Question

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)

Answer #1

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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