GCSE Maths: Error intervals and limits of accuracy (Foundation)
Part of the GCSE Maths decimals, rounding and estimation chapter (Foundation). Chapter overview · Higher version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Error intervals and limits of accuracy
When a number has been rounded, we no longer know the original number exactly. All we know is that it was somewhere in a range of values. That range is called an error interval. For example, the length of a leaf is 3.4 cm, correct to 1 decimal place. The real length could be 3.36, 3.4 or even 3.44, because all of these round to 3.4. But it cannot be 3.3 or 3.5.
Error intervals from rounding
A number rounded to a given accuracy could have been up to half a unit smaller or bigger than the rounded value. The unit is the value of the last column: for 1 d.p. the unit is 0.1, so half a unit is 0.05. So 3.4 (to 1 d.p.) lies between 3.4 − 0.05 = 3.35 and 3.4 + 0.05 = 3.45. 3.35 would be rounded up to 3.4, so the lower end is included. 3.45 would be rounded up to 3.5, so the upper end is not included. We write this with inequality signs:
3.35 ≤ x < 3.45
Here x stands for the original number. The sign ≤ means “is less than or equal to” (the end value is allowed). The sign < means “is less than” (the end value is not allowed). On a number line, a solid circle means the end is included and an open circle means it is not.
Lower bound and upper bound
The lower bound is the smallest value the number could be (3.35 here), and the upper bound is the value at the top of the range (3.45 here). Together they give the limits of accuracy of a measurement. Although the upper bound is not actually included, in calculations we use it as if it were a possible value. You will see why in section 4.9.
Error intervals from truncation
If a number has been truncated, digits have been chopped off, so the original number is at least the truncated value, and less than one whole unit more. For example 5.3 truncated to 1 d.p. means 5.3 ≤ x < 5.4. The interval is one whole unit wide and starts at the truncated value.
Other accuracies
The same idea works for other roundings. A number that is 450 to the nearest 10 has a unit of 10, so half a unit is 5: 445 ≤ x < 455. A number that is 0.062 to 2 significant figures has its last significant figure in the thousandths column, so the unit is 0.001 and the interval is 0.0615 ≤ x < 0.0625.
Key ideas
- A rounded number lies within half a unit either side: lower bound = value − half a unit, upper bound = value + half a unit.
- Error interval: lower bound ≤ x < upper bound. The upper end uses <.
- Truncated number: value ≤ x < value + one unit.
- Unit = the value of the last digit: 0.1 for 1 d.p., 0.01 for 2 d.p., 10 for the nearest 10.
- Solid circle → included. Open circle → not included.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the last place. The number is given to 1 decimal place, so the last digit is in the tenths column. One unit of that column is 0.1.unit = 0.1
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.0.1 ÷ 2 = 0.05
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 3.4 − 0.05 = 3.35
upper bound = 3.4 + 0.05 = 3.45 - Write the error interval with inequality signs. The lower bound 3.35 rounds up to 3.4, so L can equal it: use ≤. The upper bound 3.45 would round up to 3.5, so L must be less than it: use <.3.35 ≤ L < 3.45
Check. Test the ends: 3.35 rounds to 3.4 ✓; a number just below the top, such as 3.449, rounds to 3.4 ✓; and 3.45 itself rounds up to 3.5, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the rounding unit. The number was rounded to the nearest ten, so the unit is 10.unit = 10
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.10 ÷ 2 = 5
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 450 − 5 = 445
upper bound = 450 + 5 = 455 - Write the error interval with inequality signs. The lower bound 445 rounds up to 450, so m can equal it: use ≤. The upper bound 455 would round up to 460, so m must be less than it: use <.445 ≤ m < 455
Check. Test the ends: 445 rounds to 450 ✓; a number just below the top, such as 454.9, rounds to 450 ✓; and 455 itself rounds up to 460, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the last place that is shown. The number is given to 2 significant figures. The last significant figure of 0.062 is in the thousandths column. One unit of that column is 0.001.unit = 0.001
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.0.001 ÷ 2 = 0.0005
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 0.062 − 0.0005 = 0.0615
upper bound = 0.062 + 0.0005 = 0.0625 - Write the error interval with inequality signs. The lower bound 0.0615 rounds up to 0.062, so x can equal it: use ≤. The upper bound 0.0625 would round up to 0.063, so x must be less than it: use <.0.0615 ≤ x < 0.0625
Check. Test the ends: 0.0615 rounds to 0.062 ✓; a number just below the top, such as 0.06249, rounds to 0.062 ✓; and 0.0625 itself rounds up to 0.063, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number truncated to a given accuracy could have been anywhere from the truncated value up to (but not including) one whole unit more.
