The average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.38 ounces. If a random sample of eighty 16-ounce beverage cans are selected, what is the probability that mean of this sample is less than 16.1 ounces of beverage? Answer: (round to 4 decimal places)
Given, = 16.18 ounces
= 0.38 ounces
n = 80
According to Central Limit Theorem, P( < A) = P(Z < (A - )/)
= 16.18 ounces
=
=
= 0.0425
P(mean of the sample is less than 16.1 ounces) = P( < 16.1)
= P(Z < (16.1 - 16.18)/0.0425)
= P(Z < -1.88)
= 0.0301
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