Question

oats in a box' is a manufacturer of cereal boxes. the amount of oats filled in each box has been found to be normally distributed with a mean of 14.1 oz and a standard deviation of 0.3 oz. boxes that contain less than 95% of the listed net content (i.e. 13.395 oz) can result in the manufacturer being subject to a penalty by the state office of consumer affairs, whereas boxes that have a net content above 14.4 oz may cause excess spillage upon opening. what percentage of boxes will be out of the specification that a box should have at least 13.395 oz but no more than 14.4 oz?

Answer 1

To determine the percentage of cereal boxes within the specified content limits, we calculate the z-scores for 13.395 oz and 14.4 oz using the provided mean and standard deviation and then refer to a standard normal distribution table to find the corresponding probabilities.

Step-by-step explanation:

The question asks what percentage of cereal boxes will be outside the specified content limits if they are normally distributed with a mean of 14.1 oz and a standard deviation of 0.3 oz. The boxes should contain at least 95% of the listed net content (13.395 oz) and no more than 14.4 oz to avoid penalties or spillages, respectively.

To find the percentages, we need to calculate the z-scores for both 13.395 oz and 14.4 oz.

The z-score formula is (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.

For 13.395 oz: z = (13.395 - 14.1) / 0.3
For 14.4 oz: z = (14.4 - 14.1) / 0.3

Using a standard normal distribution table, we can find the probabilities corresponding to these z-scores. These probabilities represent the percentage of boxes below 13.395 oz and above 14.4 oz.

Subtracting these from 100% will give us the percentage of boxes within the specified limits.