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Additional Problems solutions

Additional Problems solutions - Stat 133 Recitation Winter...

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Stat 133 Recitation Winter 2007 Additional Problems SOLUTIONS Suppose the skid distance during an accident at this intersection has a normal distribution with mean 50 feet and standard deviation 5 feet. 1. What is the probability that one randomly selected car skids more than 51 feet in an accident at this intersection? X has a normal distribution with mean = 50, std = 5 P(X>51) = P (Z > (51-50)/5) = P (Z > 0.2) = 1- P(Z<0.2) = 1-0.5793 = 0.4207 2. What is the probability that one randomly selected car skids between 46 and 51 feet in an accident at this intersection? Plot will help a lot! P(46<X<51) = P(X<51) – P(X<46) = P(Z<(51-50)/5) – P((46-50)/5) = P(Z<0.2) – P(Z<-0.8) = 0.5793 - 0.2119 = 0.3674 3. What is the 10 th percentile for the skid distance for an accident at this intersection? Want to find P(Z<?) = 0.1 , from Table A, Z = -1.28 Unstandardize X= 50 + (-1.28)*5 = 43.6 4. What is the probability that the average skid distance for a random sample of 100 cars is more than 51 feet? First find the distribution of sample mean X-bar X-bar has a normal distribution with mean = 50, std = 5/sqrt(100)=0.5 P(X-bar>51) = P(Z > (51-50)/0.5) = P(Z>2) = 1-P(Z<2) = 1- 0.9772 = 0.0228
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