Normal distribution: tire durability

k77i
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Homework Statement



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?


Homework Equations



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


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..
 
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k77i said:

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.

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

k77i said:
For part B, z = [85000 - 75000]/5000 = 2
Then P(z > 2) = 1- P(z = 2) = 1 - 0.9772 = 0.0288
Also incorrect..

P(z>2) = 1-P(z≤2) = 1-P(z=2)-p(z<2).
 
im not sure what u mean by which region but my hart goes to a point z?
 
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.
 
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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