Unlocking Accuracy: Bounds, Error & Percentage Error

Welcome to Lesson 3 of our “First Unit in Math AI” series! In our journey through accuracy, we’ve already covered the basics of decimal places, significant figures, and rounding. Today, we’re taking a vital leap forward to truly unlock the deeper truths of accuracy. We’ll dive into three absolutely essential concepts: Lower and Upper Bounds of Accuracy, Measurement Error ($V_A​−V_E$​), and Percentage Error. These ideas are fundamental for your Math AI course, as they can appear as standalone questions or as critical components of larger problems. Our goal is to make you super comfortable with all of them, so let’s get started!

1. Lower and Upper Bounds of Accuracy

When we measure something or round a number, there’s always a tiny bit of uncertainty. The lower bound is the smallest possible value the original measurement could have been and still round to our given number. Conversely, the upper bound is the largest possible value the original measurement could have been. The key to finding these bounds lies in understanding the degree of accuracy and taking half of that value.

Definitions:

  • Lower Bound (LB): The smallest value a number could have been before rounding. Calculated by subtracting half the degree of accuracy.
  • Upper Bound (UB): The largest value a number could have been before rounding. Calculated by adding half the degree of accuracy. (Note: The actual value is less than the upper bound).
  • Degree of Accuracy: The smallest unit to which a number has been rounded (e.g., nearest whole number, 1 decimal place, nearest 10).

Example 1: Rounded to 1 Decimal Place

A length is measured as 12.3 cm, rounded to 1 decimal place. Find its lower and upper bounds.

Solution:

The degree of accuracy is 0.1.

Half the degree of accuracy is $\frac{0.1}{2} = 0.05$.

    • Lower Bound: 12.3− 0.05 = 12.25 cm
    • Upper Bound: 12.3 + 0.05 = 12.35 cm
    • So, the actual length (L) is: 12.25 ≤ L <  12.35 cm.

Remember: L can’t be 12.35 that’s why the inequality is less than strictly (<), which also means exclusive, since 12.35 rounds to 12.4 and not 12.3

Example 2: Rounded to the Nearest 10

The mass of a bag of rice is 500 g, rounded to the nearest 10 g. Find its lower and upper bounds.

Solution: The degree of accuracy is 10. Half the degree of accuracy is $\frac{10}{2} = 5$.

    • Lower Bound: 500 − 5 = 495 g
    • Upper Bound: 500 + 5 = 505 g.
    • So, the actual mass (M) is: 495 ≤ M < 505 g.

Remember: M can’t be 505 that’s why the inequality is less than strictly (<), which also means exclusive, since 505 rounds to 510 and not 500

Measurement Error $(V_A​−V_E​)$

Measurement error is a fundamental concept for understanding the precision of our data. It’s simply the difference between the actual (true) value and the estimated (measured) value. This error can be positive or negative, indicating if your measurement was an underestimate or an overestimate.

Note: in the formula booklet, actual is called exact and estimated is called approximated.

Definition:

Measurement Error = $V_A​−V_E​$ Where:

    • $V_A​$: Actual Value (the true value)
    • $V_E$​: Estimated Value (the measured or observed value)

Example 1: Underestimated Measurement

A student measures the length of a table to be 150 cm ($V_E$). The actual length of the table is 152 cm ($V_A$). Calculate the measurement error.

Solution:

Measurement Error =  $V_A − V_E$ = 152 − 150 = 2 cm. (A positive error means the measurement was an underestimate.)

Example 2: Overestimated Measurement

A balance reads 25.5 g ($V_E$) for a substance with a known mass of 25.0 g ($V_A$). Calculate the measurement error.

Solution:

Measurement Error = $V_A − V_E$ = 25.0 − 25.5 = −0.5 g. (A negative error means the measurement was an overestimate.)

Percentage Error

While simple measurement error tells us the direct difference, percentage error provides a more meaningful context. It expresses the error as a percentage of the actual value, giving us a relative measure of accuracy.

