SOLUTION: Fix-It Copiers advertises a mean time of 100 minutes for office calls with a standard deviation of 25 minutes. What percentage of calls are completed:
a. in less than 120 minut
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Question 917438: Fix-It Copiers advertises a mean time of 100 minutes for office calls with a standard deviation of 25 minutes. What percentage of calls are completed:
a. in less than 120 minutes?
b. in less than 60 minutes?
c. Twenty percent of their jobs take more than how much time?
I got for a: .7881 b: .0548 and c: .2004. I'm not sure I did it correctly because of the mirror rule (the other side on the graph).
Thanks a bunch.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Fix-It Copiers advertises a mean time of 100 minutes for office calls with a standard deviation of 25 minutes. What percentage of calls are completed:
a. in less than 120 minutes?
z(120) = (120-100)/25 = 20/25 = 0.8
P(x < 120) = P(z < 0.8) = 0.7881
b. in less than 60 minutes?
Correct = 0.0548
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c. Twenty percent of their jobs take more than how much time?
Find the z-value with an 80% left-tail::
Using my TI-84 I get invNorm(0.8) = 0.8416
Find the corresponding time value::
x = z*s + u
x = 0.8416*25 + 100 = 121.04 minutes
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Cheers,
Stan H.
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