Answer: 0.813
Step-by-step explanation:
Let A be the event describes the number of residents believed that the amount of violent television programming had increased over the past 10 years.
& B be the event describes the number of residents believed that the amount of violent television programming had decreased over the past 10 years.
Given : n(A)=721 ; n(B)=454 ; n(A∩B)=362
We know that ,
[tex]n(A\cup B)=n(A)+n(B)-n(A\cap B)[/tex]
i.e. [tex]n(A\cup B)=721+454-362=813[/tex]
Also, the total number of U.S. residents surveyed n(S)= 1,000
Then, the proportion of the 1,000 U.S. residents believed that either the amount of violent programming had increased or the overall quality of programming had decreased over the past 10 years will be :-
[tex]P(A\cup B)=\dfrac{n(A\cup B)}{n(S)}=\dfrac{813}{1000}=0.813[/tex]
Hence, the required answer = 0.813
