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Unformatted text preview: STA 100: Homework 4 Due: 5:00pm, Wed 9th February Assigned: Thursday, 3rd February For all questions you must show your working. This enables us to understand your thought process, give partial credit and prevent crude cheating. Please see the code of the conduct in the Syllabus for rules about collaborating on homeworks. You must turn in a printed version of your R output with your solutions! (If you use another piece of software then that is fine – please state this on your solutions) 1. Normal Pets: The American Veterinary Assoication claims that the annual cost of medical care for dogs in the US averages $ 100 with a standard deviation of $ 30. The annual cost for cats is reported to be $ 120 with a standard deviation of $ 40. You may assume that veterinary costs for dogs and cats are independent, and that both are normally distributed. (a) What is the probability that a dog owner will spend more than $ 130 on medical care in a given year? We can do this by using z scores: Z = 130 100 30 = 1 Pr ( X dog ≥ 130) = P X dog 100 30 ≥ 130 100 30 = P ( Z > 1) = 1 P ( Z < 1) = 0 . 16 or, plug directly in R using: "Distributions > Continuous Disributions > Normal Distribution > Normal probabilities" Selecting "variable value"= 130, mean=100, sd=30, lower.tail=TRUE, gives Pr(X<130), so Pr(X>130) = 1  Pr(X<130): > 1  pnorm(130,mean=100,sd=30,lower.tail=TRUE) [1] 0.1586553 (b) What is the probability that a cat owner will spend $ 160 or more on medical care in a given year? Again, using z scores: Z = 160 120 40 = 1 Pr ( X cat ≥ 130) = 1 P ( Z < 1) = 0 . 16 or, plug directly into R : Selecting "variable value"= 160, mean=120, sd=40, lower.tail=TRUE, gives Pr(X<160), so Pr(X>160) = 1  Pr(X<160): > 1  pnorm(160,mean=120,sd=40,lower.tail=TRUE) [1] 0.1586553 (c) Compare your answers for (1a) and (1b). Explain what you notice. The answers are the same because both are one standard deviation above the mean (they have the same z score = 1). 1 (d) What is the probability that a dog owner will spend less than $ 80 on medical care in a given year? Again, using z scores: Z = 80 100 30 = . 6667 Pr ( X dog ≤ 80) = P ( Z < . 6667) = 0 . 2525 or plugging directly into R : Selecting "variable value"= 130, mean=100, sd=30, lower.tail=TRUE, gives Pr(X<80): > pnorm(80,mean=100,sd=30,lower.tail=TRUE) [1] 0.2524925 (e) What is the probability a dog owner will spend between $ 80 and $ 130 on medical care in a given year?...
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This note was uploaded on 01/17/2012 for the course STAT 100 taught by Professor drake during the Fall '10 term at UC Davis.
 Fall '10
 DRAKE

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