# Central Limit Theorem

## Homework Statement

A machine fills cereal boxes at a factory. Due to an accumulation of small errors (different flakes sizes, etc.) it is thought that the amount of cereal in a box is normally distributed with mean 22 oz. for a supposedly 20 oz. box. Suppose the standard deviation of the amount filled is 1.3 oz. What is the probability that a federal regulatory selects four of these boxes at random and finds that the average content of these boxes is less than 18 oz?

## The Attempt at a Solution

[PLAIN]http://img508.imageshack.us/img508/3904/49450328.jpg [Broken]

From the graph, I calculated the Z score using the Z score equation for distribution of sample means.

The Z-score I got was -6.15

I then found the area from Table E which came out to be 0.4999 (Table E said any value greater than 3.09 use 0.4999. Since it's to the left of the mean, I subtracted 0.5 from the Z value (0.5-0.4999) and my answer came out to be 0.001

I'm a bit confused by the question when it says that it's normally distributed with a mean of 22 oz for a "supposedly 20 oz box." So is it implying that the mean is 22 or 20?

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## Homework Statement

A machine fills cereal boxes at a factory. Due to an accumulation of small errors (different flakes sizes, etc.) it is thought that the amount of cereal in a box is normally distributed with mean 22 oz. for a supposedly 20 oz. box. Suppose the standard deviation of the amount filled is 1.3 oz. What is the probability that a federal regulatory selects four of these boxes at random and finds that the average content of these boxes is less than 18 oz?

## The Attempt at a Solution

[PLAIN]http://img508.imageshack.us/img508/3904/49450328.jpg [Broken]

From the graph, I calculated the Z score using the Z score equation for distribution of sample means.

The Z-score I got was -6.15

I then found the area from Table E which came out to be 0.4999 (Table E said any value greater than 3.09 use 0.4999. Since it's to the left of the mean, I subtracted 0.5 from the Z value (0.5-0.4999) and my answer came out to be 0.001