The American Veterinary Association claims that the annual cost X of medical care for a cat is a random variable with expected value E(X) = $125 and standard deviation SD(X) = $30, and that the annual medical cost Y for a dog has expected value E(Y) = $100 and standard deviation SD(Y) = $25. Assume that medical costs for different dogs and cats are independent. Question 1. How much more in medical costs do you expect to pay for the cat than for the dog?

d) If the variables X and Y are not independent then we have an error in the calculation and if we don't have a normal distribution then the probability would be incorrect for this case.

Step-by-step explanation:

a) What's the expected difference in the cost of medical care for dogs and cats?

For this case we have two random variable with the following parameters:

[tex] \mu_X = 125, \sigma_X= 30[/tex]

[tex] \mu_Y =100 , \sigma_Y= 100[/tex]

And we are interested on this random variable [tex]D=Y-X[/tex]

Using propertis of expected value we got:

[tex] E(D)= E(Y)-E(X) =100-125= -25[/tex]

b) What's the standard deviation of that difference?

Since X and Y are independent variables we have that [tex] Cov(X,Y)=0[/tex] and we can find the variance with this formula:

[tex] Var(D) = Var(X) +Var(Y)-2Cov(X,Y)[/tex]

And repplacing we got:

[tex] Var(D)= 30^2 + 25^2 -0 = 1525[/tex]

And the deviation would be given by:

[tex] Sd(X) = \sqrt{1525}= 39.05[/tex]

c) If the costs can be described by Normal models, what's the probability that medical expenses are higher for someone's dog than for her cat?

For this case we can assume that the random variable D follows a normal distribution:

[tex] D\sim N(\mu_D = -25 , \sigma_D =39.05)[/tex]

And we want to find this probability:

[tex] P(Y-X >0) = P(D>0)[/tex]

And we can use the z score formula given by:

[tex] z= \frac{D-\mu_D}{\sigma_D}[/tex]

And replacing we got:

[tex] P(D > \frac{0-(-25)}{39.05}) = P(z>0.64)[/tex]

And using the complement rule we got:

[tex] P(Z>0.64) = 1-P(Z<0.64) = 1-0.739= 0.261[/tex]

d) What concerns do you have?

If the variables X and Y are not independent then we have an error in the calculation and if we don't have a normal distribution then the probability would be incorrect for this case.