Suppose that the life span of a certain tire is normally distributed, with u=25,

Question

Suppose that the life span of a certain tire is normally distributed, with u=25,000 miles and standard deviation=2,000 miles.
Find the probability that a tire will last between 28,000 and 30,000 miles.

Find the probability that a tire will last more than 29,000 miles.
Suppose that the life span of a certain tire is normally distributed, with u=25,000 miles and standard deviation=2,000 miles.
Find the probability that a tire will last between 28,000 and 30,000 miles.

Find the probability that a tire will last more than 29,000 miles.

Find the probability that a tire will last between 28,000 and 30,000 miles.

Find the probability that a tire will last more than 29,000 miles.

Explanation / Answer

We need to find P(28,000<x<30,000)

Now define variable z= (x-mean)/standard deviation = (x-25,000)/2,000 and this is standard variable for normal distribution

Now for x=28,000, z= 28000-25000/2000 = 3000/2000 = 3/2 = 1.5

For x= 30,000, z= 30000-25000/2000 = 5000/2000 = 5/2 = 2.5

So we need to find P(1.5<z<2.5) = P(z=2.5) - P(z= 1.5) = 0.994-0.933 = 0.061 or 0,060597

Now we also want P(x>29,000)

z= 29000-25000/2000 = 4000/2000 = 2

So we want P(z>2) = 0.023 or 0.0227501

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