A soft drink bottler purchases glass bottles from a vendor. The bottles are required to have an internal pressure of at least 150 pounds per square inch (psi). A prospective bottle vendor claims that its production process yields bottles with a mean internal pressure of 157 psi and a standard deviation of 3 psi. The bottler strikes an agreement with the vendor that permits the bottler to sample from the production process to verify the claim. the bottler randomly selects 50 bottles from the last 10000 produced, measures the internal pressure of each, and finds the mean pressure for the sample to be 0.6 psi below the process mean cited by the vendor.
(a) Assuming that the vendor is correct in his claim, what is the probability of obtaining a sample mean this far or farther below the process mean?
(b) If the standard deviation were 3 psi as claimed, but the mean was 151 psi, what is the probability of obtaining a sample mean of 156.4 psi or below?
(c) If the process mean were 157 psi as claimed, but the standard deviation was 2 psi, what is the probability of obtaining a sample mean of 156.4 psi or below?
(a) To find the probability of obtaining a sample mean this far or farther below the process mean, we need to calculate the z-score and then find the corresponding probability from the standard normal distribution.
Given that the process mean provided by the vendor is 157 psi, the sample mean is 0.6 psi below the process mean. We can calculate the z-score using the formula:
z = (x - μ) / (σ / √n)
Where:
x = sample mean = 0.6 psi below the process mean = 157 - 0.6 = 156.4 psi
μ = process mean = 157 psi
σ = standard deviation = 3 psi
n = sample size = 50
Substituting the values into the formula:
z = (156.4 - 157) / (3 / √50)
Simplifying:
z = -0.6 / (3 / √50)
z = -0.6 / (3 / 7.071)
z = -0.6 / 0.4243
z ≈ -1.414
Now, we can find the probability associated with this z-score using a standard normal distribution table or a calculator. The probability of obtaining a sample mean this far or farther below the process mean is the area to the left of the z-score.
P(z ≤ -1.414) ≈ 0.078
Therefore, the probability of obtaining a sample mean this far or farther below the process mean is approximately 0.078 or 7.8%.
(b) In this case, the process mean is 151 psi, and the standard deviation is 3 psi. We are interested in finding the probability of obtaining a sample mean of 156.4 psi or below.
Using the same formula as in part (a):
z = (x - μ) / (σ / √n)
Substituting the values:
z = (156.4 - 151) / (3 / √50)
Simplifying:
z = 5.4 / (3 / √50)
z = 5.4 / (3 / 7.071)
z = 5.4 / 0.4243
z ≈ 12.739
The probability of obtaining a sample mean of 156.4 psi or below is the area to the left of the z-score.
P(z ≤ 12.739) is practically 1 (close to 100%).
Therefore, the probability is approximately 1 or 100%.
(c) In this case, the process mean is 157 psi, and the standard deviation is 2 psi. We want to calculate the probability of obtaining a sample mean of 156.4 psi or below.
Using the same formula as before:
z = (x - μ) / (σ / √n)
Substituting the values:
z = (156.4 - 157) / (2 / √50)
Simplifying:
z = -0.6 / (2 / √50)
z = -0.6 / (2 / 7.071)
z = -0.6 / 0.2828
z ≈ -2.12
The probability of obtaining a sample mean of 156.4 psi or below is the area to the left of the z-score.
P(z ≤ -2.12) ≈ 0.017
Therefore, the probability is approximately 0.017 or 1.7%.
To solve these probability questions, we will use the concept of sampling distribution and the z-score.
(a) To find the probability of obtaining a sample mean that is 0.6 psi below the process mean, we need to calculate the z-score for this sample mean. The z-score measures the number of standard deviations a data point is away from the mean.
The formula to calculate the z-score is:
z = (x - μ) / (σ / √n)
Where:
x = sample mean (157 - 0.6 = 156.4 psi)
μ = population mean (157 psi as claimed by the vendor)
σ = population standard deviation (3 psi)
n = sample size (50 bottles)
Substituting the values into the formula, we get:
z = (156.4 - 157) / (3 / √50) ≈ -1.39
Next, we need to find the probability of obtaining a z-score this far or further below the process mean. We can look up this probability in a standard normal distribution table or use a statistical calculator.
Using a standard normal distribution table, we can find that the probability corresponding to a z-score of -1.39 is approximately 0.0823. Therefore, the probability of obtaining a sample mean this far or farther below the process mean is approximately 0.0823 or 8.23%.
(b) In this case, we will calculate the z-score using the formula mentioned above, but the mean will be 151 psi instead of 157 psi.
z = (x - μ) / (σ / √n)
z = (156.4 - 151) / (3 / √50) ≈ 2.05
Now, we need to find the probability of obtaining a z-score of 2.05 or below. Again, using a standard normal distribution table, we find that the probability is approximately 0.9798 or 97.98%.
(c) In this case, the mean remains 157 psi, but the standard deviation changes to 2 psi.
z = (x - μ) / (σ / √n)
z = (156.4 - 157) / (2 / √50) ≈ -0.79
Finding the probability of obtaining a z-score of -0.79 or below using the standard normal distribution table gives us approximately 0.2149 or 21.49%.
Hope this helps you understand how to solve these probability questions using sampling distributions and z-scores!