1. ## Normal Distribution Problem

If the life, in years, of a television set is normally distributed with a mean of 35 years and a standard deviation of 8 years, what should be the guarantee period if the company wants less than 3% of the television sets to fail under warranty?

I know how to set up the curve with the mean and standard deviation, but the 3% is throwing me off?

Thanks

2. the 3% would imply that it wants a 97% success rate. the Z corresponding to a value of 3% is 1.88.

to find the appropriate value corresponding to that result:

$1.88=\frac{X-\mu}{\sigma} \rightarrow 1.88=\frac{X-35}{8}$ now just isolate for X.

3. Originally Posted by lion66
If the life, in years, of a television set is normally distributed with a mean of 35 years and a standard deviation of 8 years, what should be the guarantee period if the company wants less than 3% of the television sets to fail under warranty?

I know how to set up the curve with the mean and standard deviation, but the 3% is throwing me off?

Thanks

See Rule #1 of the forum.

4. Where did you get the 1.88 from?

Originally Posted by lllll
the 3% would imply that it wants a 97% success rate. the Z corresponding to a value of 3% is 1.88.

to find the appropriate value corresponding to that result:

$1.88=\frac{X-\mu}{\sigma} \rightarrow 1.88=\frac{X-35}{8}$ now just isolate for X.

5. If you look at the the graph in the attachment, the blue area is the one associated with a 3% failure rate. So now we need to find the Z that goes with that 3%. We need to look at the normal distribution table to find the appropriate Z. So if you open the second attachment I have highlighted 3%, so we scroll left and up and realize that the value given is 1.88.