Help - I think I'm right

[DiceyPlay]

I'm a math tutor at a nearby college. I was reviewing a test with a student and came across a problem I thought the student had done right but the teacher marked it wrong. I emailed the teacher. Here is a copy of the email I sent to the teacher:

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Hi,

My name is Michael xxxxxxx. I am a tutor in the tutorial center at XYZ. I reviewed an exam or quiz with a student. The exam or quiz was given this semester in a statistics class you are teaching. In particular question #9 is worded as follows:

A used car dealership has found the length of time before a major repair is required on the cars it sells is normally distributed with mean equal to 10 months and standard deviation of 3 months. If the dealer wants only 5% of the cars to fail before the end of the guarantee period, how many months should the car guarantee be?

The student gave an answer of 5 months. The students answer was marked wrong and a correct answer of 15 was indicated.

If one recommends a quarantee period (15 months) that is greater then the mean time (10 months) until a major repair is required, we should expect more than 50% of the cars to come back for service under the guarantee.

Let x be a random variable with distribtion as described in the question above. 95.15% of the area under the curve lies to the left of x=15. This means that, on average, 95.15% of the cars will require major repair within 15 months.

The more I look at this the more I convince myself that the students answer is correct. Am I looking at this wrong somehow?

A reply is greatly appreciated.

Kind and respectful regards,

-Michael

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Here is the reply I received from the teacher:

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Hello Michael,

Yes, at first glance 5 months seems like a correct answer. Actually when I did the problem first myself I got that answer!
But let me reword the problem.
Let's say that on average it takes 10 months for an average person to finish a training program and successfully pass a test, with SD of three months. If you want to make sure(guarantee) that 95% of the people enrolled will pass the test, how many months should the program last? You see 5 months does not make sense.

You want only 5% of the cars to fail or 95% not fail. If you set up the problem the way that you have, 95% of the cars fail and only 5% will not have problems.

Hope this helps,

Kia

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To me he's changing the problem. Of course one could construct a different problem where a random variable has the same distribution and the answer is different. I don't know how to make it any more clear than with what I wrote to him.

Or maybe I'm just flat wrong. But I obviously don't think so. Can someone either explain it to me or show me a way to explain to him so he can see the error in his ways?

Thanks ... your help is greatly appreciated.

-Michael

[Victor]

so hes saying that 95% of cars will fail after 5mo? but the mean failure time is 10mo?

[Colima420]

I think that the professor made a typo and the problem should have been stated as:

"A used car dealership has found the length of time before a major repair is required on the cars it sells is normally distributed with mean equal to 10 months and standard deviation of 3 months. If the dealer wants only 5% of the cars to fail AFTER the end of the guarantee period, how many months should the car guarantee be? "

First of all, it wouldn't make sense for a dealer to want the mayority of his cars (95%) to fail when the customers don't have the guarantee. It only makes sense for the dealer to want only 5% of cars to fail after the guarantee. This way the guarantee actually works. If 95% of cars need repairs after the guarantee is over, then the guarantee has no purpose.

Secondly, if we analyze his example of his reply to you (regarding the time it takes a person to complete a course), it is almost identical to the car problem with the AFTER word in it.

This way,


z = (x - mu) / sigma

The z critical value for .05 (probability of car failure, since he wants 5%) is 1.645. This can be found in any z table.

now if we solve the previous equation with this number we get:

1.645 = (x - 10) / 3
4.935 = x - 10
14.935 = x Rounding up we get x = 15. We need to have the guarantee to be for 15 months in order to have 5% of all cars fail after that.

So my whole theory is that the original problem should have been: "How many months should the guaranteee be in order to have 5% of the cars not be covered by it?"

or conversely: "How many months should the guarantee be for in order to have 95% of the cars covered by it"?

Colima420

[ojc02]

Yeah, I agree this must have been the intention. The student answered the question correctly and frankly, the original question makes more sense. The dealership is likely to want most of the cars to fail after the end of the warranty.

It is totally unreasonable of the professor to change the wording of the problem so drastically afterwards. The student provided the correct answer to the question posed.

[madnak]

You're right, teacher is wrong.

[DiceyPlay]

Let me know if you're interested and I'll make sure I post the results or the on-going saga.

Here is my response to his reply.

Hi Kia,

I'm not sure how to respond to your reply. But stating a different problem does not help. Discussing the similarities and dis-similarities between the two problems simply would not be fruitful. Let's stick with the original problem. I thought my argument was clear and compelling. Let's try again.

1) Intuitively - longer guarantee periods result in more cars being covered under the guarantee.

2) A mean failure time of 10 months suggests 50% the cars fail before 10 months and 50% the cars fail after 10 months (normal distribution -- median = mean).

