Question

Suppose that in a survey of 1,000 U.S. residents, 721 residents believed that the amount of violent television programming had increased over the past 10 years, 454 residents believed that the overall quality of television programming had decreased over the past 10 years, and 362 residents believed both. What 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? (Round your answer to 3 decimal places.)

Answer 1

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 ,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

i.e. $n(A\cup B)=721+454-362=813$

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 :-

$P(A\cup B)=\dfrac{n(A\cup B)}{n(S)}=\dfrac{813}{1000}=0.813$

Hence, the required answer = 0.813