Probability

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Sue0113

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Replacement times for TV sets are normally distributed with a mean of 8.2 and a standard deviation of 1.1years. Find the probability that a randomly selected TV will have a replacement time less than 5 years. If you want to provide a warranty so that only 1% of the TV sets will be replaced before the warrenty expires, what is the time length of the warranty?

8.2+(-2.326)(1.1) = 5.64 years is length of warrenty

How do I find the probability that a randomly selected TV will have a replacement time less than 5 years?
 
Sue0113 said:
Replacement times for TV sets are normally distributed with a mean of 8.2 and a standard deviation of 1.1years. Find the probability that a randomly selected TV will have a replacement time less than 5 years. If you want to provide a warranty so that only 1% of the TV sets will be replaced before the warrenty expires, what is the time length of the warranty?

8.2+(-2.326)(1.1) = 5.64 years is length of warrenty

How do I find the probability that a randomly selected TV will have a replacement time less than 5 years?
Click to expand...

How did you get there? Please show work ....
 
Sue0113 said:
Replacement times for TV sets are normally distributed with a mean of 8.2 and a standard deviation of 1.1years. Find the probability that a randomly selected TV will have a replacement time less than 5 years. If you want to provide a warranty so that only 1% of the TV sets will be replaced before the warrenty expires, what is the time length of the warranty?

8.2+(-2.326)(1.1) = 5.64 years is length of warrenty

How do I find the probability that a randomly selected TV will have a replacement time less than 5 years?
Click to expand...
Assuming that you know where the value of -2.326 came from [which you haven't shown us yet!], you have found where a particular probability takes you in the normal distribution, and converted that to years.

The procedure to find the probability for a given number of years is the opposite of what you did to find years as a function of probability. If that clue is not enough, please post details of what you have done.
 
How I got there

I found the z score by z=x- µ/Ϭ =5-8.2/1.1


x-8.2/1.1 = 2.3264
x-8.2 = 2.559
x=5.64

So is this the only answers I need to provide?
 
Sue0113 said:
I found the z score by z=x- µ/Ϭ =5-8.2/1.1


x-8.2/1.1 = 2.3264
x-8.2 = 2.559
x=5.64

So is this the only answers I need to provide?
Click to expand...
For the other part of the question, use the table "backwards" from this part. Find the z score for 5 years: z = (5 - 8.2)/1.1 = ..
Then go to the table entry for that z (probably have to look for positive, but interpret result as for negative), and read off the cumulative probability (usually called "F"). What fraction of the distribution is further than z from the mean, on the negative side?
 
Therefore

Therefore the (5-8.2)/1.1 = -2.909
the z entry would be .9981
so the probability that a randomly selected tv will have a replacement time less than 5 years is .9981
Wow this is confusing!
 
Sue0113 said:
Therefore the (5-8.2)/1.1 = -2.909
the z entry would be .9981
so the probability that a randomly selected tv will have a replacement time less than 5 years is .9981
Wow this is confusing!
Click to expand...
The confusion is because the table is written for positive z, but you have negative. When you flip it around, the probability is the complement - you want to know what fraction of the distribution is more than 2.909 standard deviations from the mean - the tail of the distribution.

1 - F = 0.0019

The probability of failure before 5 years is only about 0.2%

And from the other part of the question that you already did,
probability of failure before 5.64 years is 1%.
Those two results are nicely compatible with each other.
 
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