Really simple question too hard for a simple mind

Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept coins weighing between 5.48 g and 5.82 g. What percentage of legal quarters will be rejected?

**The answer is:**

The standard score ("Z") for 5.48:

Z = (5.48 -5.67)/0.070 = -2.71

The area between the -2.71 and the mean (from a standard normal table): 0.4966

The standard score for 5.82:

Z = (5.82-5.67)/0.070 = 2.14

The area between 2.14 and the mean: 0.4838

The area between 5.48 (Z=-2.71) and 5.82 (Z=2.14) = 0.4966+0.4838 = 0.9804.

The area outside that area is 1 = 0.9804 = 0.0196.

Therefore, about 2% of legal quarters will be rejected.

My question is, the table for the *standard normal curve areas* i have is not giving me that -2.71= .4966 & 2.14=.4838.

For me my table says (-2.71<**z**<2.14) = (**z**<2.14)=.9836 and [1-(z<2.71)]=1-.9966=.0034