Error Intervals

The lower bound is the smallest value that rounds to 36. Find the halfway point below 36.

Find the halfway point between 170 and the next multiple of 10.

Use the stated accuracy to find both endpoints. Keep the inequality signs as shown.

Use the stated accuracy to find both endpoints. Keep the inequality signs as shown.

Use the stated accuracy to find both endpoints. Keep the inequality signs as shown.

Use the stated accuracy to find both endpoints. Keep the inequality signs as shown.

Part (a)

1. The second significant digit is the 4 in the hundreds column. The rounding step is 100, with half-step 50.
2. The lower endpoint is 8400-50=8350.

Part (b)

1. The half-step is 50. The upper endpoint is 8400+50=8450, which is excluded.

Use the stated accuracy to find both endpoints. Keep the inequality signs as shown.

Part (a)

1. The last significant digit is worth 0.0001, so the half-step is 0.00005.
2. The lower endpoint is 0.0062-0.00005=0.00615.

Part (b)

1. The upper endpoint is 0.0062+0.00005=0.00625. This endpoint rounds to 0.0063, so it is excluded.

Use the stated accuracy to find both endpoints. Keep the inequality signs as shown.

Part (a)

1. Every positive number starting 5.6 stays 5.6 when later decimal digits are cut off.
2. The smallest such value is 5.6, so the lower endpoint is 5.6.

Part (b)

1. The next tenth is 5.7. Values below it can start 5.6, but 5.7 itself cannot.
2. The upper endpoint is 5.7, and it is excluded.
3. Cutting off digits keeps the interval at or above the value shown.

Use the stated accuracy to find both endpoints. Keep the inequality signs as shown.

Part (a)

1. The full value can be 3.48000\ldots, so the lower endpoint is 3.48.
2. The digits after 3.48 were cut off, not rounded, so do not subtract 0.005.

Part (b)

1. As soon as the value reaches 3.49, it no longer begins 3.48.
2. The upper endpoint is 3.49, which is excluded.

A filled circle includes its endpoint; an open circle excludes its endpoint.

1. The filled circle at 12.5 means x can equal 12.5.
2. The open circle at 13.5 means x must be less than 13.5.
3. The interval is 12.5\leq x<13.5.

Use the stated accuracy to find both endpoints. Keep the inequality signs as shown.

Part (a)

1. Half of 10 cm is 5 cm.
2. The lower endpoint is 460-5=455 cm.

Part (b)

1. The upper endpoint is 460+5=465 cm.
2. A length of exactly 465 cm would round to 470 cm, so this endpoint is excluded.

Find the interval first, then remember that pupils are counted in whole numbers.

Part (a)

1. The half-step is 5, giving 345\leq n<355.
2. The lower endpoint 345 is an integer and is included, so the least possible count is 345.

Part (b)

1. The count must be less than 355. The greatest integer below 355 is 354.
2. 354 rounds to 350, while 355 rounds to 360.

Find the two boundaries, then test each listed value. Pay attention to equality.

1. The error interval is 4.75\leq x<4.85.
2. 4.75 is included. Both 4.799 and 4.849 lie inside the interval.
3. 4.74 is too small and 4.85 is excluded. The values are 4.75, 4.799, 4.849.

Compare cutting off digits with rounding to the nearest tenth.

1. The interval 2.55\leq x<2.65 is for a number rounded to 2.6. The display has cut off the later digits, not rounded them.
2. So 2.6 itself is allowed, and every value stays below 2.7.
3. The correct interval is 2.6\leq x<2.7, as the second statement explains.

Use the centre and the width of the interval.

Part (a)

1. The midpoint of 6.25 and 6.35 is (6.25+6.35)\div2=6.3.
2. The rounded result is 6.3.

Part (b)

1. The interval has width 6.35-6.25=0.10.
2. A rounding step of 0.1 means rounding to 1 decimal place.

Leaving off a digit is not rounding. Which numbers with two decimal places begin 9.1?

1. Leaving off the hundredths digit leaves 9.1, so 9.1\leq x<9.2.
2. With two digits after the decimal point, the possible values are 9.10, 9.11, \ldots, 9.19.
3. The value 9.20 would begin 9.2, so the greatest possible value is 9.19.

Use the stated accuracy to find both endpoints. Keep the inequality signs as shown.

Part (a)

1. Half of 0.01 is 0.005.
2. The lower endpoint is 6.25-0.005=6.245.

Part (b)

1. The upper endpoint is 6.25+0.005=6.255.
2. A value of exactly 6.255 would round to 6.26, so it is excluded.

Find the whole-number range that rounds to 780, then keep only multiples of 4.

1. Rounding to 780 gives 775\leq n<785.
2. The multiples of 4 in that interval are 776,\ 780 and 784.
3. The next multiples outside the interval are 772 and 788, so the list is complete.

Distinguish the upper boundary from a greatest possible value.

Part (a)

1. The half-step is 0.05. The upper bound is 2.4+0.05=2.45.
2. The upper bound is excluded; x must be less than 2.45.

Part (b)

1. 2.4492 and 2.4498 are both greater than 2.449 and less than 2.45, so both still round to 2.4.
2. 2.4501 rounds to 2.5. The value 2.44 is smaller than Rafi’s claimed greatest value, so it does not disprove the claim.
3. These two counterexamples are not an exhaustive list of possible values. There is no greatest real value below 2.45: a value can always be chosen closer to the upper bound.

Find the error interval for the length first, then use the whole-number condition.

1. The second significant figure, 4, is in the tens column, so the length is rounded to the nearest 10 mm. Half of 10 mm is 5 mm.
2. The length l mm satisfies 635\leq l<645.
3. The greatest whole number below 645 is 644. The greatest possible length is 644 mm.