Question

In: Statistics and Probability

A pet store completed a survey of customers regarding pet preferences and ownership. The results for...

A pet store completed a survey of customers regarding pet preferences and ownership. The results for dog ownership and feelings about cats are displayed in the table. Use them to answer the following questions.

Likes cats & Owns a dog =13

Likes cats & does not own a dog= 30

Total=43

Dislikes cats & owns a dog=18

Dislikes cats & does not own a dog=18

Total=36

No opinion about cats & owns a dog= 0

No opinion about cats & does not own a dog= 6

Total= 6

(a) What is the probability that a randomly selected person from the group (i) dislikes cats? (ii) likes cats and does not own a dog? (iii) dislikes cats or has no opinion about them? (iv) dislikes cats or does not own a dog? (v) dislikes cats, given that they own a dog? (vi) owns a dog, given that they dislike cats?

(b) Are the events dislikes cats and owns a dog independent? How do you know?

(c) Are the events no opinion about cats and owns a dog mutually exclusive? How do you know?

Solutions

Expert Solution

The contingency table can be written as :

Owns dog Do not own dog Total
Likes cat 13 30 43
Dislikes cat 18 18 36
No opinion about cat 0 6 6
Total 31 54 85

a) The probability that a randomly selected person from the group dislikes cats

b) The probability that a randomly selected person from the group likes cats and does not own a dog

c) The probability that a randomly selected person from the group dislikes cats or has no opinion about them

= P(a random person dislikes cats) + P(a random person has no opinion about cats)

d) The probability that a randomly selected person from the group dislikes cats or does not own a dog

= P(a random person dislikes cats) + P(a random person does not own a dog) - P(a random person dislikes cats and does not own a dog)

orchestra answered 2 years ago
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