SOLUTION: 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.

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.
a)What is the probability that a randomly selected cable repair visit will take at least 36 minutes?
b)What is the probability that a randomly selected cable repair visit will take at most 60 minutes?
c)What is the probability that a randomly selected cable repair visit will take between 36 minutes and 60 minutes?
d)5% of the cable repair visits will require more than what time?
e)What is the probability that the mean tame for 25 visits will be more than 50 minutes?
What is the probability that the mean tame for 36 visits will be more than 50 minutes

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
z=(x-mean)/sd
a. z is > (36-45)/15 or -0.6, prob. 0.7257
b. at most 60 minutes is z < 15/15 or 1, probability 0.8413.
c. between the two is -0.6 < z < 1, with probability 0.5671
d. z is 1.645 for the 95th percentile, so x-mean=24.675 and x=69.675 min. More than that.
e. this is z > (50-45)/15/sqrt(25) or z>1.6 with probability 0.0548
f. this is z > 5/15/sqrt(36) or z>2 with probability 0.0228
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