SOLUTION: A phone company has determined that the length of its customers' calls is normally with a mean of 2.87 minutes and a standard deviation of 1.51 minutes. Find the probability th

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Question 364862: A phone company has determined that the length of its customers' calls is normally with a mean of 2.87 minutes and a standard deviation of 1.51 minutes.
Find the probability that a customers' call will last for more than 4 minutes.
the probability that a customer's call will last between 1.1 and 3.2 minutes.
the 21st percentile for the length of telephone calls (a time such that 21% of calls are less than that time)
the probability that a random sample of 30 calls has a mean length of more than 3.43minutes
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A phone company has determined that the length of its customers' calls is normally with a mean of 2.87 minutes and a standard deviation of 1.51 minutes.
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Find the probability that a customers' call will last for more than 4 minutes.
z(4) = (4-2.87)/1.51 = 0.7483
P(x > 4) = P(z > 0.7483) = 0.2271
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the probability that a customer's call will last between 1.1 and 3.2 minutes.
Find the z-values of 1.1 and 3.2
Find the area under the normal curve over that z interval.
I get: 0.4659
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the 21st percentile for the length of telephone calls (a time such that 21% of calls are less than that time)
invNorm(0.21) = -0.8064
x = -0.8064*1.51+2.87 = 1.6523
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the probability that a random sample of 30 calls has a mean length of more than 3.43minutes
t(3.43) = (3.43-2.87)/[1.51/sqrt(30)] = 2.0313
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P(x-bar < 3.43) = P(t> 2.0313 when df=29) = 0.0257
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Cheers,
Stan H.