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Error Intervals

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How to write an error interval for a number that has been rounded to the nearest 10, to decimal places or to significant figures, and for a calculator answer whose last digits were left off without rounding.

What an error interval is

When a number has been rounded, you can't tell exactly what it was. If a length is 60 cm to the nearest 10 cm, it could have been 57 cm, 62.5 cm or many other lengths.

The error interval is the range of all the values the number could have been. You write it with inequalities. For that length, x cm, it is 55 \leq x < 65, which means x is 55 or more, but less than 65.

The smaller end of the interval is called the lower bound. The larger end is called the upper bound.

In the exam this is usually a 2-mark question near the end of a calculator paper. You are told how a number was rounded, such as to 1 decimal place, and asked to complete the error interval. The signs are often printed for you, and you fill in the two ends. You may instead be asked for the least or greatest possible value, or told what a calculator display began with.

Finding the two ends

First look at what the number was rounded to: the nearest 10, the nearest whole number, 1 decimal place, and so on. Call that the step. The rounded number sits in the middle of its interval, and each end is half a step away from it.

  • Nearest 10: the step is 10, so go 5 each way.
  • Nearest whole number: the step is 1, so go 0.5 each way.
  • 1 decimal place: the step is 0.1, so go 0.05 each way.
  • 2 decimal places: the step is 0.01, so go 0.005 each way.

Then write the interval with these two signs:

\text{lower bound} \leq x < \text{upper bound}

The lower bound is in the interval, so it gets \leq. The upper bound is not, so it gets <.

Worked example 1: Rounded to the nearest 10

A number n is rounded to the nearest 10. The result is 240. Write down the error interval for n.

  1. The step is 10, so each end is half of 10 away: 10 \div 2 = 5
  2. Lower bound: 240 - 5 = 235

  3. Your turn. Work out the upper bound, 240 + 5

    Upper bound: 240 + 5 = 245

  4. The lower bound gets \leq and the upper bound gets <
  5. Answer: 235 \leq n < 245

Why the top end uses <

Every number from 235 up to just below 245 rounds to 240. 235 is exactly halfway between 230 and 240, so it rounds up to 240. That is why it is in the interval.

245 is exactly halfway between 240 and 250, so it rounds up to 250. It is not in the interval, which is why the top end has < and not \leq. But 244.9, 244.99 and 244.999 all round to 240

So write the upper bound as 245, with < in front of it. Don't write 244.9 or 244.99 instead: there are always more numbers between those and 245 that still round to 240.

Error intervals for decimal places

For 1 decimal place the step is 0.1, so each end is 0.05 away. For 2 decimal places the step is 0.01, so each end is 0.005 away.

A number line helps. Mark the rounded value and the values one step either side of it. The ends of the interval are halfway between them.

Check: the bounds have one more decimal place than the rounded number. If it was rounded to 1 decimal place, the bounds have 2 decimal places.

Worked example 2: Rounded to 1 decimal place

A number x is rounded to 1 decimal place. The result is 5.3. Write down the error interval for x.

  1. One step either side of 5.3 are 5.2 and 5.4. The ends are halfway between them.
  2. Half a step is 0.1 \div 2 = 0.05, so the lower bound is 5.3 - 0.05 = 5.25

  3. Your turn. Work out the upper bound, 5.3 + 0.05

    Upper bound: 5.3 + 0.05 = 5.35

  4. Answer: 5.25 \leq x < 5.35

Worked example 3: Rounded to 2 decimal places

The length of a seed is 0.72 cm, correct to 2 decimal places. Write down the error interval for the length, s cm.

  1. 2 decimal places means the step is 0.01, so half a step is 0.005

  2. Your turn. Work out the lower bound, 0.72 - 0.005

    Lower bound: 0.72 - 0.005 = 0.715

  3. Upper bound: 0.72 + 0.005 = 0.725
  4. Check: 0.715 rounds up to 0.72, so it is in. 0.725 rounds up to 0.73, so it is left out.
  5. Answer: 0.715 \leq s < 0.725

Error intervals for significant figures

For significant figures, find the place value of the last significant figure that was kept. That place is the step.

For example, 7600 is correct to 2 significant figures. The 2nd significant figure, the 6, is in the hundreds place. So the step is 100, each end is 50 away, and the interval is 7550 \leq x < 7650

The zeros at the end of a large number only hold the place. Look at the last significant figure, not the last digit: in 7600 the last digit is a 0 in the units place, but the step is 100

Worked example 4: Rounded to 2 significant figures

The mass of a lorry is 18\,000 kg, correct to 2 significant figures. Complete the error interval for the mass, m kg.

  1. The significant figures are the 1 and the 8. The zeros after them only hold the place.
  2. The 8 is in the thousands place, so the step is 1000 and half a step is 500

  3. Your turn. Work out the lower bound, 18\,000 - 500

    Lower bound: 18\,000 - 500 = 17\,500

  4. Upper bound: 18\,000 + 500 = 18\,500
  5. Answer: 17\,500 \leq m < 18\,500

When digits are left off, without rounding

Sometimes a number is not rounded. Instead, the digits after a certain place are just left off, without rounding. A common exam question says that the answer on a calculator display began with a number, such as 4.7

If the display begins 4.7, the full number could be 4.7, 4.71, 4.756 or 4.7999. It can't be 4.8 or more, because then it would begin 4.8. It can't be less than 4.7 either.

So the interval starts at the number itself and goes one whole step up:

4.7 \leq y < 4.8

Read the question carefully. Rounded means half a step each way. Digits left off, or a display that begins with the number, means from the number up one whole step.

Worked example 5: Digits left off a calculator answer

Leon works out a positive number w on his calculator. He copies down 15.2 and leaves off the rest of the digits, without rounding. Complete the error interval for w.

  1. The digits were left off, not rounded, so w is at least 15.2. The lower bound is 15.2

  2. Your turn. The step is 0.1. Work out the upper bound, 15.2 + 0.1

    Upper bound: 15.2 + 0.1 = 15.3. A number of 15.3 or more would not begin 15.2

  3. The ends are not 15.15 and 15.25: those are the ends for rounding, not for digits left off.
  4. Answer: 15.2 \leq w < 15.3

Worked example 6: Least and greatest number of people

The number of people at a festival is 6300, correct to 2 significant figures. (a) Write down the least possible number of people. (b) Write down the greatest possible number of people.

  1. The 2nd significant figure, the 3, is in the hundreds place. So the step is 100 and half a step is 50
  2. (a) Least possible: 6300 - 50 = 6250. It rounds up to 6300, so it is possible.
  3. The upper bound is 6300 + 50 = 6350, but 6350 people would round up to 6400

  4. Your turn. (b) People are counted in whole numbers. Write down the greatest possible number of people.

    6350 is not allowed, so the greatest whole number is one less: 6349

  5. Answer: (a) 6250 (b) 6349

Common mistakes

  • Going a whole step each way. For 70 to the nearest whole number the step is 1, so the ends are 69.5 and 70.5, not 69 and 71
  • Making the upper bound a little too small. The upper bound is exactly halfway, such as 5.35, not 5.34 or 5.349
  • Using the wrong signs. The lower bound gets \leq and the upper bound gets <
  • Writing the interval back to front. The smaller number goes on the left: 5.25 \leq x < 5.35
  • Rounding when the digits were left off. If a display begins 15.2, the interval is 15.2 \leq w < 15.3, not 15.15 \leq w < 15.25
  • Using the wrong place value. Check what the number was rounded to before you halve: the nearest 10 means a step of 10, but 2 significant figures in 18\,000 means a step of 1000

Practice questions

More on this topic: Error Intervals worksheet with full solutions