Quote: Meadas company makes shocks. The currently guarantee their shocks for 50,000 miles. The distance traveled before the shocks wear out is normally distributed with mean 63,000 and standard deviation 10,000.
What % of shocks will be replaced at no charge to the customer?
We're looking the the number of shocks that wear out before 50000 miles
Shocks wear out according to a Normal(63000,10000) distribution.
First we normalize 50000 according to the distribution.
z = (50000 - 63000) / 10000 = -1.3
Now we look that up in the table or however you usually deal with the CDF of the Normal distribution.
Pr[-1.3] = 0.0968 = 9.68%
+6251 
