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Mileage of five different vehicles as reported by Jack
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25 Aug 2021, 04:45
25 Aug 2021, 04:45
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Question Stats:
40% (01:20) correct
60% (01:15) wrong based on 20 sessions
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Mileage of five different vehicles as reported by Jack were 16, 18, 11, 13, and 19. Two of the mileages were wrongly reported by Jack, so that one was 5 units more than the correct value, and other was 5 units less than the correct value.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Standard Deviation of reported mileages |
Standard Deviation of actual mileages |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Mileage of five different vehicles as reported by Jack
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25 Aug 2021, 14:31
25 Aug 2021, 14:31
1
Expert Reply
The average of the reported mileage and the correct mileage will be the same i.e. \( /mu \), since it is stated that the error in the reported measurements was +5 and -5.
Average of reported measurements: \( \frac{ 16 + 18 + 11 + 13 + 19 }{ 5 } = \frac{ 77 }{5 } = 15.4 \)
Average of actual measurements: \( \frac{ 16 + 18 + 11 + 13 + 19 +5 -5 }{ 5 } = \frac{ 77 }{5 } = 15.4 \)
The standard deviation, however, is a different story. We cannot assume anything about the SD since SD measures the spread from the average.
As an example, having a set of 5 elements where each element is 15.4, the average would be the same as the given set but the SD would be 0.
\( \sigma = \sqrt{\frac{ \Sigma_{i =1}^{5}(x_i - \mu)^2 }{ 5}}\)
We can solve this problem by visual inspection.
Since the measurements are 16, 18, 11, 13, and 19, assume that 16 was overstated by 5 and 11 was understated by 5 i.e. 16 was actually 11 and 11 was actually 16.
In this case, the standard deviation does not change as the values do not change, just when they were recorded.
On the other hand, assume, that 11 was overstated by 5 and 19 was understated by 5 i.e 11 was actually 6 and 19 was actually 24.
Our list now becomes 16, 18, 6, 13, and 24. While the average is still the same, we can see that \((6 -15.4)^2 + (24 -15.4)^2\) will be greater than \((11 -15.4)^2 + (19 -15.4)^2\), all else being the same. Thus in this case, the SD will be greater.
Using these two scenarios, we have seen two contradictory situations.
The answer thus is Option D
Average of reported measurements: \( \frac{ 16 + 18 + 11 + 13 + 19 }{ 5 } = \frac{ 77 }{5 } = 15.4 \)
Average of actual measurements: \( \frac{ 16 + 18 + 11 + 13 + 19 +5 -5 }{ 5 } = \frac{ 77 }{5 } = 15.4 \)
The standard deviation, however, is a different story. We cannot assume anything about the SD since SD measures the spread from the average.
As an example, having a set of 5 elements where each element is 15.4, the average would be the same as the given set but the SD would be 0.
\( \sigma = \sqrt{\frac{ \Sigma_{i =1}^{5}(x_i - \mu)^2 }{ 5}}\)
We can solve this problem by visual inspection.
Since the measurements are 16, 18, 11, 13, and 19, assume that 16 was overstated by 5 and 11 was understated by 5 i.e. 16 was actually 11 and 11 was actually 16.
In this case, the standard deviation does not change as the values do not change, just when they were recorded.
On the other hand, assume, that 11 was overstated by 5 and 19 was understated by 5 i.e 11 was actually 6 and 19 was actually 24.
Our list now becomes 16, 18, 6, 13, and 24. While the average is still the same, we can see that \((6 -15.4)^2 + (24 -15.4)^2\) will be greater than \((11 -15.4)^2 + (19 -15.4)^2\), all else being the same. Thus in this case, the SD will be greater.
Using these two scenarios, we have seen two contradictory situations.
The answer thus is Option D
General Discussion
Re: Mileage of five different vehicles as reported by Jack
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25 Aug 2021, 15:30
25 Aug 2021, 15:30
Expert Reply
AL88 wrote:
The average of the reported mileage and the correct mileage will be the same i.e. \( /mu \), since it is stated that the error in the reported measurements was +5 and -5.
Average of reported measurements: \( \frac{ 16 + 18 + 11 + 13 + 19 }{ 5 } = \frac{ 77 }{5 } = 15.4 \)
Average of actual measurements: \( \frac{ 16 + 18 + 11 + 13 + 19 +5 -5 }{ 5 } = \frac{ 77 }{5 } = 15.4 \)
The standard deviation, however, is a different story. We cannot assume anything about the SD since SD measures the spread from the average.
As an example, having a set of 5 elements where each element is 15.4, the average would be the same as the given set but the SD would be 0.
\( \sigma = \sqrt{\frac{ \Sigma_{i =1}^{5}(x_i - \mu)^2 }{ 5}}\)
We can solve this problem by visual inspection.
Since the measurements are 16, 18, 11, 13, and 19, assume that 16 was overstated by 5 and 11 was understated by 5 i.e. 16 was actually 11 and 11 was actually 16.
In this case, the standard deviation does not change as the values do not change, just when they were recorded.
On the other hand, assume, that 11 was overstated by 5 and 19 was understated by 5 i.e 11 was actually 6 and 19 was actually 24.
Our list now becomes 16, 18, 6, 13, and 24. While the average is still the same, we can see that \((6 -15.4)^2 + (24 -15.4)^2\) will be greater than \((11 -15.4)^2 + (19 -15.4)^2\), all else being the same. Thus in this case, the SD will be greater.
Using these two scenarios, we have seen two contradictory situations.
The answer thus is Option D
Average of reported measurements: \( \frac{ 16 + 18 + 11 + 13 + 19 }{ 5 } = \frac{ 77 }{5 } = 15.4 \)
Average of actual measurements: \( \frac{ 16 + 18 + 11 + 13 + 19 +5 -5 }{ 5 } = \frac{ 77 }{5 } = 15.4 \)
The standard deviation, however, is a different story. We cannot assume anything about the SD since SD measures the spread from the average.
As an example, having a set of 5 elements where each element is 15.4, the average would be the same as the given set but the SD would be 0.
\( \sigma = \sqrt{\frac{ \Sigma_{i =1}^{5}(x_i - \mu)^2 }{ 5}}\)
We can solve this problem by visual inspection.
Since the measurements are 16, 18, 11, 13, and 19, assume that 16 was overstated by 5 and 11 was understated by 5 i.e. 16 was actually 11 and 11 was actually 16.
In this case, the standard deviation does not change as the values do not change, just when they were recorded.
On the other hand, assume, that 11 was overstated by 5 and 19 was understated by 5 i.e 11 was actually 6 and 19 was actually 24.
Our list now becomes 16, 18, 6, 13, and 24. While the average is still the same, we can see that \((6 -15.4)^2 + (24 -15.4)^2\) will be greater than \((11 -15.4)^2 + (19 -15.4)^2\), all else being the same. Thus in this case, the SD will be greater.
Using these two scenarios, we have seen two contradictory situations.
The answer thus is Option D
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