# Thread: Statistics- Z-test Worded problem

1. ## Statistics- Z-test Worded problem

Here's the problem::

"A certain kind of tire is known to begin showing wear when it has been driven for 16,500 miles on the average with a standard deviation of 3,300 miles. If a car rental agency buys 36 of these tires for replacement purposes and puts each one on a different car, what are the probabilities that the 36 tires will average."

i couldn't really get the question... it's either i'm very bad at english or i'm very bad at analyzing. In any case, I also find it quite confusing having 2 units (miles and tires)

In urgent need... Help greatly appreciated!

2. As I read the question, the last sentence doesn't really make sense...

Have you missed out some words from the question, perhaps?

3. Originally Posted by Pterid
As I read the question, the last sentence doesn't really make sense...

Have you missed out some words from the question, perhaps?
nope... that's exactly what the question states...

maybe it means that the buyer buys 36 tires on the average(??)

4. OK. It's strange, because the final sentence looks like the beginning of a question - but it doesn't finish with a question mark. ("?")

Is it possible that the question continues on the next page of your book? Or maybe it looks like this:

A certain kind of tire is known to begin showing wear when it has been driven for 16,500 miles on the average with a standard deviation of 3,300 miles. If a car rental agency buys 36 of these tires for replacement purposes and puts each one on a different car, what are the probabilities that the 36 tires will average

a) X miles
b) Y miles
c) Z miles

before showing signs of wear?

If you are right, and there is no more to the question, then it does not give you enough information to solve it - so there's no answer!

5. Originally Posted by Pterid
OK. It's strange, because the final sentence looks like the beginning of a question - but it doesn't finish with a question mark. ("?")

Is it possible that the question continues on the next page of your book? Or maybe it looks like this:

If you are right, and there is no more to the question, then it does not give you enough information to solve it - so there's no answer!
Hmm... you think it can be reverse engineered..?? coz our professor gave the answer... which is 0.0146... maybe it's sort of a trick question (coz its a plus point on the exam)

6. I just think - actually, I know for sure - that the question is not finished. You need more information before you can solve it.

7. ok.. but i'm still curious, if for example it was complete... what are the ways to solve this problem?

8. Well, here's how I would approach it:

Let's call $N$ the number of hours that a tyre lasts, before it starts to show wear. Obviously, $N$ is different for every tyre. So, we say that $N$ is a random variable - it can take any value, according to a distribution.

In this case, we assume it obeys the normal distribution, with a mean average value $\bar{N} = 16500$ and a standard deviation $\sigma_{N} = 3300$. I assume you know what these things mean.

For your question, you are interested in the average lifetime of 36 tyres. I will call this $N_{36}$. Now, it's possible to show that $N_{36}$ also follows the normal distribution. Just like $N$, it has mean $\bar{N_{36}} = \bar{N}$. But it has a different standard deviation: $\sigma_{N_{36}} = \sigma_{N} / \sqrt{36} = \sigma_{N} / 6 = 550$. (Did you see something like this in your course?)

Now, to find out the probability of the average lifetime of 36 tyres, $N_{36}$ being greater than a certain number of miles - say $X$ miles - you can use the following formula:

$P(N_{36}>X) = \text{erf}\left(\frac{X-\bar{N_{36}} }{\sigma_{N_{36}}}\right)$

Tthe function $\text{erf}(x)$ is called the error function. You might have access to it - or something smiilar - in the form of a big table of values.

- - -

If you would like me to give more detail on part of this, reply and let me know.

9. if you want to convert tire into miles youve got the formula L=2piR (ouarf ouarf)
to be serious i think its what you call the standard deviation combined whith the information off the averrage duration of a tire thats give you the information of the probability that one tire is 'normal'
as i dont know how exactly is calculated the standard deviation (i've got an idea but i have no more time) i can't help you
i think normal most mean that a tire doesnt break before the average-standart deviation but can we be sure
once you would have the probability for on tire it will be easy to answer the question (there is only one tire per car)
if it was in french i would look at 'variance' in a encyclopedia to refresh my ideas

10. i think it is higly probable that the term 'standard deviation' or the chapter you picked your exo off is refering to a percular type of probability law (whith high probability for the gausian) otherwise something would be missing somewhere!