Potatoes - Samples: Assume the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 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 42 ounces. Here we determine how lucky or unlucky you are.
(a) What is the mean potato weight in your bag of 6? Enter your answer to 1 decimal place.
ounces
(b) If 6 potatoes are randomly selected, find the probability that the mean weight is less than the mean found in your bag. Round your answer to 4 decimal places.
(c) Which statement best describes your situation?
This is not particularly unusual because the mean weight in your bag is only 1 ounce below the mean of all Carl's potatoes.
This is an unusually small amount of potatoes. You are unlucky because the probability of getting a bag weighing less than yours is only about 3.0%.
You got lucky with such a generous amount of potato your bag.
You got extremely unlucky because the probability of getting a bag weighing less than yours is about 0.01%.
erhqwektgyfwdvw ery uqr3 oq3yeqjgq6q vqer ivfe rgw ergygugqrg
(a) The mean potato weight in your bag of 6 can be calculated by dividing the total weight (42 ounces) by the number of potatoes (6):
Mean potato weight = Total weight / Number of potatoes
= 42 ounces / 6 potatoes
= 7 ounces (rounded to 1 decimal place)
Therefore, the mean potato weight in your bag of 6 is 7 ounces.
(b) To find the probability that the mean weight of 6 randomly selected potatoes is less than the mean weight in your bag (7 ounces), we can use the concept of the sampling distribution of the sample mean. Given that the population follows a normal distribution and the sample size is sufficiently large (n=6 is relatively small but we'll assume it's large enough), the sampling distribution of the sample mean will also be normally distributed with the same mean as the population mean (8 ounces) but with a standard deviation equal to the population standard deviation divided by the square root of the sample size (1.3 ounces / √6).
Now, we need to find the probability that the mean weight of 6 randomly selected potatoes from this distribution is less than 7 ounces.
Using a standard normal distribution table or a calculator, we can find the probability associated with a z-score (standard score) of (7-8) / (1.3 / √6):
z-score = -0.963
Looking up -0.963 in the standard normal distribution table or using a calculator, we find the probability to be approximately 0.1660 (rounded to 4 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 0.1660.
(c) Based on the given probability, the statement that best describes your situation is:
This is not particularly unusual because the mean weight in your bag is only 1 ounce below the mean of all Carl's potatoes.
To find the answers to these questions, we can use the concept of sampling distributions. The mean weight of a bag of 6 potatoes can be determined by dividing the total weight of the bag by the number of potatoes.
(a) To find the mean potato weight in your bag of 6, we divide the total weight (42 ounces) by the number of potatoes (6):
Mean potato weight = Total weight / Number of potatoes
Mean potato weight = 42 ounces / 6 potatoes
Mean potato weight = 7 ounces
So, the mean potato weight in your bag of 6 is 7 ounces.
(b) To find the probability that the mean weight of 6 randomly selected potatoes is less than the mean found in your bag (7 ounces), we can use the concept of the sampling distribution of the mean. We need to calculate the z-score and then use a standard normal distribution table or calculator to find the probability.
The z-score formula is given by:
z = (x - μ) / (σ / √n)
where:
x is the mean weight of your bag (7 ounces),
μ is the mean weight of all Carl's potatoes (8 ounces),
σ is the standard deviation of all Carl's potatoes (1.3 ounces), and
n is the number of potatoes in your bag (6).
Substituting the given values into the formula, we get:
z = (7 - 8) / (1.3 / √6)
z = -1 / (1.3 / √6)
z ≈ -1.96 (rounded to 2 decimal places)
Now, we need to find the probability that the z-score is less than -1.96. This probability can be looked up in a standard normal distribution table or using a calculator. The probability is approximately 0.025 (rounded to 4 decimal places).
So, the probability that the mean weight of 6 randomly selected potatoes is less than the mean weight in your bag (7 ounces) is 0.025.
(c) Now, let's determine which statement best describes your situation based on the probability calculated in part (b).
The probability of getting a bag weighing less than yours is 0.025, which is about 2.5%. Therefore, the correct statement is:
This is not particularly unusual because the mean weight in your bag is only 1 ounce below the mean of all Carl's potatoes.
This means that getting a bag with a weight of 42 ounces is not significantly different from the average weight of all Carl's potatoes. You are not particularly lucky or unlucky in this situation.
a is 4.5
b is 8.542
figure out c