Assume the weights of Farmer Carl's potatoes are normally distributed with a mean of 7.0 ounces and a standard deviation of 1.3 ounces. He bags his potatoes in groups of 6. You buy a bag and the total weight is 36 ounces. Here we determine how lucky or unlucky you are.

(a) What is the mean potato weight in your bag of 6?

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is less than the mean found in your bag.

In "YOUR BAG" there is no mean. It weighs 36 ounces. If you are talking about each individual potato, 36/6 = 6

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

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If 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/probability related to the Z score.

A manufacturer claims that the mean amount of juice in its 16 ounce bottles is 16.2 ounces. A consumer advocacy group wants to determine whether the mean amount is actually less than this. The mean volume of juice for a random sample of 70 bottles was 16.04 ounces with a standard deviation of 0.9 ounces. Find a 95% confidence interval to estimate the true mean amount of juice in these bottles.

A) What is the sample mean?

B) Standard deviation?

C) How many standard deviations do we need to capture 95% estimate?

D) Margin of error?

To find the mean potato weight in your bag of 6, we divide the total weight of the bag (36 ounces) by the number of potatoes (6).

(a) Mean potato weight in your bag = Total weight / Number of potatoes
= 36 ounces / 6 potatoes
= 6 ounces

So, the mean potato weight in your bag of 6 is 6 ounces.

To find the probability that the mean weight of 6 randomly selected potatoes is less than the mean weight in your bag (6 ounces), we need to calculate the z-score and then find the corresponding probability using the standard normal distribution.

First, let's calculate the standard error of the mean (SEM), which is equal to the standard deviation divided by the square root of the sample size.

Standard Error of the Mean (SEM) = Standard Deviation / √(Sample Size)
= 1.3 ounces / √6
≈ 0.5307 ounces (rounded to four decimal places)

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

z = (X - Mean) / SEM

Here, X is the mean weight in your bag (6 ounces), Mean is the mean weight of the population (7 ounces), and SEM is the standard error of the mean.

z = (6 - 7) / 0.5307
≈ -1.88 (rounded to two decimal places)

Using a standard normal distribution table or a calculator, we can find that the probability that a z-score is less than -1.88 is approximately 0.0307 (rounded to four decimal places).

Therefore, the probability that the mean weight of 6 randomly selected potatoes is less than the mean weight in your bag is approximately 0.0307 or 3.07%.

To answer these questions, we need to calculate the mean and standard deviation of the distribution of the mean weight of the potatoes in your bag.

(a) The mean potato weight in your bag of 6 can be calculated by dividing the total weight of the bag by the number of potatoes in the bag. In this case, the total weight of the bag is 36 ounces, and there are 6 potatoes in the bag. Therefore, the mean potato weight in your bag is 36 ounces divided by 6 potatoes, which is equal to 6 ounces.

(b) To find the probability that the mean weight is less than the mean found in your bag, we need to determine the z-score for your bag's mean weight and then use the z-table to find the corresponding probability.

The z-score is calculated using the formula:
z = (x - μ) / (σ / √n)
where x is the mean weight in your bag, μ is the population mean weight (7.0 ounces), σ is the population standard deviation (1.3 ounces), and n is the number of potatoes in your bag (6).

Using the values given:
x = 6 ounces
μ = 7.0 ounces
σ = 1.3 ounces
n = 6

Calculating the z-score, we have:
z = (6 - 7.0) / (1.3 / √6) ≈ -1.88

Now, we can look up the z-score in the z-table to find the corresponding probability. The table will give us the area under the standard normal distribution curve to the left of the z-score. In this case, we want the probability of the mean weight being less than your bag's mean weight, so we look for the probability to the left of the z-score.

The z-table gives us the probability of about 0.0307. This means that there is a 0.0307 probability that the mean weight of another bag of 6 potatoes randomly selected from the population is less than the mean weight of your bag.