- Find the size of the last place. The number is given to 1 decimal place, so the last digit is in the tenths column. One unit of that column is 0.1.unit = 0.1
- Think about what truncating does. Truncating cuts the digits off without rounding, so the original number is never smaller than the truncated value, and is less than one whole unit bigger.5.3 ≤ y < 5.3 + 0.1
- Add one unit to get the upper end. If the number were one unit more it would have a bigger digit in the last column.5.3 + 0.1 = 5.4
- Write the error interval. The lower end 5.3 is possible (the digits after it could all be zero), so use ≤. The upper end 5.4 is not possible, so use <.5.3 ≤ y < 5.4
Check. Truncate 5.3 to the same place: 5.3 ✓. Truncate a value just below the top, 5.399: 5.3 ✓. Truncating 5.4 gives 5.4, which is different. ✓
Watch out. For truncation the interval starts at the given value and goes up by one whole unit, not half a unit.
Method. Compare the number with the lower end and with the upper end of the error interval.
- Write the error interval and the number side by side. The number is possible only if it lies inside the interval.2.45 ≤ x < 2.55
number: 2.55 - Check the lower end. The number must be at least the lower bound (it can equal it).2.55 ≥ 2.45: true
- Check the upper end. The number must be less than the upper bound (it cannot equal it).2.55 < 2.55: false
- Decide. Both checks must be true.No, 2.55 is not a possible value.
Check. Picture it on a number line: 2.55 is outside the interval. ✓
Watch out. An open circle (a < sign) means the end value itself is not allowed.
Watch out
The upper end of an error interval uses <, not ≤. A number exactly half-way rounds up, so the upper bound is not a possible value of the original number. Also, the interval is half a unit either side for rounding, but a whole unit for truncation.Exercise 4I Error intervals and limits of accuracy
Non-calculatorKey words. The lower bound is the smallest value a measurement could be. The upper bound is the limit above which it would round to a different value.
Method. The lower bound is half a unit below the measurement and the upper bound is half a unit above it.
- Find the unit of the last digit. The measurement is given to 1 decimal place, so the last digit is in the tenths column. One unit of that column is 0.1.unit = 0.1
- Halve it. Any value within half a unit of the measurement would round to it.0.1 ÷ 2 = 0.05
- Subtract and add half a unit. The lower bound is the smallest possible value and the upper bound is the top end of the range.lower bound = 7.3 − 0.05 = 7.25 cm
upper bound = 7.3 + 0.05 = 7.35 cm
Check. Both bounds are exactly half a unit (0.05) from 7.3, and 7.25 to 7.35 is a width of one whole unit (0.1). ✓
Watch out. Half of the unit, not the whole unit: 7.3 (1 d.p.) goes from 7.25 to 7.35, not 7.2 to 7.4.
Key words. The lower bound is the smallest value a measurement could be. The upper bound is the limit above which it would round to a different value.
Method. The lower bound is half a unit below the measurement and the upper bound is half a unit above it.
- Find the unit of the last digit. The measurement is given to the nearest ten, so the last digit is in the tens column. One unit of that column is 10.unit = 10
- Halve it. Any value within half a unit of the measurement would round to it.10 ÷ 2 = 5
- Subtract and add half a unit. The lower bound is the smallest possible value and the upper bound is the top end of the range.lower bound = 45 − 5 = 40 kg
upper bound = 45 + 5 = 50 kg
Check. Both bounds are exactly half a unit (5) from 45, and 40 to 50 is a width of one whole unit (10). ✓
Watch out. Half of the unit, not the whole unit: 7.3 (1 d.p.) goes from 7.25 to 7.35, not 7.2 to 7.4.
Key words. The lower bound is the smallest value a measurement could be. The upper bound is the limit above which it would round to a different value.
