Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.2 ounces.

(a) If 4 potatoes are randomly selected, find the probability that the mean weight is less than 9.8 ounces.

(b) If 8 potatoes are randomly selected, find the probability that the mean weight is more than 9.1 ounces.

Z = (score-mean)/SEm

SEm = SD/√n

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

To solve these probability problems, we will use the concept of the sampling distribution of the sample mean, assuming that the weight of each potato is independent and identically distributed.

(a) To find the probability that the mean weight of 4 randomly selected potatoes is less than 9.8 ounces, we need to calculate the z-score and use the standard normal distribution table.

First, we calculate the standard error of the sample mean, which is the standard deviation of the population divided by the square root of the sample size (n):

Standard Error (SE) = Standard Deviation (SD) / √n

In this case, the standard deviation (SD) is given as 1.2 ounces, and the sample size (n) is 4. So, the standard error is:

SE = 1.2 / √4 = 1.2 / 2 = 0.6 ounces

Next, we calculate the z-score using the formula:

z = (x - μ) / SE

Here, x represents the desired mean weight (9.8 ounces), μ represents the population mean (8.0 ounces), and SE represents the standard error (0.6 ounces).

z = (9.8 - 8.0) / 0.6 = 1.8 / 0.6 = 3

Now, referring to the standard normal distribution table, we can find the probability corresponding to a z-score of 3, which represents the probability of obtaining a mean weight less than 9.8 ounces.

From the table, we find that the probability for a z-score of 3 is approximately 0.9987.

Therefore, the probability that the mean weight of 4 randomly selected potatoes is less than 9.8 ounces is 0.9987 (or approximately 99.87%).

(b) To find the probability that the mean weight of 8 randomly selected potatoes is more than 9.1 ounces, we follow the same steps as in part (a).

First, we calculate the standard error, which is the standard deviation of the population divided by the square root of the sample size:

SE = 1.2 / √8 = 1.2 / 2.83 ≈ 0.424 ounces

Next, we calculate the z-score using the formula:

z = (x - μ) / SE

Here, x represents the desired mean weight (9.1 ounces), μ represents the population mean (8.0 ounces), and SE represents the standard error (0.424 ounces).

z = (9.1 - 8.0) / 0.424 ≈ 2.62

Referring to the standard normal distribution table, we can find the probability corresponding to a z-score of 2.62, which represents the probability of obtaining a mean weight more than 9.1 ounces.

From the table, we find that the probability for a z-score of 2.62 is approximately 0.9952.

Therefore, the probability that the mean weight of 8 randomly selected potatoes is more than 9.1 ounces is 0.9952 (or approximately 99.52%).