The weights of ice cream cartons produced by a manufacturer are normally distributed with a mean weight of 10 ounces and a standard deviation of 0.5 ounce.(a) What is the probability that a randomly selected carton has a weight greater than 10.21 ounces?(b) You randomly select 25 cartons. What is the probability that their mean weight is greater than 10.21 ounces?(c)Compare the probabilities from parts (a) and (b). (50 points)
(a) Well, to find the probability that a randomly selected carton has a weight greater than 10.21 ounces, we can use the z-score formula. First, we calculate the z-score:
z = (x - μ) / σ
where x is the value we're interested in, μ is the mean, and σ is the standard deviation. Substituting the values, we have:
z = (10.21 - 10) / 0.5
z = 0.42
Now, we consult a standard normal distribution table or use a calculator to find the area under the curve to the right of 0.42. Guess what? It's approximately 0.3365. So, there's a 33.65% chance of randomly selecting a carton with a weight greater than 10.21 ounces.
(b) Oh, we can use the Central Limit Theorem here! Since the sample size is 25, we know that the distribution of the sample mean will approach a normal distribution with the same mean as the population and a standard deviation of σ/√n. In this case, σ is 0.5 (as given) and n is 25, so the standard deviation of the sample mean is 0.5/√25 = 0.1.
Now, we can use the z-score formula again, but this time, we'll calculate the z-score for the sample mean:
z = (x - μ) / (σ/√n)
z = (10.21 - 10) / (0.1)
z = 2.1
We go back to our trusty standard normal distribution table (or calculator) and find the area to the right of 2.1, which is approximately 0.0179. So, there's a 1.79% chance that the mean weight of the 25 randomly selected cartons is greater than 10.21 ounces.
(c) To compare the probabilities, let's take a step back and reflect. In part (a), we were only looking at one carton, so the probability was relatively higher at 33.65%. However, in part (b), we looked at a sample of 25 cartons, so we zoomed in on the mean weight and found a lower probability of 1.79%.
So, in a nutshell, the probability of randomly selecting one carton with a weight greater than 10.21 ounces is higher than the probability of randomly selecting a sample of 25 cartons with a mean weight greater than 10.21 ounces. Now that's quite a difference!
To answer part (a) of the question, we need to use the concept of the standard normal distribution. Since the weights of ice cream cartons are normally distributed, we can standardize the value of 10.21 ounces using the given mean and standard deviation.
(a) To calculate the probability that a randomly selected carton has a weight greater than 10.21 ounces, we need to find the area under the standard normal curve to the right of 10.21.
1. Standardize the value:
z = (x - μ) / σ
where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.
z = (10.21 - 10) / 0.5
z = 0.42
2. Find the probability using a standard normal distribution table or calculator:
P(Z > 0.42)
Using a standard normal distribution table or calculator, we find that the probability P(Z > 0.42) is approximately 0.3372.
Therefore, the probability that a randomly selected carton has a weight greater than 10.21 ounces is approximately 0.3372.
(b) To calculate the probability that the mean weight of 25 randomly selected cartons is greater than 10.21 ounces, we need to use the concept of the sampling distribution of the sample mean.
1. Since the sample size is larger than 30 and we know the population standard deviation, we can use the normal approximation to the sampling distribution of the sample mean.
2. Find the standard deviation of the sampling distribution (also known as the standard error):
σᵢ = σ / sqrt(n)
where σ is the population standard deviation and n is the sample size.
σᵢ = 0.5 / sqrt(25)
σᵢ = 0.5 / 5
σᵢ = 0.1
3. Standardize the value:
z = (x̄ - μ) / σᵢ
where x̄ is the sample mean, μ is the population mean, and σᵢ is the standard error.
z = (10.21 - 10) / 0.1
z = 2.1
4. Find the probability using a standard normal distribution table or calculator:
P(Z > 2.1)
Using a standard normal distribution table or calculator, we find that the probability P(Z > 2.1) is approximately 0.0179.
Therefore, the probability that the mean weight of 25 randomly selected cartons is greater than 10.21 ounces is approximately 0.0179.
(c) Comparing the probabilities from parts (a) and (b):
The probability from part (a) (0.3372) represents the probability of selecting a single randomly chosen carton with a weight greater than 10.21 ounces.
The probability from part (b) (0.0179) represents the probability of obtaining a sample mean greater than 10.21 ounces from 25 randomly selected cartons.
The probability from part (b) is smaller than the probability from part (a). This is because the larger sample size reduces the variability and provides a more precise estimation of the population mean. The smaller probability from part (b) indicates that it is less likely to obtain a sample mean greater than 10.21 ounces compared to selecting a single carton with a weight greater than 10.21 ounces.