Measurement Error & Bounds

Week 1, Day 2: Absolute Error, Percentage Error, and Compounding.

Made by Mr. Shen

Absolute Error (Measurement Bounds)

In mathematics, the term "Absolute Error" changes slightly depending on the situation. We must clearly define the context.

Context 1: Continuous Measurement (True Value Unknown)

When you measure a physical object (length, mass, time), no tool is perfect. Because the exact true value is unknown, we calculate the Maximum Absolute Error based on the tool's precision limits.

The Golden Rule of Measurement Bounds

The Maximum Absolute Error is always HALF of the unit of precision.

$$ \text{Max Absolute Error} = \frac{\text{Unit of Precision}}{2} $$ $$ \text{Upper Bound} = \text{Measured} + \text{Max Absolute Error} $$ $$ \text{Lower Bound} = \text{Measured} - \text{Max Absolute Error} $$

Example: Whole Numbers

A desk is measured as $15 \text{ cm}$ to the nearest cm. Find the limits of accuracy.

1. Unit of Precision: $1 \text{ cm}$

2. Max Absolute Error: $1 \div 2 = 0.5 \text{ cm}$

3. Bounds:

$$ \text{Upper Bound} = 15 + 0.5 = 15.5 \text{ cm} $$ $$ \text{Lower Bound} = 15 - 0.5 = 14.5 \text{ cm} $$

The true length $L$ lies in the inequality: $14.5 \le L < 15.5$

Example: Decimals

A bag weighs $4.2 \text{ kg}$ to the nearest 0.1 kg. Find the limits.

1. Unit of Precision: $0.1 \text{ kg}$

2. Max Absolute Error: $0.1 \div 2 = 0.05 \text{ kg}$

3. Bounds:

$$ \text{Upper Bound} = 4.2 + 0.05 = 4.25 \text{ kg} $$ $$ \text{Lower Bound} = 4.2 - 0.05 = 4.15 \text{ kg} $$

The true weight $W$ lies in: $4.15 \le W < 4.25$

Knowledge Check 1

1 Find the Upper Bound of $24 \text{ m}$ (measured to the nearest m): m
2 Find the Lower Bound of $3.2 \text{ kg}$ (measured to nearest 0.1 kg): kg

Marking Schedule:

Q1: Precision is $1$. Max absolute error is $1 \div 2 = 0.5$. Upper bound = $24 + 0.5 = 24.5$

Q2: Precision is $0.1$. Max absolute error is $0.1 \div 2 = 0.05$. Lower bound = $3.2 - 0.05 = 3.15$

Relative & Percentage Error

Absolute error tells us the total amount we are off by, but it doesn't give scale. Being off by 1 cm when measuring a pencil is a big deal; being off by 1 cm when measuring a football field is excellent. Percentage Error gives us that scale.

Which "Absolute Error" do I use?

Before calculating percentage error, you must determine which definition of Absolute Error applies to your specific problem:

  • Context 1 (Measurement): If you are only given a measured value and its precision, the true value is unknown. You must calculate the Maximum Absolute Error $= \text{Precision} \div 2$.
  • Context 2 (Estimation): If you are given an exact True Value and an Estimated Value, the Absolute Error is simply the exact mathematical difference: $|\text{True Value} - \text{Estimated Value}|$.
$$ \text{Percentage Error} = \frac{\text{Absolute Error}}{\text{True (or Measured) Value}} \times 100\% $$

Example: Context 2 (Estimation)

You estimate a crowd at $50$ people. The actual count is $40$ people. What is the percentage error?

1. Absolute Error: $|40 - 50| = 10$

2. True Value: $40$

3. Calculate:

$$ \frac{10}{40} \times 100\% = 0.25 \times 100\% = 25\% $$

Example: Context 1 (Measurement)

A line is measured as $12 \text{ cm}$ to the nearest cm. What is the maximum percentage error?

1. Max Absolute Error: nearest cm $\to \pm 0.5 \text{ cm}$

2. Measured Value: $12 \text{ cm}$

3. Calculate:

$$ \frac{0.5}{12} \times 100\% \approx 4.17\% $$

Knowledge Check 2

1 You guess $80$ jellybeans, actual is $100$. Find the % error: %
2 Measure $200 \text{ g}$ to nearest $10 \text{ g}$. Find Max % error: %

Marking Schedule:

Q1 (Context 2): Error = $|100 - 80| = 20$. Percentage = $(20 \div 100) \times 100 = 20\%$

Q2 (Context 1): Precision is $10$. Max Abs Error = $5$. Percentage = $(5 \div 200) \times 100 = 2.5\%$

Compounding Errors in Area/Volume

When you multiply measurements (to find Area or Volume), the small absolute errors compound. To find the true maximum area, you must multiply the Upper Bounds together. To find the minimum area, multiply the Lower Bounds.

Target Problem Analysis

"A square's side is measured as $1449 \pm 1 \text{ mm}$. Calculate the absolute maximum and minimum possible areas."

1. Identify the Bounds of the Length
Upper Bound = $1449 + 1 = 1450 \text{ mm}$
Lower Bound = $1449 - 1 = 1448 \text{ mm}$
2. Calculate Max Area (Multiply Upper Bounds)
$$ \text{Max Area} = 1450 \times 1450 = 2,102,500 \text{ mm}^2 $$
3. Calculate Min Area (Multiply Lower Bounds)
$$ \text{Min Area} = 1448 \times 1448 = 2,096,704 \text{ mm}^2 $$

Notice the gap! The measured area is $1449^2 = 2,099,601$. The difference between the measured area and the Max area is nearly $3000 \text{ mm}^2$!

Knowledge Check 3

A rectangle is measured as $10 \text{ cm}$ by $8 \text{ cm}$ (both to the nearest cm).

1 Calculate the Maximum Area: cm²
2 Calculate the Minimum Area: cm²

Marking Schedule:

Upper Bounds: $10.5 \text{ cm}$ and $8.5 \text{ cm}$.
Lower Bounds: $9.5 \text{ cm}$ and $7.5 \text{ cm}$.

Q1 (Max Area): $10.5 \times 8.5 = 89.25 \text{ cm}^2$

Q2 (Min Area): $9.5 \times 7.5 = 71.25 \text{ cm}^2$

Synthesis Task

Combine all skills. Complete 18 mixed-context questions. You have 2 attempts. (Round to 2 d.p. if needed)

Section A: Limits of Accuracy (Bounds)

Section B: Percentage Error

Section C: Compounding Errors

Section D: Rich Context and Harder Multi-Step Tasks

Student Mastery Report

Total Score
/18
Sec A (Limits):
/6
Sec B (% Error):
/4
Sec C (Compounding):
/4
Sec D (Hard Context):
/4

Take a screenshot of this report to submit to Mr. Shen.

Full Marking Schedule & Step-by-Step Solutions