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 Post subject: Empirical RulePosted: Mon, 4 Sep 2006 05:02:07 UTC
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Joined: Fri, 3 Mar 2006 01:07:07 UTC
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A certain brand of automobile tire has a mean life span of 35,000 miles and has a standard deviation of 2250 miles. (Assume the life spans of the tires have a bell-shaped distribution).

The life spans of three randomly selected tires are 30,500 miles, 37,250 miles, and 35,000 miles. Using the Empirical Rule, find the percentile that corresponds to each life span.

I understand how to solve using the Empirical Rule, but I don't understand how to use that answer to figure out the percentile it goes with. I'm hoping someone can explain that to me.

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 Post subject: Re: Empirical RulePosted: Mon, 4 Sep 2006 09:01:21 UTC
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brightdim wrote:
A certain brand of automobile tire has a mean life span of 35,000 miles and has a standard deviation of 2250 miles. (Assume the life spans of the tires have a bell-shaped distribution).

The life spans of three randomly selected tires are 30,500 miles, 37,250 miles, and 35,000 miles. Using the Empirical Rule, find the percentile that corresponds to each life span.

I understand how to solve using the Empirical Rule, but I don't understand how to use that answer to figure out the percentile it goes with. I'm hoping someone can explain that to me.

You are to convert the percentages in the Empirical Rule to percentiles. In the following graph of a bell-shaped curve, the mean is and the standard deviation is

What percentile is at ? This percentile is the percentage of the area under the curve to the left of Using the Empircal Rule, 95% of the area under the curve is between the two lines at and so the area outside those two lines is Half of this 5% is to the left of the line at which means the percentile at is

The percentile at is the sum of the areas to the left the line at which is

Last edited by Bilbo on Mon, 28 Feb 2011 04:43:58 UTC, edited 1 time in total.

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 Post subject: Posted: Tue, 5 Sep 2006 09:44:42 UTC
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Joined: Fri, 3 Mar 2006 01:07:07 UTC
Posts: 39
This may be a stupid question, but how did you know they were two standard deviations away from the mean? I couldn't find anything in the textbook about figuring that out. Is it just a constant or is there some way to figure it out from the z-scores?

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 Post subject: Posted: Tue, 5 Sep 2006 16:09:34 UTC
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brightdim wrote:
This may be a stupid question, but how did you know they were two standard deviations away from the mean? I couldn't find anything in the textbook about figuring that out. Is it just a constant or is there some way to figure it out from the z-scores?

I'm not sure how to interpret your question. The Empirical Rule says 95% of the area under the curve is within 2 standard deviations of the mean. Are you asking where the rule comes from? The rule is composed of approximations derived from the Normal distribution. For that distribution, 95% of the area under the curve is actually within 1.96 standard deviations from the mean.

Last edited by Bilbo on Mon, 28 Feb 2011 04:44:03 UTC, edited 1 time in total.

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 Post subject: Posted: Tue, 5 Sep 2006 18:40:02 UTC
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brightdim wrote:
This may be a stupid question, but how did you know they were two standard deviations away from the mean? I couldn't find anything in the textbook about figuring that out. Is it just a constant or is there some way to figure it out from the z-scores?

Bilbo was just using +/- 2 standard deviations as an example. For this problem, you are first supposed to solve for the number of standard deviations away from the mean and then translate this into a percentile. You have to think about whether the one-tailed or two-tailed test is applicable to this problem.

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 Post subject: Posted: Fri, 8 Sep 2006 02:30:34 UTC
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Joined: Fri, 3 Mar 2006 01:07:07 UTC
Posts: 39
Well, I wasn't understanding why you said they were both two standard deviations away. When I did the math, I got two different z-scores: -2 and 1. That's where I was confused. I double checked the math and the answers in the back of the book and it turned out it should have been -2 and 1 which gave me 2.5% and 84%.

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