I'm confused. With such a low standard deviation, practically all cans should contain between 15.8 and 16.2 ounces.
Q. If all the 1 pound cans of beans filled by a food processor have a mean weight of 16.00 ounces with a standard deviation of 0.02 ounces, at least what percentage of the cans must contain between 15.80 and 16.20 ounces of beans, assuming the weights are normally distributed?
A.
(16.20-15.80) x SQRT 16 / 0.02
= (0.4)(4)/0.02
= 1.6/0.02
= 80%
Is this correct or have i used the wrong equation?
I don't know if I'm heading in the right direction and if there isn't an easier, perhaps approximate, way to do this, but as far as I see it, you would have to look at the (cumulative) distribution function of the normal distribution.
Then compute , which is approximately 1, i.e. 100%.
You're meant to calculate Pr(15.80 < X < 16.20). Convert each value of X to a z-value and use your tables. As has already been implied, the interval is 10 standard deviations either side of the mean so for all practical purposes the answer is 100%. This should be obvious (and the z-values would leave no room for doubt had you calculated them).