Question 1011521
As estimate of the mean time of continuous use until an
answering machine will first require service is desired. 
If it can be assumed that α = 60 days, how large a sample
is needed so that one will be able to assert with 90%
confidence that the sample mean is off by at most 10
days?
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E = z*s/sqrt(n)
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Therefore::
n = [z*s/E]^2
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Your Problem::
n = [1.654*60/10] = 9.924^2 = 99 when rounded up
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Cheers,
Stan H.
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