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 can use the concept of the standard normal distribution, which allows us to standardize any normally distributed variable.
To find the percentage of boxes that weigh more than 16 ounces (question a), we can calculate the z-score for 16 ounces and then use a standard normal distribution table or a calculator to find the corresponding percentage.
The formula for calculating the z-score is:
z = (x - μ) / σ
where:
- x is the value we want to find the probability for (in this case, 16 ounces)
- μ is the mean of the distribution (16.5 ounces in this case)
- σ is the standard deviation of the distribution (0.4 ounces in this case)
Substituting the values, we have:
z = (16 - 16.5) / 0.4 = -1.25
Using a standard normal distribution table or calculator, we can find the percentage corresponding to a z-score of -1.25. This percentage represents the proportion of boxes that weigh more than 16 ounces.
For question b, we need to find the percentage of boxes that weigh between 15.5 and 16.5 ounces. To do this, we need to find the z-scores for both values and then calculate the area between these two z-scores.
For 15.5 ounces:
z1 = (15.5 - 16.5) / 0.4 = -2.5
For 16.5 ounces:
z2 = (16.5 - 16.5) / 0.4 = 0
Now, we can use the standard normal distribution table or calculator to find the area between these two z-scores. This represents the percentage of boxes that weigh between 15.5 and 16.5 ounces.
For question c, we need to find the number of boxes that weigh less than 15.5 ounces. Since we have the mean and standard deviation, we can calculate the z-score for 15.5 ounces using the formula mentioned earlier (z = (x - μ) / σ).
Once we find the z-score, we can use it to calculate the proportion of boxes that weigh less than 15.5 ounces. We can then multiply this proportion by the total number of boxes (500 in this case) to find the actual number of boxes that weigh less than 15.5 ounces.