Question

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 72000 miles and a standard deviation of 7000 miles.A. What is the probability that the tire wears out before 60000 miles?Probability = What is the probability that a tire lasts more than 80000 miles? Probability=

Answer 1

a. 0.0436

b. 0.1271

We are given the following:

Distance (x) = 60,000

Mean (u) = 72,000

Standard Deviation(s) = 7,000

We are also told that it is a normal disribution relationship. The formula for ND is as follows:

z = (x - u) / s

Now we can continue with part a and b as follows:

a) P (x < 60,000)

= P (z < (60000 - 72000) / 7000)

= P (z < -1.714)

We can find the corresponding z score by looking at a z score table, and we find th probability to be 0.0436

b) P ( x > 80,000)

= P(z > (80000 - 72000) / 7000)

= P( z > 1.143)

We find the corresponding z score to be 0.8729, now we can substract this from 1 sinsce our probability is larger than the given distance (meaning we are trying to find the area to the right of the z score) to find our final answer:

1 - 0.8729 = 0.1271