Cartons of milk from a particular supermarket are advertised as containing 1 liter, but in fact the volume of the contents is normally distributed with a mean of 1012 ml and astandard deviation of 5 ml .ln a batch of1000 cartons estimate the number of cartons that contain less than the advertised volume of milk
I don't know may you kindly help
You can play around with Z table stuff at
davidmlane.com/hyperstat/z_table.html
To estimate the number of cartons that contain less than the advertised volume of milk, we can use the concept of standard deviation and the normal distribution.
Step 1: Calculate the z-score
The z-score measures how many standard deviations a value is away from the mean. We can calculate the z-score using the formula:
z = (x - μ) / σ
where z is the z-score, x is the observed value (1000 ml in this case), μ is the mean (1012 ml), and σ is the standard deviation (5 ml).
For the advertised volume of 1 liter (1000 ml), the z-score is:
z = (1000 - 1012) / 5 = -2.4
Step 2: Find the cumulative probability
We need to find the cumulative probability associated with the calculated z-score. This probability represents the proportion of values that fall below the advertised volume.
Using a standard normal distribution table or calculator, we find that the cumulative probability for a z-score of -2.4 is approximately 0.0082.
Step 3: Calculate the estimated number of cartons
To estimate the number of cartons that contain less than the advertised volume, we multiply the cumulative probability by the total number of cartons in the batch.
Estimated number of cartons = cumulative probability * total number of cartons
= 0.0082 * 1000
≈ 8.2
Therefore, we estimate that approximately 8 cartons out of the 1000 cartons in the batch contain less than the advertised volume of milk (1 liter).
To estimate the number of cartons that contain less than the advertised volume of milk, we can use the concept of the standard normal distribution.
First, let's calculate the z-score for the advertised volume of milk, which is 1 liter or 1000 ml. The z-score formula is given by:
z = (x - μ) / σ
where x is the observed value, μ is the mean, and σ is the standard deviation.
In this case, the observed value x is 1000 ml, the mean μ is 1012 ml, and the standard deviation σ is 5 ml.
z = (1000 - 1012) / 5
z = -12 / 5
z = -2.4
Next, we need to find the cumulative probability associated with this z-score. We can use a standard normal distribution table or a statistical calculator to find this value.
After finding the cumulative probability, we can estimate the number of cartons that contain less than the advertised volume by multiplying the cumulative probability by the total number of cartons in the batch (1000).
Let's assume that the cumulative probability for a z-score of -2.4 is 0.0082.
Number of cartons containing less than the advertised volume = cumulative probability * total number of cartons
Number of cartons = 0.0082 * 1000
Number of cartons = 8.2
Therefore, we can estimate that approximately 8 cartons out of the 1000 cartons in the batch will contain less than the advertised volume of milk.