Originally Posted by
Aquameatwad 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