I am presented with the question:
The fill volume of an automated filling machine used for filling cans of carbonated beverage is normally distributed with a mean of 12.4 fluid ounces and a standard deviation of 0.1 fluid ounce. Round the answers to 3 significant digits.
b) If all cans less than 12.1 or greater than 12.6 ounces are scrapped, what proportion of cans is scrapped?
I was able to figure out how to get the rest of the question, but for some reason I can't seem to get this answer. Can anyone explain how I might solve this problem? BTW, I am able to use a TI 84, so an explanation using normalcdf() would be fine.
The sum of the proportion of cans that are scrapped and cans that are not scrapped is 1. These are complementary probabilities, because together they cover every possible scenario. A can is either within the range, or outside of it. There is no chance that a can is neither in the range nor outside of it, so the probabilities of these two situations add to 1.
You can use normalcdf to figure out the proportion of cans that are not scrapped.
Now you can subtract this from 1 to figure out the proportion of cans that are scrapped.