Putting (1) and (2) together a guarantee period of 15 (15 > 10) months results in more than 50% of the cars coming back for service under the guarantee.

It should now be clear 15 couldn't possibly be the correct answer.

You wrote:

"You want only 5% of the cars to fail or 95% not fail."

All the cars eventually fail. The question is when. We don't know when any one car will fail. We can only make inferences about when a proportion of the cars will fail.

Let x be a rv with distribution as described in the question. x denotes the number of months until a car fails. 95.15% of the area under the curve lies to the left of x=15. Therefore, in the first 15 months ~95% of the cars have failed and ~5% have not failed. A symmetry argument shows the correct answer to the problem to be 5.

I hope this is clear. I am open to being wrong and easily accept it when deonstrated.

Thank you for your time. Again, a reply is greatly appreciated.

Respectfully,

-Michael

[Lestat]

I'm following this thread, but staying out of it, because my math skills are terrible. I have to be in the exact right mood to figure stuff like this out, because I don't have the formal education. But...

<font color="blue">2) A mean failure time of 10 months suggests 50% the cars fail before 10 months and 50% the cars fail after 10 months (normal distribution -- median = mean). </font>

Are you sure this is what it means? To me, it means if I buy any given car out of 100 on the lot, I could "expect" it to last 10 months. And the difference (the times it will last longer or fail before 10 months), is the SD. I wouldn't "expect" every 2nd car I buy to NOT last 10 months or last longer than 10 months, even though I realize that it's unlikely to last "exactly" 10 months.

I'm sure I'm making a fool of myself, because you know much more about this than I do. I only know about SD as it pertains to my poker results. Every hour I play, I expect to earn my mean hourly rate, even though it's unlikely that I'll earn "exactly" my hourly rate in any given hour or series of hours.

And to the person who said, "First of all, it wouldn't make sense for a dealer to want the mayority of his cars (95%) to fail when the customers don't have the guarantee."

Of course it does! It COSTS dealers money to fix cars when they're still under warranty. They MAKE money fixing cars when they're out of warranty.

I know math problems in schoolbooks aren't always realistic, but instinct says that 5 months is too low. 15 months seems more appropriate. But that has nothing to do with the math and I haven't taken the time to work the problem out myself. That's just what my common sense would tell me.

[jogsxyz]

Dealerships want no cars to fail during the guarantee period. You buy an appliance. It's guaranteed for one year. It will fail at one year and one day.
The mean is ten months. Half the repairs will fail in less than ten months. No way does the repair shop wish to guarantee more than ten months. The answer must be less than ten months.
At a quick glance the answer should be four months, two standard deviations. In the real world the normal distribution only approximates the fail rate. Five months seems like a reasonable period for the guarantee.

[DiceyPlay]

You're not making a fool of yourself in my opinion.

Yes, I'm sure that's what it means. This particular dealer has observed that on average the cars they sell break down in 10 months. So the expectation for any car they sell from this point forward or until they recalculate their statistics is for that car to last 10 months before it needs a major service. An sd of three months means the standard deviation if 3 months. In a normal distribution 68% of the observations fall within 1 sd of the mean (average). An observation is noticing how long it took for a car they sold to require major service. So the dealer expects 68% of the cars they sell to require service anywhere from 7 to 13 months from the date of sale. 7 = 10 - 3 and 13 = 10 + 3. Likewise, in a normal distribution 95% of the observations fall within 2 sd's of the mean. So the dealer should expect 95% of the cars they sell to require major service between 4 and 16 months from the date of sale. 4 = 10 - 2 * 3 and 16 = 10 + 2 * 3.

I agree with you on your comment about what one of the other posters suggested.

Do you still think 15 months is more appropriate?

[Lestat]

I missed that it was a "used" car dealership! Sorry. So 15 months isn't what my instinct tells me anymore. -lol


<font color="blue">So the dealer should expect 95% of the cars they sell to require major service between 4 and 16 months from the date of sale. </font>

I agree. This seems right. But here's what's buggin me:

The mean is 10 and the dealer doesn't want to service more than 5% BEFORE the warranty expires. "Before" is the key word, because now you're only working with the TOP half of the bell curve, no? We can expect any given car to last 10 months. So how many MORE months are needed before we expect to be servicing greater than 5% of the cars we sell?

Seriously, I'm an idiot with math. This should be really simple for anyone that knows the calculations. DS, Bruce Z, Borodog, and many others could do this in their sleep. And no offense to madmak, yourself, and others who are probably just as capable. You guys could be right. If necessary, I can break out Malmuth's GTAOT and look up the correct calculations. But there are so many others on here who could do this off the top of their head. This is why I stay out of these threads, although I do find them interesting to read.