Question 1208762: 1. 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.
Answer by ikleyn(52847)   (Show Source):
You can put this solution on YOUR website! .
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.
Probably, it is all that you need to know to get the answer for this school assignment.
But I want to present you an alternative, much more attractive way to solve this and
many other similar problems.
Go to web-site https://onlinestatbook.com/2/calculators/normal_dist.html
and use free of charge calculator there.
It allows to do all these calculations, and every time it shows you
the area of interest under the normal curve.
Its interface is very friendly, very simple, very intuitive and very convenient.
It will help you to learn the entire subject in 5 - 10 minutes, better than any tutor/teacher will do it.
You may play with this calculator as long as you need, until everything
in the subject becomes clear to you.
After that, you may return to your regular hand calculator TI-83 / TI-84
enriched with full understanding of the subject.
Solved.
Using this technique, you can easily solve part (c) on your own.
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