A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 122 ounces and a standard deviation of 0.30 ounce. You randomly select 50 cans and carefully measure the contents. The sample mean of the cans is 121.9 ounces. Does the machine need to be reset? Explain your reasoning. ▼, it is within the range of a usual that you would have randomly sampled 50 cans with a mean equal to 121.9 ounces, because it of the mean of the sample means. event, namely within
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 122 ounces and a standard deviation of 0.30 ounce. You randomly select 50 cans and carefully measure the contents. The sample mean of the cans is 121.9 ounces. Does the machine need to be reset? Explain your reasoning. ▼, it is within the range of a usual that you would have randomly sampled 50 cans with a mean equal to 121.9 ounces, because it of the mean of the sample means. event, namely within
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Understanding Sample Means and Standard Deviation in Quality Control**
When regulating a machine used for filling gallon-sized paint cans, it is essential to ensure the amount of paint dispensed aligns closely with the desired mean. For this scenario, the machine is set to fill the cans with an average of 122 ounces (mean) with a standard deviation of 0.30 ounces.
**Scenario:**
You randomly select 50 cans and measure their contents. The sample mean of these 50 cans is 121.9 ounces. To determine whether the machine needs to be reset, consider the following:
**Steps to determine if the machine needs resetting:**
1. **Calculation of the Standard Error (SE):**
\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \]
Where:
- \(\sigma\) is the population standard deviation (0.30 ounces)
- \(n\) is the number of samples (50 cans)
Plugging in the values:
\[ \text{SE} = \frac{0.30}{\sqrt{50}} \approx 0.0424 \]
2. **Determine the Z-score:**
The Z-score will help you understand how many standard errors away your sample mean is from the population mean.
\[ Z = \frac{\bar{X} - \mu}{\text{SE}} \]
Where:
- \(\bar{X}\) is the sample mean (121.9 ounces)
- \(\mu\) is the population mean (122 ounces)
- \(\text{SE}\) is the standard error calculated above
Plugging in the values:
\[ Z = \frac{121.9 - 122}{0.0424} \approx -2.36 \]
3. **Interpret the Z-score:**
With a Z-score of -2.36, compare this with the standard normal distribution. Typically, a Z-score between -2 and 2 falls within the range of usual events.
**Conclusion:**
It is (likely/unlikely/very likely) that you would have randomly sampled 50 cans with a mean equal to 121.9 ounces, because it (is/is not/may be) within the range of a usual event, namely within (2 standard deviations/0.30 ounces/standard mean deviation) of the mean of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fac4255ba-2787-4b8b-b63d-664d949df058%2F27c5999f-4053-4f3e-8fdb-2806783ead27%2Fzn29uey_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Sample Means and Standard Deviation in Quality Control**
When regulating a machine used for filling gallon-sized paint cans, it is essential to ensure the amount of paint dispensed aligns closely with the desired mean. For this scenario, the machine is set to fill the cans with an average of 122 ounces (mean) with a standard deviation of 0.30 ounces.
**Scenario:**
You randomly select 50 cans and measure their contents. The sample mean of these 50 cans is 121.9 ounces. To determine whether the machine needs to be reset, consider the following:
**Steps to determine if the machine needs resetting:**
1. **Calculation of the Standard Error (SE):**
\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \]
Where:
- \(\sigma\) is the population standard deviation (0.30 ounces)
- \(n\) is the number of samples (50 cans)
Plugging in the values:
\[ \text{SE} = \frac{0.30}{\sqrt{50}} \approx 0.0424 \]
2. **Determine the Z-score:**
The Z-score will help you understand how many standard errors away your sample mean is from the population mean.
\[ Z = \frac{\bar{X} - \mu}{\text{SE}} \]
Where:
- \(\bar{X}\) is the sample mean (121.9 ounces)
- \(\mu\) is the population mean (122 ounces)
- \(\text{SE}\) is the standard error calculated above
Plugging in the values:
\[ Z = \frac{121.9 - 122}{0.0424} \approx -2.36 \]
3. **Interpret the Z-score:**
With a Z-score of -2.36, compare this with the standard normal distribution. Typically, a Z-score between -2 and 2 falls within the range of usual events.
**Conclusion:**
It is (likely/unlikely/very likely) that you would have randomly sampled 50 cans with a mean equal to 121.9 ounces, because it (is/is not/may be) within the range of a usual event, namely within (2 standard deviations/0.30 ounces/standard mean deviation) of the mean of
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