137 C1.08: Interval Part 4
Another method of reporting it would be [latex]52700_{+50}^{-50}[/latex] feet, which is usually only used if the distances on the two sides are not equal.
Example 10: If a measured number is reported as 0.30 meters, what is the rounding precision, underline the significant digits, and find the interval of possible actual values.
Solution: The rounding precision is 0.01, which is one-hundredth. The significant digits are underlined: 0.30 The next-smaller rounded number is 0.29 and the next-larger rounded number is 0.31. We’ll write those as 0.290 and 0.300 and 0.310. (All numbers here are in meters.)
This might also be reported as [latex]0.30\pm0.005[/latex] meters
Another method of reporting it would be [latex]0.30_{+0.005}^{-0.005}[/latex] meters, which is usually only used if the distances on the two sides are not equal.
Example 11: If a measured number is reported as 12 meters,
- what is the interval of actual values consistent with that,
- what is the maximum amount that the actual value could differ from the reported value
- what is the maximum amount of error as a percentage of the reported number?
Solution:
- The interval of actual values is between 11.5 and 12.5 meters.
- The maximum amount that the actual value could differ from the reported value is 0.5 meters.
- To find the percentage of the maximum error as a percentage of the reported value, we must write it as a fraction, and then do the division and get a decimal fraction. Then convert that decimal to a percentage. [latex]\frac{0.5\text{m}}{12\text{m}}=0.041666667[/latex], which is 4.16666667%.
So the maximum error is about 4% of the reported value. This is called the relative error.
Example 12: If a measured number is reported as 12.0 meters,
- what is the interval of actual values consistent with that,
- what is the maximum amount that the actual value could differ from the reported value
- what is the maximum amount of error as a percentage of the reported number?
Solution:
- The interval of actual values is between 11.95 and 12.05 meters.
- The maximum amount that the actual value could differ from the reported value is 0.05 meters.
- To find the percentage of the maximum error as a percentage of the reported value, we must write it as a fraction, and then do the division and get a decimal fraction. Then convert that decimal to a percentage. [latex]\frac{0.05\text{m}}{12\text{m}}=0.0041666667[/latex], which is 0.416666667%.
So the maximum error is about 0.4% of the reported value. This relative error is one-tenth of the relative error in Example 11, reflecting the increased precision of the reported number.