The weights of boxes of Brand Z cereal were found to be normally distrubted with a mean of 16.5 ounces and a standard deviation of 0.4 ounces.

a. what percentage of the boxes will weigh more than 16 ounces?

b. what percentage of the boxes will weigh between 15.5 and 16.5 ounces?

c. if a store has 500 boxes of cereal, how many of them will weigh less than 15.5 ounces?

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.

To solve these questions, we need to use the properties of the normal distribution and z-scores. A z-score measures the number of standard deviations a data point is from the mean.

a) To find the percentage of boxes weighing more than 16 ounces, we first need to calculate the z-score corresponding to 16 ounces using the formula:

z = (x - μ) / σ

where x is the given value, μ is the mean, and σ is the standard deviation. In this case:

z = (16 - 16.5) / 0.4 = -1.25

Next, we find the percentage of values greater than this z-score using a standard normal distribution table or calculator. In this case, we are interested in the percentage of values in the right tail (values greater than the z-score).

Using a standard normal distribution table, a z-score of -1.25 corresponds to a cumulative area of 0.1056 (or 10.56%). So, approximately 10.56% of the boxes will weigh more than 16 ounces.

b) To find the percentage of boxes weighing between 15.5 and 16.5 ounces, we need to calculate the z-scores for both values.

For 15.5 ounces:

z = (15.5 - 16.5) / 0.4 = -2.5

For 16.5 ounces:

z = (16.5 - 16.5) / 0.4 = 0

Next, we find the area between these two z-scores using the standard normal distribution table or calculator. This area represents the percentage of values between the given range.

Using the standard normal distribution table, we can find the area corresponding to a z-score of -2.5, which is approximately 0.0062. Similarly, the area corresponding to a z-score of 0 is 0.5 (50%). To find the area between these two z-scores, subtract the smaller area from the larger area:

0.5 - 0.0062 = 0.4938 or 49.38%

Therefore, approximately 49.38% of the boxes will weigh between 15.5 and 16.5 ounces.

c) To find the number of boxes that will weigh less than 15.5 ounces out of 500 boxes, we need to find the z-score for 15.5 ounces using the formula mentioned earlier:

z = (15.5 - 16.5) / 0.4 = -2.5

Next, we need to find the area to the left of this z-score value using the standard normal distribution table or calculator. This area represents the percentage of values less than the given value.

Using the standard normal distribution table, the area corresponding to a z-score of -2.5 is approximately 0.0062. This means that approximately 0.62% of the boxes will weigh less than 15.5 ounces.

To find the number of boxes out of 500, we multiply the percentage by the total number of boxes:

0.0062 * 500 ≈ 3.1

Therefore, approximately 3 of the 500 boxes will weigh less than 15.5 ounces.