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 answer these questions, we will use the properties of normal distribution and calculate the z-scores.
A. What percentage of the boxes will weigh more than 16 ounces?
To find the percentage of boxes that weigh more than 16 ounces, we need to calculate the area under the normal distribution curve to the right of 16 ounces. We can do this by calculating the z-score for 16 ounces and then finding the area to the right of that z-score.
The z-score formula is:
z = (x - μ) / σ
where:
- x is the value we want to find the area for (16 ounces in this case),
- μ is the mean of the distribution (16.5 ounces), and
- σ is the standard deviation of the distribution (0.4 ounces).
Plugging in the values, we get:
z = (16 - 16.5) / 0.4
= -0.5 / 0.4
= -1.25
Using a standard normal distribution table (or a calculator), we can find the area to the right of this z-score. The z-score of -1.25 corresponds to an area of approximately 0.1056.
To find the percentage, we multiply the area by 100:
Percentage = 0.1056 * 100
= 10.56%
Therefore, approximately 10.56% of the boxes will weigh more than 16 ounces.
B. What percentage of the boxes will weigh between 15.5 and 16.5 ounces?
To find the percentage of boxes that weigh between 15.5 and 16.5 ounces, we need to calculate the area under the normal distribution curve between these two values. We can do this by calculating the z-scores for both values and then finding the difference between the two areas.
First, let's find the z-score for 15.5 ounces:
z1 = (15.5 - 16.5) / 0.4
= -1 / 0.4
= -2.5
Next, let's find the z-score for 16.5 ounces:
z2 = (16.5 - 16.5) / 0.4
= 0 / 0.4
= 0
Using the standard normal distribution table (or a calculator), we can find the areas corresponding to these z-scores. For z1 = -2.5, the area is approximately 0.0062, and for z2 = 0, the area is 0.5.
To find the percentage, we subtract the smaller area from the larger area and then multiply by 100:
Percentage = (0.5 - 0.0062) * 100
= 0.4938 * 100
= 49.38%
Therefore, approximately 49.38% 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?
To find the number of boxes that will weigh less than 15.5 ounces, we need to calculate the percentage of boxes that fall under this weight and then multiply it by the total number of boxes (500).
From part B, we found that approximately 0.62% of the boxes weigh less than 15.5 ounces. So, we calculate:
Number of boxes = 0.0062 * 500
= 3.1
Therefore, approximately 3 boxes out of the 500 will weigh less than 15.5 ounces.