The answer is 502.4 ml. How to get the answer?

Question

A company making cans for lemonade wishes to print ‘Average content x ml’ on their cans, and to be 99% confident that the true mean volume is greater than x ml. The volume of lemonade in a can is known to have a standard deviation of 3.2 ml and a random sample of 50 cans contained an average volume of 503.6 ml. What volume of x should be stated?

Answer 1

the LOWER confidence limit is the only one of interest...

99% confidence corresponds to a z-score whose probability

is 0.995 because it is TWO SIDED , the lower of which we

are interested.

So Z_limit_99 = Normal_inverse( prob=0.995, mean=0, stddev=1) = 2.65

the lower limit has formula:

mean - Z_limit_99 * stndDev / sqrt(N)

503.6 - 2.65 * 3.2 / sqrt(50) = 502.4

Therefore we are 99% certain that the product contains AT LEAST that much.