Question 1178022: I'm working on this problem:
The lifetime of a certain car battery is reported to be 154 weeks with a standard deviation of 8 weeks and is normally distributed. If the company offers a 3-year guarantee, compute the percentage of batteries that will be returned for refund for this battery (52 weeks = 1 year).
I don't think I did it correctly at all.... do I calculate the P(z<156-154/(8/100^.5)?
Please help. I think I just need a nudge in the right direction.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the lifetime of a battery is reported to be 154 weeks with a standard deviation of 8 weeks and is normally distributed.
that would be the average lifetime.
the company offers a 3 year guarantee.
3 years * 52 weeks a year is equal to 156 weeks.
if the battery dies in less than 156 weeks, then the battery can be returned for a refund.
you are looking for the probability that the life of the battery will be less than 156 weeks.
you want to find the z-score.
z-score formula is z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation in this case.
with the numbers you have, the formula becomes:
z = (156 - 154) / 8.
solve for z to get z = 2/8 = .25.
find the area to the left of a z-score of .25 is equal to .5987062744.
based on the data, it is expected that approximately 60% of the batteries will be returned for a refund.
it looks like you were calculating a standard error rather than just using the standard deviation.
standard error is used when you have a sample of a certain size.
you then take the standard deviation and divide it by the square root of the sample size.
in this problem, you don't have a sample as far as i can see.
it looks like you are dealing with the population.
in that case, you just use the standard deviation rather than the standard error.
your calculation is p(z < (156-154)/(8/sqrt(100)).
it looks like you think you are dealing with a sample of 100 batteries.
i didn't see that in the problem statement.
based on what i saw, your formula should have been p(z < (156-154)/8.
if in fact, you did have a sample of size 100, then i think your formula is correct.
the standard error would be 8 / sqrt(100) = 8/10 = .8
the formula would become z = (156-154)/.8 = 2.5
the probability of getting less than a z-score of 2.5 would be much higher.
it would be .99379...
pretty much all of the batteries would need to be returned for a refund.
that doesn't sound right.
if they were looking for a small percentage of refunds to be given, i would think that the average life of the batteries would need to be much higher, or the guarantee would need to be much lower.
for example, if they were looking for a refund percentage of 5%, with a guarantee of 3 years, then the average life of the battery should have been much higher.
5% on the left side of the normal distribution curve would need a z-score of -1.645.
z-score formula is z = (x-m)/s
using the standard deviation of 8, and an x of 156, the formula would become:
-1.645 = (156 - m) / 8
solve for m to get m = 169.16.
the average life of the battery would have had to be 169.16 if you were dealing with the population.
in that case, z would be equal to (156 - 169.16) / 8 = -1.645 and the area to the left of that z-score would be equal to .-5, meaning that 5% of the batteries would need to be returned for a refund.
if you are talking about the average life of a sample of 100, then you would get a different result.
in that case, you would use the standard error rather than the standard deviation.
the standard error would be 8/sqrt(100) = 8/10 = .8
your desired z-score is still -1.645 if you only want to give a refund to 5% of the customers.
the z-score formula would become -1.645 = (156 - m) / .8
solve for m to get m = 157.316.
the average life of the sample of 100 would need to be 157.316 weeks.
your z-score formula would becomes z = (156 - 157.216) / .8 = -1.645.
bottom line is you need to check your problem statement again to see if something is missing.
let me know what you find and what you think your calculations need to be if the problem statement is different than the one i'm seeing.
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