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Normal distribution: tire durability

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data

    The top-selling Red and Voss tire is rated 60000 miles, which means nothing. In fact, the distance the tires can run until wear-out is a normally distributed random variable with a mean of 75000 miles and a standard deviation of 5000 miles.

    A. What is the probability that the tire wears out before 60000 miles?

    B. What is the probability that a tire lasts more than 85000 miles?


    2. Relevant equations

    z = [X - (mean of X)]/Standard deviation


    3. The attempt at a solution

    This is another simple question that I can't seem to be getting correct.

    For part A, I did z = (60000 - 75000)/5000 = -3
    Then P(z < -3) = 0.0013 according to the chart.. well 0.0013 is P(z= -3.0).. I tried entering the answer for P( z= -2.99) = 0.0014 since it's the last value on the chart lower than -3 also but that was also incorrect.

    For part B, z = [85000 - 75000]/5000 = 2
    Then P(z > 2) = 1- P(z = 2) = 1 - 0.9772 = 0.0288
    Also incorrect..
     
  2. jcsd
  3. Jun 28, 2011 #2

    rock.freak667

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    What region does you chart provide z-values for? Some charts will directly give you the value of P(z<a).

    P(z>2) = 1-P(z≤2) = 1-P(z=2)-p(z<2).
     
  4. Jun 28, 2011 #3
    im not sure what u mean by which region but my hart goes to a point z?
     
  5. Jun 28, 2011 #4

    rock.freak667

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    Forget my initial posts with P(z=a), you can't have P(z=a) for a normal distribution. I would think your answers are correct, I am not sure why exactly it is saying is incorrect.
     
  6. Jun 29, 2011 #5

    I like Serena

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    Your first answer should be correct as rf already indicated.
    Perhaps you need to find a certain input format?
    Like present it as a percentage, or with a different number of significant digits?

    You made a calculation error in the second. It should be 0.0228 or 2.28%.
     
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