Method. The lower bound is half a unit below the measurement and the upper bound is half a unit above it.
- Find the unit of the last digit. The measurement is given to 2 decimal places, so the last digit is in the hundredths column. One unit of that column is 0.01.unit = 0.01
- Halve it. Any value within half a unit of the measurement would round to it.0.01 ÷ 2 = 0.005
- Subtract and add half a unit. The lower bound is the smallest possible value and the upper bound is the top end of the range.lower bound = 0.82 − 0.005 = 0.815
upper bound = 0.82 + 0.005 = 0.825
Check. Both bounds are exactly half a unit (0.005) from 0.82, and 0.815 to 0.825 is a width of one whole unit (0.01). ✓
Watch out. Half of the unit, not the whole unit: 7.3 (1 d.p.) goes from 7.25 to 7.35, not 7.2 to 7.4.
Key words. The lower bound is the smallest value a measurement could be. The upper bound is the limit above which it would round to a different value.
Method. The lower bound is half a unit below the measurement and the upper bound is half a unit above it.
- Find the unit of the last digit. The measurement is given to the nearest hundred, so the last digit is in the hundreds column. One unit of that column is 100.unit = 100
- Halve it. Any value within half a unit of the measurement would round to it.100 ÷ 2 = 50
- Subtract and add half a unit. The lower bound is the smallest possible value and the upper bound is the top end of the range.lower bound = 2400 − 50 = 2350 people
upper bound = 2400 + 50 = 2450 people
Check. Both bounds are exactly half a unit (50) from 2400, and 2350 to 2450 is a width of one whole unit (100). ✓
Watch out. Half of the unit, not the whole unit: 7.3 (1 d.p.) goes from 7.25 to 7.35, not 7.2 to 7.4.
Key words. The lower bound is the smallest value a measurement could be. The upper bound is the limit above which it would round to a different value.
Method. The lower bound is half a unit below the measurement and the upper bound is half a unit above it.
- Find the unit of the last digit. The measurement is given to 2 significant figures, so the last digit is in the tenths column. One unit of that column is 0.1.unit = 0.1
- Halve it. Any value within half a unit of the measurement would round to it.0.1 ÷ 2 = 0.05
- Subtract and add half a unit. The lower bound is the smallest possible value and the upper bound is the top end of the range.lower bound = 3.5 − 0.05 = 3.45 m
upper bound = 3.5 + 0.05 = 3.55 m
Check. Both bounds are exactly half a unit (0.05) from 3.5, and 3.45 to 3.55 is a width of one whole unit (0.1). ✓
Watch out. Half of the unit, not the whole unit: 7.3 (1 d.p.) goes from 7.25 to 7.35, not 7.2 to 7.4.
Key words. The lower bound is the smallest value a measurement could be. The upper bound is the limit above which it would round to a different value.
Method. The lower bound is half a unit below the measurement and the upper bound is half a unit above it.
- Find the unit of the last digit. The measurement is given to the nearest whole number, so the last digit is in the units column. One unit of that column is 1.unit = 1
- Halve it. Any value within half a unit of the measurement would round to it.1 ÷ 2 = 0.5
- Subtract and add half a unit. The lower bound is the smallest possible value and the upper bound is the top end of the range.lower bound = 6 − 0.5 = 5.5 s
upper bound = 6 + 0.5 = 6.5 s
Check. Both bounds are exactly half a unit (0.5) from 6, and 5.5 to 6.5 is a width of one whole unit (1). ✓
Watch out. Half of the unit, not the whole unit: 7.3 (1 d.p.) goes from 7.25 to 7.35, not 7.2 to 7.4.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the last place. The number is given to 1 decimal place, so the last digit is in the tenths column. One unit of that column is 0.1.unit = 0.1
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.0.1 ÷ 2 = 0.05
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 3.4 − 0.05 = 3.35
upper bound = 3.4 + 0.05 = 3.45 - Write the error interval with inequality signs. The lower bound 3.35 rounds up to 3.4, so x can equal it: use ≤. The upper bound 3.45 would round up to 3.5, so x must be less than it: use <.3.35 ≤ x < 3.45
Check. Test the ends: 3.35 rounds to 3.4 ✓; a number just below the top, such as 3.449, rounds to 3.4 ✓; and 3.45 itself rounds up to 3.5, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the last place. The number is given to the nearest whole number, so the last digit is in the units column. One unit of that column is 1.unit = 1
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.1 ÷ 2 = 0.5
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 52 − 0.5 = 51.5
