document.write( "Question 1170606: Suppose that the times required for a cable company to fix cable problems in its customers' homes are normally distributed with mean 45 minutes and standard deviation 15 minutes.
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document.write( "a)What is the probability that a randomly selected cable repair visit will take at least 36 minutes?
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document.write( "b)What is the probability that a randomly selected cable repair visit will take at most 60 minutes?
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document.write( "c)What is the probability that a randomly selected cable repair visit will take between 36 minutes and 60 minutes?
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document.write( "d)5% of the cable repair visits will require more than what time?
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document.write( "e)What is the probability that the mean tame for 25 visits will be more than 50 minutes?
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document.write( "What is the probability that the mean tame for 36 visits will be more than 50 minutes\r
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Algebra.Com's Answer #795459 by Boreal(15235) You can put this solution on YOUR website! z=(x-mean)/sd \n" ); document.write( "a. z is > (36-45)/15 or -0.6, prob. 0.7257 \n" ); document.write( "b. at most 60 minutes is z < 15/15 or 1, probability 0.8413. \n" ); document.write( "c. between the two is -0.6 < z < 1, with probability 0.5671 \n" ); document.write( "d. z is 1.645 for the 95th percentile, so x-mean=24.675 and x=69.675 min. More than that. \n" ); document.write( "e. this is z > (50-45)/15/sqrt(25) or z>1.6 with probability 0.0548 \n" ); document.write( "f. this is z > 5/15/sqrt(36) or z>2 with probability 0.0228 \n" ); document.write( " |