Question

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 = –$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? Group of answer choices

Answer 1

Answer:

The probability that the cost for someone's dog is higher than for the cat is 33.2%.

Step-by-step explanation:

With the data given, we know that the difference in cost of medical care of dogs and cats have a normal distribution with μ=-20 and σ=46.

To know the probability of the cost for the dog is higher than the cat, we have to calculate the probability of P(d>0).

Then we have to calculate z, and look up in a standarized normal distribution table.

Calculate z:

$z=\frac{X-\mu}{\sigma}=\frac{0-(-20)}{46}=\frac{20}{46}=   0.4348$

The probability of the difference being higher than 0, we have:

$P(d>0)=P(z>0.4348)=0.33185$

We can say that the probability that the cost for someone's dog is higher than for the cat is 33.2%.