upper bound = 52 + 0.5 = 52.5 - Write the error interval with inequality signs. The lower bound 51.5 rounds up to 52, so x can equal it: use ≤. The upper bound 52.5 would round up to 53, so x must be less than it: use <.51.5 ≤ x < 52.5
Check. Test the ends: 51.5 rounds to 52 ✓; a number just below the top, such as 52.49, rounds to 52 ✓; and 52.5 itself rounds up to 53, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the last place. The number is given to 2 decimal places, so the last digit is in the hundredths column. One unit of that column is 0.01.unit = 0.01
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.0.01 ÷ 2 = 0.005
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 0.85 − 0.005 = 0.845
upper bound = 0.85 + 0.005 = 0.855 - Write the error interval with inequality signs. The lower bound 0.845 rounds up to 0.85, so x can equal it: use ≤. The upper bound 0.855 would round up to 0.86, so x must be less than it: use <.0.845 ≤ x < 0.855
Check. Test the ends: 0.845 rounds to 0.85 ✓; a number just below the top, such as 0.8549, rounds to 0.85 ✓; and 0.855 itself rounds up to 0.86, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the rounding unit. The number was rounded to the nearest hundred, so the unit is 100.unit = 100
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.100 ÷ 2 = 50
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 6700 − 50 = 6650
upper bound = 6700 + 50 = 6750 - Write the error interval with inequality signs. The lower bound 6650 rounds up to 6700, so x can equal it: use ≤. The upper bound 6750 would round up to 6800, so x must be less than it: use <.6650 ≤ x < 6750
Check. Test the ends: 6650 rounds to 6700 ✓; a number just below the top, such as 6749, rounds to 6700 ✓; and 6750 itself rounds up to 6800, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the last place that is shown. The number is given to 2 significant figures. The last significant figure of 4.5 is in the tenths column. One unit of that column is 0.1.unit = 0.1
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.0.1 ÷ 2 = 0.05
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 4.5 − 0.05 = 4.45
upper bound = 4.5 + 0.05 = 4.55 - Write the error interval with inequality signs. The lower bound 4.45 rounds up to 4.5, so x can equal it: use ≤. The upper bound 4.55 would round up to 4.6, so x must be less than it: use <.4.45 ≤ x < 4.55
Check. Test the ends: 4.45 rounds to 4.5 ✓; a number just below the top, such as 4.549, rounds to 4.5 ✓; and 4.55 itself rounds up to 4.6, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the last place that is shown. The number is given to 2 significant figures. The last significant figure of 0.062 is in the thousandths column. One unit of that column is 0.001.unit = 0.001
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.0.001 ÷ 2 = 0.0005
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 0.062 − 0.0005 = 0.0615
upper bound = 0.062 + 0.0005 = 0.0625 - Write the error interval with inequality signs. The lower bound 0.0615 rounds up to 0.062, so x can equal it: use ≤. The upper bound 0.0625 would round up to 0.063, so x must be less than it: use <.0.0615 ≤ x < 0.0625
Check. Test the ends: 0.0615 rounds to 0.062 ✓; a number just below the top, such as 0.06249, rounds to 0.062 ✓; and 0.0625 itself rounds up to 0.063, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number truncated to a given accuracy could have been anywhere from the truncated value up to (but not including) one whole unit more.
- Find the size of the last place. The number is given to 1 decimal place, so the last digit is in the tenths column. One unit of that column is 0.1.unit = 0.1
- Think about what truncating does. Truncating cuts the digits off without rounding, so the original number is never smaller than the truncated value, and is less than one whole unit bigger.5.3 ≤ x < 5.3 + 0.1
- Add one unit to get the upper end. If the number were one unit more it would have a bigger digit in the last column.5.3 + 0.1 = 5.4
- Write the error interval. The lower end 5.3 is possible (the digits after it could all be zero), so use ≤. The upper end 5.4 is not possible, so use <.5.3 ≤ x < 5.4
Check. Truncate 5.3 to the same place: 5.3 ✓. Truncate a value just below the top, 5.399: 5.3 ✓. Truncating 5.4 gives 5.4, which is different. ✓
Watch out. For truncation the interval starts at the given value and goes up by one whole unit, not half a unit.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number truncated to a given accuracy could have been anywhere from the truncated value up to (but not including) one whole unit more.
