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A company ships 5500 cell phones. They are expected to last an average of 10,000 hours
before needing repair; with a standard deviation of 500 hours.
Assume the survival times of the phones are normally distributed.
If a phone is randomly selected to be tracked for repairs find the probability
of cell phones that need repair,
a) after 11,500 hours
b) before 8900 hours
c) between 8,200 hours and 11,300 hours
d) exactly equal to 9,878 hours
d) Find the 20th percentile survival time of these phones.
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(a) Notice that in case (a), 11,500 hours is 3 times standard deviation of 500 hours from the mean.
So, you can use the empirical rule for normal distribution, which says in this case that
+----------------------------------------+
| "99.7% of the data fall within three |
| standard deviations from the mean". |
+----------------------------------------+
So, the probability to be out of 3 standard deviations from the mean is 100% - 99.7% = 0.3%.
This probability is the area of two tails under the normal curve out of the 3 standard deviations
from the mean. These two tails are symmetric - so we need only half of 0.3%, which is 0.15%.
This value of 0.15%, or 0.0015 decimal, is the answer for case (a).
(b) In part (b), the probability is the area under the normal curve
on the left of the raw mark of 8900 hours.
You may use the standard function "normalcdf" in your trgular calculator TI-83 or TI-84.
z1 z2 mean SD <<<---=== formatting pattern
P = normalcfd(-9999, 8900, 10000, 500)
The calculator gives the answer P = 0.0139.
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Solved.
Using this technique, you can easily solve part (c) on your own.