Definition:

Percentage Error = $\frac{ | \;V_A \;​− \;V_E \;| }{V_A} ​× 100\%$

    • Important: We use the absolute value (|…|) of the difference between $V_A$​ and $V_E$​ to ensure the percentage error is always positive, representing the magnitude of the error.

Example 1: Table Length

Using the table example: Actual length ($V_A$) = 152 cm, Measured length ($V_E$) = 150 cm. Calculate the percentage error.

Solution:

Percentage Error =$\frac{|152−150|}{152}×100\%=\frac{|2|}{152}×100\%=\frac{2}{152}×100%≈1.32%$

Exam Tip: Always keep your final answer exact or round it to 3 s.f. unless the question specifies otherwise.

Example 2: Beads in a Jar

A student estimates a jar contains 200 beads ($V_E$). The actual number of beads is 210 ($V_A$). Calculate the percentage error.

Solution:

Percentage Error =$\frac{|210−200|}{210}×100\%=\frac{|10|}{210}×100\%=\frac{10}{210}×100%≈4.76%$

Exam Tip: Always keep your final answer exact or round it to 3 s.f. unless the question specifies otherwise.

GDC Calculations for Accuracy

As a math Application and Interpretation student your Graphic Display Calculator (GDC) is an indispensable tool for these calculations. While specific button presses may vary by model, practicing these on your own GDC will significantly boost your efficiency during exams.

    • For Bounds: Input the simple addition/subtraction.
    • For Measurement Error: Directly subtract $V_E$​ from $V_A$​.
    • For Percentage Error: Utilize the formula from the formula booklet, ensure correct use of parentheses, and remember the absolute value function (usually ABS or |x|).
    • Always double-check your input, especially parentheses and absolute value signs, to avoid calculation errors!

Important Note: The percentage error formula in the formula booklet uses the variables names “exact” instead of “actual” AND “approximate” instead of “estimate”.

Quiz Time! Test Your Understanding

Ready to see how much you’ve absorbed? Try these quick quiz questions. Work them out, then click to reveal the solutions!

Question 1:

A weight is recorded as 15.0 kg to 1 decimal place. What are its lower and upper bounds?

Degree of accuracy: 0.1 kg.

Half degree of accuracy: 0.05 kg.

Lower Bound: 15.0 − 0.05 = 14.95 kg

Upper Bound: 15.0 + 0.05 = 15.05 kg

So, 14.95 ≤ Weight < 15.05 kg.

Question 2:

The expected temperature of a solution is 25.0∘C ($V_A​$), but the thermometer reads 25.3∘C ($V_E​$). Calculate the percentage error.

Percentage Error = $\frac{|V_A – V_E|}{V_A}\times 100\%$

Percentage Error =

$\frac{|25.0-25.3|}{25.0}\times 100\% = \frac{|-0.3|}{25.0}\times 100\% = \frac{0.3}{25.0}\times 100\% = 1.2\%$

External Exam Question: Put Your Knowledge to the Test!

Here’s an external exam question that combines the concepts we’ve explored. Take a moment to read it carefully, identify the keywords, and decide which formulas you’ll need. Try to solve it before looking at the step-by-step solution below!

Question:

A rectangular room has a measured length of 8.5 m (to the nearest 0.1 m) and a measured width of 6.2 m (to the nearest 0.1 m).

a) Calculate the lower bound for the area of the room.

b) If the actual length of the room is 8.48 m and the actual width is 6.19 m, calculate the percentage error for the measured area.

Part a) Calculate the lower bound for the area of the room.

Degree of accuracy for length and width is 0.1 m.

Half degree of accuracy is 0.05 m.

Lower Bound for Length (LBL): 8.5 − 0.05 = 8.45 m

Lower Bound for Width (LBW): 6.2 −  0.05 = 6.15 m

Lower Bound for Area = LBL × LBW = 8.45 × 6.15 = 51.9675 $m^2$. (This is an exact answer)

Part b) If the actual length of the room is 8.48 m and the actual width is 6.19 m, calculate the percentage error for the measured area.

