Limits of Accuracy

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What Are Limits of Accuracy?

When we measure something in real life, we almost never get a perfectly exact answer. We round our measurements to a certain level of accuracy. For example, if you measure the length of a desk and say it's 120 cm to the nearest centimetre, the actual length isn't exactly 120 cm – it's somewhere close to 120 cm, but we've rounded it.

Limits of accuracy (also called bounds) tell us the range of possible values that the actual measurement could be.

There are two bounds for any rounded measurement:

• Lower bound – the smallest value the actual measurement could be
• Upper bound – the largest value the actual measurement could be

Understanding Bounds Through an Example

Imagine you measure a length as 8 metres to the nearest metre.

The actual length could be anything that rounds to 8 metres. Let's think about what values round to 8:

• 7.5 m would round up to 8 m
• 7.8 m would round up to 8 m
• 8.0 m stays as 8 m
• 8.4 m would round down to 8 m
• 8.5 m would round up to 9 m (not 8!)

So the actual length is at least 7.5 m but less than 8.5 m.

• Lower bound = 7.5 m
• Upper bound = 8.5 m

We write this as: 7.5 m ≤ actual length < 8.5 m

Notice the symbols: ≤ means "less than or equal to" and < means "less than" (but not equal to). The actual length can equal 7.5, but it cannot equal 8.5 (because 8.5 would round to 9, not 8).

How to Find Bounds – The Rule

Here's a simple rule to find bounds:

1. Work out what you're rounding to (is it the nearest 10? nearest whole number? nearest 0.1? 1 decimal place? 2 significant figures?)

2. Find the "gap" – this is half of the unit you're rounding to

   • Rounding to nearest 10 → gap is 5
   • Rounding to nearest 1 → gap is 0.5
   • Rounding to nearest 0.1 → gap is 0.05
   • Rounding to 1 decimal place → gap is 0.05
   • Rounding to nearest 0.01 → gap is 0.005

3. Calculate the bounds:

   • Lower bound = rounded value − gap
   • Upper bound = rounded value + gap

Worked Examples: Finding Bounds

Example 1: A length is measured as 15 cm to the nearest centimetre. Find the upper and lower bounds.

Solution:

• Rounding to nearest 1 cm
• Gap = 0.5 cm
• Lower bound = 15 − 0.5 = 14.5 cm
• Upper bound = 15 + 0.5 = 15.5 cm
• We can write: 14.5 cm ≤ actual length < 15.5 cm

Example 2: A mass is 7.3 kg to 1 decimal place. Find the upper and lower bounds.

Solution:

• Rounding to 1 decimal place (same as nearest 0.1)
• Gap = 0.05 kg
• Lower bound = 7.3 − 0.05 = 7.25 kg
• Upper bound = 7.3 + 0.05 = 7.35 kg
• We can write: 7.25 kg ≤ actual mass < 7.35 kg

Example 3: A distance is 400 m to the nearest 10 metres. Find the upper and lower bounds.

Solution:

• Rounding to nearest 10 m
• Gap = 5 m
• Lower bound = 400 − 5 = 395 m
• Upper bound = 400 + 5 = 405 m
• We can write: 395 m ≤ actual distance < 405 m

Example 4: A time is 8.0 seconds to 2 significant figures. Find the upper and lower bounds.

Solution:

• The number 8.0 has 2 significant figures
• We're rounding to the nearest 0.1 (first decimal place)
• Gap = 0.05 seconds
• Lower bound = 8.0 − 0.05 = 7.95 seconds
• Upper bound = 8.0 + 0.05 = 8.05 seconds
• We can write: 7.95 s ≤ actual time < 8.05 s

Worked Examples: Bounds in Calculations

Example 5: A rectangle has length 12 cm and width 8 cm, both measured to the nearest centimetre. Find the upper bound of the perimeter.

Solution:

Step 1: Find bounds for length

• Length = 12 cm (to nearest cm)
• Lower bound = 11.5 cm
• Upper bound = 12.5 cm

Step 2: Find bounds for width

• Width = 8 cm (to nearest cm)
• Lower bound = 7.5 cm
• Upper bound = 8.5 cm

Step 3: Calculate upper bound of perimeter

• Perimeter = 2 × length + 2 × width
• For maximum perimeter, use upper bounds
• Upper bound = 2 × 12.5 + 2 × 8.5
• Upper bound = 25 + 17
• Upper bound = 42 cm

Example 6: A rectangle has length 15 cm and width 9 cm, both measured to the nearest centimetre. Find the upper bound of the area.

Solution:

Step 1: Find bounds for length

• Length = 15 cm (to nearest cm)
• Upper bound = 15.5 cm

Step 2: Find bounds for width

• Width = 9 cm (to nearest cm)
• Upper bound = 9.5 cm

Step 3: Calculate upper bound of area

• Area = length × width
• For maximum area, use upper bounds
• Upper bound = 15.5 × 9.5
• Upper bound = 147.25 cm²

Example 7: A car travels 240 km (to the nearest 10 km) in 3.5 hours (to 1 decimal place). Find the lower bound of the average speed.

Solution:

Step 1: Find bounds for distance

• Distance = 240 km (to nearest 10 km)
• Lower bound = 235 km

Step 2: Find bounds for time

• Time = 3.5 hours (to 1 d.p.)
• Upper bound = 3.55 hours

Step 3: Calculate lower bound of speed

• Speed = distance ÷ time
• For minimum speed, use lower bound of distance and upper bound of time
• Lower bound = 235 ÷ 3.55
• Lower bound = 66.2 km/h (to 1 d.p.)

Why use lower distance and upper time? Because speed = distance ÷ time. To make this as small as possible, we want to divide a small number by a large number. So we use the smallest possible distance and the largest possible time.

Quick Summary of Rules

(UB = upper bound, LB = lower bound)