Physics » The Nature of Physics » Accuracy, Precision and Significant Figures

# Precision and Significant Figures

## Precision of Measuring Tools and Significant Figures

An important factor in the accuracy and precision of measurements involves the precision of the measuring tool. In general, a precise measuring tool is one that can measure values in very small increments. For example, a standard ruler can measure length to the nearest millimeter, while a caliper can measure length to the nearest 0.01 millimeter. The caliper is a more precise measuring tool because it can measure extremely small differences in length. The more precise the measuring tool, the more precise and accurate the measurements can be.

When we express measured values, we can only list as many digits as we initially measured with our measuring tool. For example, if you use a standard ruler to measure the length of a stick, you may measure it to be $$36.7\;\text{cm}.$$ You could not express this value as $$36.71\;\text{cm}$$ because your measuring tool was not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement.

For example, the person measuring the length of a stick with a ruler notices that the stick length seems to be somewhere in between $$36.6\;\text{cm}$$ and $$36.7\;\text{cm},$$ and he or she must estimate the value of the last digit. Using the method of significant figures, the rule is that the last digit written down in a measurement is the first digit with some uncertainty.

In order to determine the number of significant digits in a value, start with the first measured value at the left and count the number of digits through the last digit written on the right. For example, the measured value $$36.7\;\text{cm}$$ has three digits, or significant figures. Significant figures indicate the precision of a measuring tool that was used to measure a value.

### Zeros

Special consideration is given to zeros when counting significant figures. The zeros in 0.053 are not significant, because they are only placekeepers that locate the decimal point. There are two significant figures in 0.053. The zeros in 10.053 are not placekeepers but are significant—this number has five significant figures. The zeros in 1300 may or may not be significant depending on the style of writing numbers. They could mean the number is known to the last digit, or they could be placekeepers. So 1300 could have two, three, or four significant figures. (To avoid this ambiguity, write 1300 in scientific notation.) Zeros are significant except when they serve only as placekeepers.

### Example

Determine the number of significant figures in the following measurements:

1. 0.0009
2. 15,450.0
3. 6 × 103
4. 87.990
5. 30.42

#### Solution

(a) 1; the zeros in this number are placekeepers that indicate the decimal point

(b) 6; here, the zeros indicate that a measurement was made to the 0.1 decimal point, so the zeros are significant

(c) 1; the value 103 signifies the decimal place, not the number of measured values

(d) 5; the final zero indicates that a measurement was made to the 0.001 decimal point, so it is significant

(e) 4; any zeros located in between significant figures in a number are also significant

## Significant Figures in Calculations

When combining measurements with different degrees of accuracy and precision, the number of significant digits in the final answer can be no greater than the number of significant digits in the least precise measured value. There are two different rules, one for multiplication and division and the other for addition and subtraction, as you will see below.

### 1. For multiplication and division:

The result should have the same number of significant figures as the quantity having the least significant figures entering into the calculation. For example, the area of a circle can be calculated from its radius using $$A = \pi r^2.$$ Let us see how many significant figures the area has if the radius has only two—say, $$r = 1.2\;\text{m}.$$ Then,

$$A = \pi r^2$$$$= (3.1415927…) × (1.2\;\text{m})^2$$$$= 4.5238934\;\text{m}^2$$

is what you would get using a calculator that has an eight-digit output. But because the radius has only two significant figures, it limits the calculated quantity to two significant figures or

$$A = 4.5\;\text{m}^2$$

even though $$\pi$$ is good to at least eight digits.

### 2. For addition and subtraction:

The answer can contain no more decimal places than the least precise measurement. Suppose that you buy 7.56-kg of potatoes in a grocery store as measured with a scale with precision 0.01 kg. Then you drop off 6.052-kg of potatoes at your laboratory as measured by a scale with precision 0.001 kg. Finally, you go home and add 13.7 kg of potatoes as measured by a bathroom scale with precision 0.1 kg. How many kilograms of potatoes do you now have, and how many significant figures are appropriate in the answer? The mass is found by simple addition and subtraction:

$$\begin{array}{cccccccc} & & 7 & . & 5 & 6 & & & \text{kg} \\ – & & 6 & . & 0 & 5 & 2 & & \text{kg} \\ + & 1 & 3 & . & 7 & & & & \text{kg} \\ \hline & 1 & 5 & . & 2 & 0 & 8 & & \text{kg} \\ \end{array} = 15.2\;\text{kg}$$

Next, we identify the least precise measurement: 13.7 kg. This measurement is expressed to the 0.1 decimal place, so our final answer must also be expressed to the 0.1 decimal place. Thus, the answer is rounded to the tenths place, giving us 15.2 kg