- Find the size of the last place. The number is given to the nearest whole number, so the last digit is in the units column. One unit of that column is 1.unit = 1
- Think about what truncating does. Truncating cuts the digits off without rounding, so the original number is never smaller than the truncated value, and is less than one whole unit bigger.12 ≤ x < 12 + 1
- Add one unit to get the upper end. If the number were one unit more it would have a bigger digit in the last column.12 + 1 = 13
- Write the error interval. The lower end 12 is possible (the digits after it could all be zero), so use ≤. The upper end 13 is not possible, so use <.12 ≤ x < 13
Check. Truncate 12 to the same place: 12 ✓. Truncate a value just below the top, 12.99: 12 ✓. Truncating 13 gives 13, which is different. ✓
Watch out. For truncation the interval starts at the given value and goes up by one whole unit, not half a unit.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number truncated to a given accuracy could have been anywhere from the truncated value up to (but not including) one whole unit more.
- Find the size of the last place. The number is given to 2 decimal places, so the last digit is in the hundredths column. One unit of that column is 0.01.unit = 0.01
- Think about what truncating does. Truncating cuts the digits off without rounding, so the original number is never smaller than the truncated value, and is less than one whole unit bigger.0.47 ≤ x < 0.47 + 0.01
- Add one unit to get the upper end. If the number were one unit more it would have a bigger digit in the last column.0.47 + 0.01 = 0.48
- Write the error interval. The lower end 0.47 is possible (the digits after it could all be zero), so use ≤. The upper end 0.48 is not possible, so use <.0.47 ≤ x < 0.48
Check. Truncate 0.47 to the same place: 0.47 ✓. Truncate a value just below the top, 0.4799: 0.47 ✓. Truncating 0.48 gives 0.48, which is different. ✓
Watch out. For truncation the interval starts at the given value and goes up by one whole unit, not half a unit.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number truncated to a given accuracy could have been anywhere from the truncated value up to (but not including) one whole unit more.
- Find the size of the rounding unit. The number was rounded to the nearest hundred, so the unit is 100.unit = 100
- Think about what truncating does. Truncating cuts the digits off without rounding, so the original number is never smaller than the truncated value, and is less than one whole unit bigger.2400 ≤ x < 2400 + 100
- Add one unit to get the upper end. If the number were one unit more it would have a bigger digit in the last column.2400 + 100 = 2500
- Write the error interval. The lower end 2400 is possible (the digits after it could all be zero), so use ≤. The upper end 2500 is not possible, so use <.2400 ≤ x < 2500
Check. Truncate 2400 to the same place: 2400 ✓. Truncate a value just below the top, 2499: 2400 ✓. Truncating 2500 gives 2500, which is different. ✓
Watch out. For truncation the interval starts at the given value and goes up by one whole unit, not half a unit.
Method. Read the interval from the number line: find the ends and see which circles are solid and which are open.
- Read the end values. The ends are the numbers where the circles are drawn.left end 2.35, right end 2.45
- Look at the circles. A solid circle means the value is included (use ≤). An open circle means it is not included (use <).left circle solid: ≤, right circle open: <
- Write the inequality. Put x in the middle.2.35 ≤ x < 2.45
Check. Half way between 2.35 and 2.45 is 2.4 (the given value), so the number was 2.4 to 1 d.p. ✓
Watch out. Solid circle = included. Open circle = not included.
Method. Compare the number with the lower end and with the upper end of the error interval.
- Write the error interval and the number side by side. The number is possible only if it lies inside the interval.3.35 ≤ x < 3.45
number: 3.4 - Check the lower end. The number must be at least the lower bound (it can equal it).3.4 ≥ 3.35: true
- Check the upper end. The number must be less than the upper bound (it cannot equal it).3.4 < 3.45: true
- Decide. Both checks must be true.Yes, 3.4 is a possible value.
