The amount of cereal in a box is normally distributed with a mean of 24oz and a standard deviation of 0.72oz. If 17500 boxes of cereal are packaged in one day, approximately how many boxes would contain less than 23.1oz
First, we need to standardize the value of 23.1oz using the formula:
z = (x - μ) / σ
where
x = 23.1 (the value we want to convert)
μ = 24 (mean)
σ = 0.72 (standard deviation)
z = (23.1 - 24) / 0.72
z = -0.97
Next, we can use a standard normal distribution table or calculator to find the proportion of values that are less than -0.97. This area represents the proportion of boxes that contain less than 23.1oz.
Using a calculator, the proportion is calculated as 0.1664.
Finally, we can multiply this proportion by the total number of boxes to get an estimate of the number of boxes that would contain less than 23.1oz:
17500 x 0.1664 = 2912
Therefore, approximately 2912 boxes would contain less than 23.1oz.
To calculate the approximate number of boxes that would contain less than 23.1oz, we need to use the concept of standard deviation and the z-score.
The z-score formula is: z = (x - μ) / σ
Where:
x = 23.1oz (given value)
μ = mean = 24oz
σ = standard deviation = 0.72oz
Let's calculate the z-score for 23.1oz:
z = (23.1 - 24) / 0.72
z = -0.9 / 0.72
z ≈ -1.25
Now, we need to find the corresponding cumulative probability associated with this z-score from the standard normal distribution table. The table gives us the probability of having a value less than the z-score.
From the standard normal distribution table, the cumulative probability for a z-score of -1.25 is approximately 0.1056.
To calculate the number of boxes that would contain less than 23.1oz, we multiply this probability by the total number of boxes of cereal packaged in one day (17,500):
Number of boxes = Probability * Total number of boxes
Number of boxes ≈ 0.1056 * 17,500
Number of boxes ≈ 1,848
Therefore, approximately 1,848 boxes would contain less than 23.1oz of cereal.