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DarkendSinz30

The mean lifetime of a certain tire is 30,000 miles and the standard deviation is 2500 miles. If we assume the mileages are normally distributed, approximately what percentage of all such tires will last between 22,500 and 37,500 miles?

  • one year ago
  • one year ago

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  1. DarkendSinz30
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    I need to use the empyrical rule

    • one year ago
  2. tkhunny
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    (22500-30000)/2500 (37500-30000)/2500 What are these and what do these values tell us?

    • one year ago
  3. DarkendSinz30
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    Um they tell us the mileage of the tire?

    • one year ago
  4. DarkendSinz30
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    or they tell us the dev

    • one year ago
  5. tkhunny
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    z-scores, standard deviations. What are the values? Do the calculations

    • one year ago
  6. DarkendSinz30
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    -3 and 3

    • one year ago
  7. tkhunny
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    That's it. And what does the Empirical Rule say about that -- 3 standard deviations each way?

    • one year ago
  8. DarkendSinz30
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    do we place the percentages?

    • one year ago
  9. DarkendSinz30
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    The empirical rule says that if you go 1 standard deviation out from the mean that you will capture 68 % of your data, Go out one more stan dev (thats 2 so far) and you will capture 95% of your data, go out one more (thats 3 so far) and you will capture 99.7% of your data.

    • one year ago
  10. DarkendSinz30
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    So how do we place this in this problem could you give me an example?

    • one year ago
  11. tkhunny
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    It's not a matter of placing or calculating. This is the point of the "Empirical Rule". You just read off the result. +/- 1 Standard Deviation is 68% +/- 2 Standard Deviations is 95% +/- 3 Standard Deviations is 99.7% In this case, pick the third one. Done.

    • one year ago
  12. DarkendSinz30
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    So the answer to my question is 99.7% right

    • one year ago
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