Check. Picture it on a number line: 3.4 is inside the interval. ✓
Watch out. An open circle (a < sign) means the end value itself is not allowed.
Method. Compare the number with the lower end and with the upper end of the error interval.
- Write the error interval and the number side by side. The number is possible only if it lies inside the interval.3.35 ≤ x < 3.45
number: 3.45 - Check the lower end. The number must be at least the lower bound (it can equal it).3.45 ≥ 3.35: true
- Check the upper end. The number must be less than the upper bound (it cannot equal it).3.45 < 3.45: false
- Decide. Both checks must be true.No, 3.45 is not a possible value.
Check. Picture it on a number line: 3.45 is outside the interval. ✓
Watch out. An open circle (a < sign) means the end value itself is not allowed.
Method. Compare the number with the lower end and with the upper end of the error interval.
- Write the error interval and the number side by side. The number is possible only if it lies inside the interval.3.35 ≤ x < 3.45
number: 3.35 - Check the lower end. The number must be at least the lower bound (it can equal it).3.35 ≥ 3.35: true
- Check the upper end. The number must be less than the upper bound (it cannot equal it).3.35 < 3.45: true
- Decide. Both checks must be true.Yes, 3.35 is a possible value.
Check. Picture it on a number line: 3.35 is inside the interval. ✓
Watch out. An open circle (a < sign) means the end value itself is not allowed.
Method. Compare the number with the lower end and with the upper end of the error interval.
- Write the error interval and the number side by side. The number is possible only if it lies inside the interval.3.35 ≤ x < 3.45
number: 3.3499 - Check the lower end. The number must be at least the lower bound (it can equal it).3.3499 ≥ 3.35: false
- Check the upper end. The number must be less than the upper bound (it cannot equal it).3.3499 < 3.45: true
- Decide. Both checks must be true.No, 3.3499 is not a possible value.
Check. Picture it on a number line: 3.3499 is outside the interval. ✓
Watch out. An open circle (a < sign) means the end value itself is not allowed.
Method. Compare the number with the lower end and with the upper end of the error interval.
- Write the error interval and the number side by side. The number is possible only if it lies inside the interval.3.35 ≤ x < 3.45
number: 3.5 - Check the lower end. The number must be at least the lower bound (it can equal it).3.5 ≥ 3.35: true
- Check the upper end. The number must be less than the upper bound (it cannot equal it).3.5 < 3.45: false
- Decide. Both checks must be true.No, 3.5 is not a possible value.
Check. Picture it on a number line: 3.5 is outside the interval. ✓
Watch out. An open circle (a < sign) means the end value itself is not allowed.
Method. Compare the number with the lower end and with the upper end of the error interval.
- Write the error interval and the number side by side. The number is possible only if it lies inside the interval.1.45 ≤ x < 1.55
number: 1.46 - Check the lower end. The number must be at least the lower bound (it can equal it).1.46 ≥ 1.45: true
- Check the upper end. The number must be less than the upper bound (it cannot equal it).1.46 < 1.55: true
- Decide. Both checks must be true.Yes, 1.46 is a possible value.
Check. Picture it on a number line: 1.46 is inside the interval. ✓
Watch out. An open circle (a < sign) means the end value itself is not allowed.
Method. Compare the number with the lower end and with the upper end of the error interval.
- Write the error interval and the number side by side. The number is possible only if it lies inside the interval.1.45 ≤ x < 1.55
number: 1.55 - Check the lower end. The number must be at least the lower bound (it can equal it).1.55 ≥ 1.45: true
- Check the upper end. The number must be less than the upper bound (it cannot equal it).1.55 < 1.55: false
- Decide. Both checks must be true.No, 1.55 is not a possible value.
Check. Picture it on a number line: 1.55 is outside the interval. ✓
Watch out. An open circle (a < sign) means the end value itself is not allowed.
Method. Compare the number with the lower end and with the upper end of the error interval.
- Write the error interval and the number side by side. The number is possible only if it lies inside the interval.1.45 ≤ x < 1.55
number: 1.449 - Check the lower end. The number must be at least the lower bound (it can equal it).1.449 ≥ 1.45: false
- Check the upper end. The number must be less than the upper bound (it cannot equal it).1.449 < 1.55: true
- Decide. Both checks must be true.No, 1.449 is not a possible value.
