Question
Two machines in a factory are supposed to work at the same speed to pass inspection. On five random days, the number of items built by each machine is recorded in the table. The inspector believes that the machines should not pass inspection because the mean speed of Machine X is much faster than the mean speed of Machine Y. Which measure of center and variability should be used to compare the performances of each machine? Then explain why the inspector is correct or incorrect with her decision.
Number of Items Built
Machine X
19
15
20
16
17
Machine Y
20
3
17
18
17
Question content area bottom
Part 1
Which measure of center and variability should be used to compare the performances of each machine? Select all that apply.
A.
Range
B.
Interquartile range
Your answer is correct.C.
Median
This is the correct answer.D.
Mean
Your answer is not correct.
Part 2
Is the inspector correct or incorrect with her decision?
A.
The inspector is incorrect. The mean speeds of both machines are equal and the ranges are the same or about the same.
B.
The inspector is correct. The data shows no unusual values and Machine X always runs 1.2 times faster than Machine Y.
C.
The inspector is incorrect. The data shows the median speeds are equal for both machines. The mean should not be used because there is one usually low value for Machine Y.
Number of Items Built
Machine X
19
15
20
16
17
Machine Y
20
3
17
18
17
Question content area bottom
Part 1
Which measure of center and variability should be used to compare the performances of each machine? Select all that apply.
A.
Range
B.
Interquartile range
Your answer is correct.C.
Median
This is the correct answer.D.
Mean
Your answer is not correct.
Part 2
Is the inspector correct or incorrect with her decision?
A.
The inspector is incorrect. The mean speeds of both machines are equal and the ranges are the same or about the same.
B.
The inspector is correct. The data shows no unusual values and Machine X always runs 1.2 times faster than Machine Y.
C.
The inspector is incorrect. The data shows the median speeds are equal for both machines. The mean should not be used because there is one usually low value for Machine Y.
Answers
Answered by
GPT-4o mini
### Part 1: Which measure of center and variability should be used to compare the performances of each machine?
To compare the performances of Machine X and Machine Y, we should consider the following measures:
1. **Median**: The median is a good measure of center to use, especially when the data may contain outliers or skewness. In this case, Machine Y has a significantly lower number of items built on one of the days (3 items), which could skew the mean.
2. **Interquartile Range (IQR)**: The IQR is also a useful measure of variability since it provides an understanding of the spread of the middle 50% of the data, thus reducing the impact of outliers.
### Correct Selections for Part 1:
- **B. Interquartile range**
- **C. Median**
### Part 2: Is the inspector correct or incorrect with her decision?
To determine whether the inspector is correct, we need to analyze the data provided:
- Machine X: 19, 15, 20, 16, 17
- Mean = (19 + 15 + 20 + 16 + 17) / 5 = 17.4
- Median = 17 (when ordered: 15, 16, 17, 19, 20)
- Machine Y: 20, 3, 17, 18, 17
- Mean = (20 + 3 + 17 + 18 + 17) / 5 = 15
- Median = 17 (when ordered: 3, 17, 17, 18, 20)
From this analysis, we can see that:
- Machine X has a higher mean than Machine Y, and the mean is influenced by the outlier (3) in Machine Y's data.
- The medians of both machines are equal (17).
### Conclusion for Part 2:
C. **The inspector is incorrect. The data shows the median speeds are equal for both machines. The mean should not be used because there is one usually low value for Machine Y.**
The inspector's conclusion based on the means is flawed because it does not consider the influence of outliers and does not take the median into account, which is a more appropriate measure in this case.
To compare the performances of Machine X and Machine Y, we should consider the following measures:
1. **Median**: The median is a good measure of center to use, especially when the data may contain outliers or skewness. In this case, Machine Y has a significantly lower number of items built on one of the days (3 items), which could skew the mean.
2. **Interquartile Range (IQR)**: The IQR is also a useful measure of variability since it provides an understanding of the spread of the middle 50% of the data, thus reducing the impact of outliers.
### Correct Selections for Part 1:
- **B. Interquartile range**
- **C. Median**
### Part 2: Is the inspector correct or incorrect with her decision?
To determine whether the inspector is correct, we need to analyze the data provided:
- Machine X: 19, 15, 20, 16, 17
- Mean = (19 + 15 + 20 + 16 + 17) / 5 = 17.4
- Median = 17 (when ordered: 15, 16, 17, 19, 20)
- Machine Y: 20, 3, 17, 18, 17
- Mean = (20 + 3 + 17 + 18 + 17) / 5 = 15
- Median = 17 (when ordered: 3, 17, 17, 18, 20)
From this analysis, we can see that:
- Machine X has a higher mean than Machine Y, and the mean is influenced by the outlier (3) in Machine Y's data.
- The medians of both machines are equal (17).
### Conclusion for Part 2:
C. **The inspector is incorrect. The data shows the median speeds are equal for both machines. The mean should not be used because there is one usually low value for Machine Y.**
The inspector's conclusion based on the means is flawed because it does not consider the influence of outliers and does not take the median into account, which is a more appropriate measure in this case.