4) The American Veterinary Association claims that the annual cost of medical ca

Question

4) The American Veterinary Association claims that the annual cost of medical care for dogs averages $100 with a standard deviation of $30. The cost for cats averages $120 with a standard deviation of $35. Some basic algebraic and statistical steps show us that the average of the difference in the cost of medical care for dogs and cats is then $100 $120 X-$20. The standard deviation of that same difference equals $46. If the difference in costs follows a Normal distribution, what is the probability that the cost for someone's dog is higher than for the cat? A) 0.2839 B) 0.3336 C) 0.6664 D) 0.7161 Answer: B

Explanation / Answer

Given to us the mean of differences is -20

and the standard devaition of the differences is 46

P(dog > cat) = P(dog - cat > 0)

= P{ z > [0 - (-20)] / 46 } = P(z > 0.43) = 0.3336 From Z table

option B

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