A carton of soy milk states on the label that it contains 956 mL.

The volume of soy milk in 9 randomly chosen cartons is measured and the mean found to be 958.72 mL.
Assume that the cartons are filled so that the actual amounts are normally distributed with a mean of 956 mL and a standard deviation of 10.2 mL. Find the probability that a sample of 9 cans will have a mean amount of less than 958.72 mL.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since only one SD is provided, you can use just that to determine SEdiff.

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

To find the probability that a sample of 9 cans will have a mean amount of less than 958.72 mL, we need to use the concept of the sampling distribution of the mean.

The sampling distribution of the mean is a probability distribution that represents the possible values of the sample mean when repeated random samples are drawn from a population. It is commonly approximated by a normal distribution when the sample size is large enough (usually considered as n ≥ 30) or when the population distribution is approximately normal.

In this case, the population standard deviation is given as 10.2 mL. Since the sample size is 9 (which is less than 30), we need to use a different distribution called the t-distribution. However, for a large enough sample size (approximated by n ≥ 30), we can use the normal distribution.

To calculate the probability, we need to find the z-score, which represents the number of standard deviations the sample mean (958.72 mL) is away from the population mean (956 mL). The formula for calculating the z-score is:

z = (sample mean - population mean) / (population standard deviation / √sample size)

Plugging in the values,
z = (958.72 - 956) / (10.2 / √9)
z = 2.72 / (10.2 / 3)
z = 2.72 / 3.4
z = 0.8

Now, we need to find the probability of getting a z-score less than 0.8. This can be done using a z-table or statistical software.

Using a z-table, we can find that the probability corresponding to a z-score of 0.8 is approximately 0.7881.

Therefore, the probability that a sample of 9 cans will have a mean amount of less than 958.72 mL is approximately 0.7881, or 78.81%.