First, the mean (for a normal distribution it is the 'expected' value). 0.8 eggs per dozen, 3 dozen. Hmm...0.8 x 3 = ? (get out the calculator?) expected result of buying a random 3 dozen is 2.4 broken eggs. But not really. That is like saying the following: "I have an important job interview, court date, or test, etc. at 4 o'clock tomorrow. I know on average it takes 20 minutes to get there, so I'll leave at 3:40." Wow! That is pretty stupid. You need to include some idea of variance (fudge factor, slack, unusual circumstances, tolerance, wiggle-room, etc.). Same with the eggs, especially if you are making your sister's wedding cake for tomorrow's wedding. So, (you should ask yourself), what is the likely range of values of breakage? You want to be 'sure' to have enough. Of course, "sure" is not fixed, not absolute. It is relative to your risk tolerance or just the foibles and flaws of the real world. "Sure" for the statistician means" "x times out of y, I want to succeed in attaining what I expect." Usually you'll see 50%, 80%, 90%, 95% probability. There is no 100% certainty with a normal distribution, the tails extend to ± infinity. Most hard sciences are ok with 95% certainty. One of the important reasons for this is that it is very close to 4 x s.d.. (They call 4x s.d. "two" s.d.'s because 4 s.d. is + or - 2 x s.d. from the mean. Assuming that 1 chance of being disappointed (or pleasantly surprised) out of 20 is your idea of "sure", then if you bought a dozen eggs you could be "sure" that you will have fewer than 0.8 + 2 x 0.5 = 1.8 eggs broken. Or the range of 0 to 1.8 at 95% (and 0 to 2.3 eggs at 99.7% = 3 s.d.) Note that I could have written the range as -0.2 to 1.8. But its hard to have a negative egg, and I hate looking stupid.
