The mean lifetime of a certain tire is 30,000 miles and the standard deviation is 2500 miles. if we assume nothing about the shape of the distribution, approximately what percentage of all such tires will last between 25,000 and 37,500 miles?

A standard deviation is 2500 miles, so the range of 25,000 to 37,500 is 2 standard deviations below, and 3 standard deviation above the mean of 30,000. 
We aren't assuming anything about the distribution, but we sort of have to assume a normal distribution in order to answer. When you don't know anything about a distribution, this is generally what you do.
The general rule for standard deviations is:
+/- 1 SD: +/- 34% of population, for a total of 68%
+/- 2 SD: Another 14% in either direction, for a total of 96% of population.
+/- 3 SD, Another 2% in either direction, which pretty much captures the entire distribution (4 standard deviations from the mean is incredibly rare/unlikely) 
We have a range of +/- 1 SD, so we know that 68% percent o the tire 'population' is captured in the range. We also have one more standard deviation in both directions, so we add 28% to this number to get 96% of the population. We have a third standard deviation above the mean, so we add another 2% to arrive at 98% of the population. 

Answer is 98% of all such tires. 

4+2=-2. 1+2=3. y=-2/3x+b. 2=4/3+b. 2/3=b. the answer is y= -2/3x+2/3