First, find the Measured Area ($V_E$ ): 8.5 × 6.2 m = 52.7 $m^2$.

Next, find the Actual Area ($V_A$): 8.48 m × 6.19 m = 52.5712 $m^2$.

Now, calculate the Percentage Error so

Percentage Error = $\frac{|V_A − V_E|}{V_A}×100\%$

Percentage Error = $\frac{|52.5712 − 52.7|}{52.5712}×100\%$

Percentage Error = $\frac{|−0.1288|}{52.5712}×100%$

Percentage Error = $\frac{0.1288}{52.5712} × 100\% ≈ 0.245\%$

Very Important Note: Don’t ever round an intermediate answers then use that answer to find a final answer. In most cases this will lead to a wrong final answer and you’ll lose the accuracy mark.

In case you need to use an intermediate answer in a formula, use the full answer on the calculator or use at least 5 to 6 decimal places to make sure your final answer is correct and accurate.

Extra Practice Problems: Sharpen Your Skills!

Want to really master lower/upper bounds, measurement error, and percentage error? Work through these additional practice problems! Use your GDC and your knowledge, then check your answers below.

Problem 1:

A packet of sugar is labelled as 1 kg (accurate to the nearest 0.1 kg). What are the lower and upper bounds of its mass?

Degree of accuracy: 0.1 kg.

Half degree of accuracy: 0.05 kg.

Lower Bound: 1 − 0.05 = 0.95 kg

Upper Bound: 1 + 0.05 = 1.05 kg

So, 0.95 ≤ Mass < 1.05 kg.

Problem 2:

A cyclist estimates their journey will take 30 minutes ($V_E$​). The actual time taken was 28.5 minutes ($V_A$​). Calculate:

a) The measurement error.

b) The percentage error.

a) Measurement Error = $V_A​−V_E​$ = 28.5 − 30 = −1.5 minutes.

b) Percentage Error = $\frac{|28.5−1.5|}{28.5} \times 100\% ≈ 5.26\%$ (3 s.f.)

Problem 3: (Challenge)

A circle has a measured circumference of 25.1 cm, given to 3 significant figures. Find the lower and upper bounds for its circumference.

The measurement 25.1 cm is accurate to 1 decimal place (since it’s given to 3 s.f. and has 1 d.p. after the whole number, its precision is to the nearest 0.1 cm).

Degree of accuracy: 0.1 cm.

Half degree of accuracy: 0.05 cm.

Lower Bound: 25.1 − 0.05 = 25.05 cm

Upper Bound: 25.1 + 0.05 = 25.15 cm

So, 25.05 ≤ Circumference < 25.15 cm.

Degree of Accuracy

Very Important Note: 

If a number is given correct to to a certain precision then its degree of accuracy is the place value of that accuracy.

For example:

    • 347000 is correct to 3 s.f. (precision) Now, 7 is the third significant figure (digit representing this precision) and its place value is 1000 so 347000 has a degree of accuracy 1000.
    •  1254.93 is correct to 6 s.f. (precision) Now, 3 is the 6th significant figure (digit representing this precision) and its place value is 0.01 since it comes after a decimal point so 1254.93 has a degree of accuracy 0.01
    • 12.0034 is correct to 4 d.p. (precision). Now, 4 is the fourth decimal place and its place value is 0.0001 so 12.0034 has a degree of accuracy 0.001

Thus, to find the degree of accuray of a number, do the following:

  • Identify the precision given in the question.
  • Find the digit that represents that precision.
  • Degree of accuracy = place value of that digit.

Conclusion & Your Next Steps!

You’ve successfully navigated the important world of accuracy bounds, measurement error, and percentage error! These concepts are crucial stepping stones for more advanced topics in Math AI and will strengthen your overall numerical reasoning. Remember, understanding these nuances is key to reliable data analysis and problem-solving. To continue your learning journey and keep building your Math AI skills, make sure to:

Finally, if you have any comments or have anything to tell us, please feel free to add a comment below, or on our video, or just contact us by email via the Contact Us page.

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