of column X
- [tex]\( 68 \)[/tex] is the total of column Y
- [tex]\( 82 \)[/tex] is the total of row A
- [tex]\( 78 \)[/tex] is the total of row B
- The total number of observations is 160
Thus the table is complete and consistent with the given probabilities and values.
Let's place these numbers step-by-step.
First, we have:
[tex]\[ P(A \mid X) = \frac{40}{92} \][/tex]
This information tells us that [tex]\( A \cap X \)[/tex] (the number of outcomes that are both A and X) is 40 and the total number of outcomes in X is 92. Thus:
- [tex]\( A_X = 40 \)[/tex]
- [tex]\( X_{total} = 92 \)[/tex]
Next, we know:
[tex]\[ P(B) = \frac{78}{160} \][/tex]
We are given the total number of outcomes is 160.
So:
- [tex]\( B_{total} = 78 \)[/tex]
- [tex]\( A_{total} = 160 - 78 = 82 \)[/tex]
Similarly, because the total is divided into X and Y:
- [tex]\( Y_{total} = 160 - 92 = 68 \)[/tex]
Now, let's find other missing values:
1. Since [tex]\( A_X = 40 \)[/tex] and [tex]\( A_{total} = 82 \)[/tex]:
- [tex]\( A_Y = A_{total} - A_X = 82 - 40 = 42 \)[/tex]
2. We know [tex]\( B_{total} = 78 \)[/tex] and [tex]\( X_{total} = 92 \)[/tex]:
- [tex]\( B_X = X_{total} - A_X = 92 - 40 = 52 \)[/tex]
3. Finally, knowing [tex]\( B_{total} = 78 \)[/tex] and [tex]\( B_X = 52 \)[/tex]:
- [tex]\( B_Y = B_{total} - B_X = 78 - 52 = 26 \)[/tex]
Now, let's place all these values into the table.
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & X & Y & Total \\ \hline A & 40 & 42 & 82 \\ \hline B & 52 & 26 & 78 \\ \hline Total & 92 & 68 & 160 \\ \hline \end{tabular} \][/tex]
Here's how the numbers are distributed into the table:
- [tex]\( 40 \)[/tex] is placed at [tex]\( A \cap X \)[/tex]
- [tex]\( 42 \)[/tex] is placed at [tex]\( A \cap Y \)[/tex]
- [tex]\( 52 \)[/tex] is placed at [tex]\( B \cap X \)[/tex]
- [tex]\( 26 \)[/tex] is placed at [tex]\( B \cap Y \)[/tex]
- [tex]\( 92 \)[/tex] is the total of column X
- [tex]\( 68 \)[/tex] is the total of column Y
- [tex]\( 82 \)[/tex] is the total of row A
- [tex]\( 78 \)[/tex] is the total of row B
- The total number of observations is 160
Thus the table is complete and consistent with the given probabilities and values.
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