Check. Picture it on a number line: 1.449 is outside the interval. ✓
Watch out. An open circle (a < sign) means the end value itself is not allowed.
Method. Find the error interval and see whether 254 is inside it.
- Find the unit. Nearest 10 cm, so the unit is 10 and half a unit is 5.250 ± 5
- Write the interval. 245 ≤ L < 255
- Check 254. 245 ≤ 254 is true and 254 < 255 is true.245 ≤ 254 < 255
- Answer. Yes, 254 cm is inside the interval, so it is possible.
Check. 254 rounds to 250 to the nearest 10 (decider 4). ✓
Watch out. Check both ends of the interval.
Key words. An error interval shows all the values a number could have been before rounding or truncating. The lower bound is the smallest possible value and the upper bound is the limit at the top.
Method. A number rounded to a given accuracy could have been anywhere within half a unit of the rounded value. The error interval says exactly where.
- Find the size of the last place that is shown. The number is given to 2 significant figures. The last significant figure of 24 000 is in the thousands column. One unit of that column is 1000.unit = 1000
- Halve the unit. A number rounds to the given value if it is within half a unit of it. Anything further away would round to a different value.1000 ÷ 2 = 500
- Subtract and add half a unit to find the two bounds. The lowest value that rounds to the given number is the lower bound. The upper bound is the point where it would round up to the next value.lower bound = 24 000 − 500 = 23 500
upper bound = 24 000 + 500 = 24 500 - Write the error interval with inequality signs. The lower bound 23 500 rounds up to 24 000, so N can equal it: use ≤. The upper bound 24 500 would round up to 25000, so N must be less than it: use <.23 500 ≤ N < 24 500
Check. Test the ends: 23 500 rounds to 24 000 ✓; a number just below the top, such as 24 490, rounds to 24 000 ✓; and 24 500 itself rounds up to 25 000, so it is not included. ✓
Watch out. The top end uses < (not ≤). A number exactly half way rounds up, so the upper bound itself is not a possible value.
Method. The greatest possible error is half a unit of the last decimal place.
- Find the unit. 1 decimal place means the unit is 0.1 tonne.0.1
- Halve it. The true value could be up to half a unit away.0.1 ÷ 2 = 0.05 tonnes
Check. The error interval is 4.45 ≤ m < 4.55, which is 0.05 either side of 4.5. ✓
Watch out. It is half a unit, not a whole unit.
Method. Round 3.45 to 1 decimal place and see what you get.
- Round 3.45. The last digit kept is 4 and the decider is 5, so round up.3.45 → 3.5
- Compare. The number must round to 3.4.3.5 ≠ 3.4
- Answer. No. 3.45 rounds to 3.5, not 3.4. The upper end of the interval is 3.45 but it is not included: 3.35 ≤ x < 3.45.
Check. 3.449 rounds to 3.4, so numbers just below 3.45 are allowed. ✓
Watch out. An exact half rounds up.
Method. Find the unit and the half unit, then think about whether each end is included.
- Find the half unit. Nearest whole number means the unit is 1, so half a unit is 0.5.6 ± 0.5
- Write the correct interval. 5.5 would round up to 6 (included). 6.5 would round up to 7 (not included).5.5 ≤ x < 6.5
- Describe Joe’s mistakes. Mistake 1: he used a whole unit (1) instead of half a unit. Mistake 2: he used ≤ at the top, but 6.5 rounds up to 7, so it must be <.
Check. 5.5 rounds to 6 and 6.4999 rounds to 6, but 6.5 rounds to 7. ✓
Watch out. Half a unit either side; < at the top.
Method. Work backwards: the rounded value is halfway between the two ends, and the width of the interval is one unit.
- Find the width. Upper end minus lower end.4.55 − 4.45 = 0.1
- Say what the width means. The width is one unit, so the last column is worth 0.1: tenths, which is 1 decimal place.unit = 0.1
- Find the number. The rounded number is halfway between the ends.(4.45 + 4.55) ÷ 2 = 4.5
- Write the answer. 4.5, correct to 1 decimal place
Check. 4.5 ± 0.05 gives 4.45 and 4.55. ✓
Watch out. The width of a rounding interval is a whole unit, and the number is in the middle.