In the Health ABC Study, 543 subjects owned a pet and 1977 subjects did not. Among the pet owners, there were 293 women; 975 of the non-pet owners were women. Find the proportion of pet owners who were women. Do the same for the non-pet owners. (Be sure to let Population 1 correspond to the group with the higher proportion so that the difference will be positive. Round your answers to three decimal places.)

The proportion of pet owners who were women is 0.540.

The proportion of non-pet owners who were women is 0.493.

Step-by-step explanation:

The data provided is summarized in the table below.

               Pet Owners      Non-Pet Owners    TOTAL

Men              250                       1002                1252

Women         293                        975                1268

TOTAL          543                       1977                2520

Denote the events as follows:

W = a subject is a women

P = a subject is a pet owner

The probability of an event, say E, is defined as the ration of the favorable outcomes of E to the the total number of outcomes of the experiment.

P (E) = n (E) ÷ N

Here,

n (E) = favorable outcome

N = total number of outcomes

(1)

The conditional probability of an event B given that another events A has already occurred is:

P (B | A) = n (A ∩ B) ÷ n (A)

Compute the probability of a pet owner being a women as follows:

P (W | P) = n (W ∩ P) ÷ n (P)

              = 975 ÷ 1977

              = 0.5396

              ≈ 0.540

Thus, the proportion of pet owners who were women is 0.540.

(2)

Compute the probability of a non-pet owner being a women as follows:

P (W | P') = n (W ∩ P') ÷ n (P')

              = 293 ÷ 543

              = 0.4932

              ≈ 0.493

Thus, the proportion of non-pet owners who were women is 0.493.

the answer is 9.8975

explanation:

when you multiply the "39.59" with the 25% it reduces it and shows its final discounted price. hope this !

step-by-step explanation:

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