Rounding Accuracy

Rounding Basics: Whole Numbers and Decimals

• Every time you estimate the cost of a full supermarket trolley, you naturally round prices to the nearest pound to make the mental arithmetic easier.
• Rounding simplifies numbers while keeping their value close to the original, but you must follow specific rules to maintain accuracy.
• The position you are rounding to is called the rounding digit, and the number immediately to its right is the check digit (or decider).
• The "5 or More" Rule: If the check digit is $5$, $6$, $7$, $8$, or $9$, you increase the rounding digit by $1$ (round up). If it is $0$, $1$, $2$, $3$, or $4$, the rounding digit stays the same (round down).
• When rounding to the nearest whole number, you are rounding to zero decimal places; the first digit after the decimal point (the tenths column) acts as your check digit.
• For whole numbers, any digits to the right of the rounding digit are replaced with placeholder zeros to ensure the number keeps its correct overall magnitude.
• For decimals, digits to the right of the target decimal place are completely discarded, not replaced with zeros.
• However, if rounding to a specific number of decimal places results in a zero at the end, you must keep these trailing zeros to prove the required level of accuracy.

Worked Example: Rounding Numbers

Calculate $14,682$ to the nearest $1000$.

Step 1: Identify the target place value and check digit (Rule)

• The target place value (thousands column) contains the rounding digit $4$.
• The check digit to its immediate right is $6$.

Step 2: Apply the rounding rule to the value (Substitution)

• Since the check digit $6 \ge 5$, it influences the rounding digit to increase by $1$.
• The $4$ becomes a $5$, giving $15$.

Step 3: Provide the final rounded result (Answer)

• Replace the digits after the thousands column with placeholder zeros to maintain the number's magnitude.
• Final answer: $15,000$

Calculate $8.996$ to $2$ decimal places (dp).

Step 1: Identify the target place value and check digit (Rule)

• The target place value ($2$nd decimal place) contains the rounding digit $9$.
• The check digit to its immediate right is $6$.

Step 2: Apply the rounding rule to the value (Substitution)

• Since $6 \ge 5$, round the $9$ up by $1$.
• This causes a carry-over effect: $9 + 1 = 10$, which increases the tenths digit to $10$, carrying over a $1$ to the units digit ($8 + 1 = 9$).

Step 3: Provide the final rounded result (Answer)

• You must keep the trailing zeros to prove the number is accurate to $2$ decimal places; discard any further digits.
• Final answer: $9.00$

Significant Figures

• A significant figure (s.f.) is a digit that contributes to the precision of a number.
• The first significant figure is always the first non-zero digit when reading the number from left to right.
• Zeros sandwiched between non-zero digits are always significant, but leading zeros (zeros to the left of the first non-zero digit) are never significant.
• Trailing zeros in large whole numbers act as placeholders to maintain the number's size, whereas trailing zeros in decimals are used to show the exact degree of accuracy.
• By default, if an OCR exam question does not specify a level of accuracy, you should give your final answer to $3$ significant figures.
• To avoid premature rounding errors, keep your intermediate working to at least $4$ significant figures.

Worked Example: Significant Figures

Calculate $0.0073519$ to $2$ significant figures (s.f.).

Step 1: Identify the target place value and check digit (Rule)

• The first significant figure is $7$.
• The second significant figure is $3$. This is the rounding digit.
• The check digit to the right of $3$ is $5$.

Step 2: Apply the rounding rule to the value (Substitution)

• Since $5 \ge 5$, the check digit influences the $3$ to round up to $4$.

Step 3: Provide the final rounded result (Answer)

• Discard the remaining decimal digits.
• Final answer: $0.0074$

Limits of Accuracy: Error Intervals and Bounds

• Whenever a measurement is rounded, the true value is hidden within a range of possibilities called an error interval.
• The precision of the initial measurement determines this error interval because any value within a specific range would automatically round to the given number.
• To find the limits of accuracy, identify the degree of accuracy (e.g., nearest $10$, nearest $0.1$) and halve it.
• By subtracting this half-value from the rounded number, you find the lower bound (the absolute smallest value that rounds up).
• By adding this half-value to the rounded number, you find the upper bound (the boundary limit the true value must be strictly less than).
• For a truncated number (where digits are simply cut off without rounding), the given value is the lower bound, and the upper bound is exactly one full unit of precision larger.
• Error intervals are expressed using inequality notation:

$$\text{Lower Bound} \le x < \text{Upper Bound}$$

• This causal mechanism explains the notation: the lower bound uses $\le$ because that exact value rounds up to the stated number, while the upper bound uses $<$ because reaching that exact value would cause the number to round up to the next increment.

Worked Example: Determining Error Intervals

Explain the error interval for a mass, $m$, given as $47 \text{ kg}$ correct to the nearest kilogram.

Step 1: Determine the degree of accuracy and halve it.

• The accuracy is to the nearest $1 \text{ kg}$.
• Half of this unit is $0.5 \text{ kg}$.

Step 2: Calculate the lower and upper bounds.

• Lower bound: $47 - 0.5 = 46.5 \text{ kg}$. This is the minimum value that rounds to $47$.
• Upper bound: $47 + 0.5 = 47.5 \text{ kg}$. This is the limit before the number would round up to $48$.

Step 3: Write the interval using inequality notation.

• The mass could be exactly $46.5$, but must be strictly less than $47.5$.
• $46.5 \le m < 47.5$