Question

A veterinary association claims that the annual cost of medical care for dogs averages $200 with...

A veterinary association claims that the annual cost of medical care for dogs averages $200 with a standard deviation of $60, and for cats averages $240 with a standard deviation of $70. Assume the costs follow a normal distribution.
a) What is the probability the annual cost of medical care for a dog is greater than $220?
b) What is the probability the average annual cost of medical care for 16 randomly chosen dogs is greater than $220?
c) How much do the most expensive 5% of cats cost?
d) What is the probability that the annual medical cost for a randomly chosen dog is higher than that for a randomly chosen cat? (Hint: calculate the mean and standard deviation for the difference between the costs)

Homework Answers

Answer #1

a)

P( annual cost of medical care for a dog is greater than $220)=P(X>220)=P(Z>(220-200)/60)=P(Z>0.33)=0.3707

b)

std error =std deviation/(16)1/2 =15

P( annual cost of medical care for a dog is greater than $220)=P(X>220)=P(Z>(220-200)/15)=P(Z>1.33)=0.0918

c)

for top 5 % ; critical z =1.645

hence coresponding cost =mean+z*std deviation =240+1.645*70=355.15

d)

mean difference of cost of dog and cat =200-240=-40

std deviaiton=(602+702)1/2 =92.195

hence probability that the annual medical cost for a randomly chosen dog is higher than that for a randomly chosen cat =P(X>0)=P(Z>(0-(-40))/92.195)=P(Z>0.43)=0.3336

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