Respuesta :
Answer with explanation:
We know that a two-way frequency table is constructed when we have to deal with two variables ( which is also known as a bivariate data)
Here we have two variables as wore a tie or not and the second variable is: Received compliment or not.
Hence, on the basis of the information provided to us in the question we see that the two-way frequency table is obtained as follows:
Wore tie Do not wore tie Total
Receive compliment 25 5 30
Do not receive compliment 9 11 20
Total 34 16 50
The considered data given about the frequencies of events can be used to fill the two-way frequency table. The values will be 25 and 5 in first row, and 9 and 11 in second row.
How to form two-way frequency table?
Suppose two dimensions are there, viz X and Y. Some values of X are there as [tex]X_1, X_2, ... , X_n[/tex] and some values of Y are there as [tex]Y_1, Y_2, ... , Y_n[/tex]
List them in title of the rows and left to the columns. There will be [tex]n \times k[/tex] table of values will be formed(excluding titles and totals), such that:
Value(ith row, jth column) = Frequency for intersection of [tex]X_i[/tex] and [tex]Y_j[/tex] (assuming X values are going in rows, and Y values are listed in columns).
Then totals for rows, columns, and whole table are written on bottom and right margin of the final table.
For n = 2, and k = 2, the table would look like:
[tex]\begin{array}{cccc}&Y_1&Y_2&\rm Total\\X_1&n(X_1 \cap Y_1)&n(X_1\cap Y_2)&n(X_1)\\X_2&n(X_2 \cap Y_1)&n(X_2 \cap Y_2)&n(X_2)\\\rm Total & n(Y_1) & n(Y_2) & S \end{array}[/tex]
where S denotes total of totals, also called total frequency.
n is showing the frequency of the bracketed quantity, and intersection sign in between is showing occurrence of both the categories together.
We're given that:
- There were 11 days where she neither wore a tie nor received a compliment.
- There were 30 days in total where she received a compliment.
- There were 16 days in total where she did not wear a tie.
If we take:
A = event of Lea wearing a tie on a day
A' = event of Lea not wearing a tie on a day
B = event of Lea getting a compliment on a day
B' = event of Lea not getting a compliment on a day
then, we're given that:
[tex]n(A' \cap B') = 11\\n(B) = 30\\n(A') = 16[/tex]
Also, total number of days = S = 50
On each of those 50 days, she either wore a tie or not.
If she didn't put on a tie on 16 out of 50 days, then she wore tie for 50-16=34 days.
Thus, we get n(A) = 34
Similarly, we get n(B') = 50 - n(B) = 50 - 30 = 20
[tex]\begin{array}{cccc}&A&A'&\rm Total\\B&n(B\cap A)&n(B\cap A')&n(B)=30\\B'&n(B' \cap A)&n(B' \cap A')=11&n(B')=20\\\rm Total & n(A)=34 & n(A')=16 & S=50 \end{array}[/tex]
Using the total of the third row, we get:
[tex]n(B' \cap A) + 11 = 20\\n(B' \cap A) = 9[/tex]
Similarly, now using second column's total, we get:
[tex]n(B \cap A) + n(B' \cap A) = 34\\n(B \cap A) + 9 = 34\\n(B \cap A) = 25[/tex]
Now using second row's total, we get:
[tex]25 + n(B \cap A') = 30\\n(B\cap A') = 5[/tex]
Thus, we get:
[tex]\begin{array}{cccc}&A&A'&\rm Total\\B&n(B\cap A)=25&n(B\cap A')=5&n(B)=30\\B'&n(B' \cap A)=9&n(B' \cap A')=11&n(B')=20\\\rm Total & n(A)=34 & n(A')=16 & S=50 \end{array}[/tex]
Thus, finally, we can say that:
Wore a tie, Didn't wore a tie
Got a compliment 25 5
Didn't got a compliment 9 11
Thus, the values will be 25 and 5 in first row, and 9 and 11 in second row.
Learn more about two-way table here:
https://brainly.com/question/26788374
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