1.07 Significant figures and accuracy in measurement

Significant figures

The Liverpool friendly soccer game at the MCG in Melbourne, Australia in 2013 had a record setting crowd.  It was the second largest on record reaching $95446$95446.  Soon after the game, a newspaper article had reported a crowd size of $95000$95000.  

Is the newspaper right?  Is $95000$95000 accurately reported?  It actually is if we consider significant figures.  In fact, $95400$95400 and $100000$100000 are also correct depending on the level of accuracy we want to use.

Aren't all numbers significant? Well yes, all the digits in numbers have a particular role to play. They tell us the value of the number, the size of the number, or individual place values. Zeros hold the place value in numbers so that we can read and understand them.  

The concept of significant figures is about representing a number with a certain level of accuracy.  

Significant figures and rounding

Digits that are significant are:

• all nonzero digits
• zeros appearing between two nonzero digits (holding the place value)
• trailing zeros at the end of a number containing a decimal point
• all digits in a number written using scientific notation ($a\times10^n$a×10n)

Here is a visual of the first three points:

So let's go back to the crowd size at the soccer game. We can round the crowd size of $95446$95446 to a variety of significant figures depending on the level of accuracy we are interested in.

Round to	Result	Process
$1$1 significant figure	$100000$100000	the first significant figure in the number is $9$9, so we are rounding to the nearest ten thousand
$2$2 significant figures	$95000$95000	the second significant figure in the number is $5$5, so we are rounding to the nearest thousand
$3$3 significant figures	$95400$95400	the third significant figure in the number is $4$4, so we are rounding to the nearest hundred
$4$4 significant figures	$95450$95450	the fourth significant figure in the number is $4$4, so we are rounding to the nearest ten
$5$5 significant figures	$95446$95446	the fifth significant figure in the number is $6$6, so we are rounding to the nearest one

So for numbers larger than $0$0, using significant figures is about rounding

Worked examples

Question 1

The width of a human hair is known to be about $0.023$0.023 cm.  Round to each significant figure.

Think: First we need to identify how many significant figures are in the number $0.023$0.023. The $2$2 and the $3$3 are significant figures but the zeros are not because they are not located between two non-zero numbers and they are not trailing at the end of the number.

Do: Since there are $2$2 significant figures we will round to each.

The first significant figure is $2$2, so we are rounding to the nearest hundredth. Rounding to $1$1 significant figure we get $0.02$0.02 as a result.

The second significant figure is $3$3, so we are rounding to the nearest thousandth.  Rounding to $2$2 significant figures we get 0.023 as a result.

Reflect: See how in this case the leading zeros do not count.

Question 2

Write the following numbers to the indicated level of significant figures

a) Round $10432$10432 to $3$3 significant figures.

Think: Starting from the left we find that the third significant figure is $4$4, so we need to round to the hundreds place.

Do: $10432$10432 to $3$3 significant figures = $10400$10400   

Reflect: The zero between the $1$1 and the $4$4 is counted as significant.

b) Round $1.040052$1.040052 to $3$3 significant figures.

Think: Starting from the left we find that the third significant figure is $4$4, so we need to round to the hundredths place.

Do: $1.040052$1.040052 to $3$3 significant figures = $1.04$1.04     

Reflect: The zero between the $1$1 and $4$4 is counted as significant.

c) Round $6.53126\times10^7$6.53126×107 to $4$4 significant figures.

Think: Starting from the left we find that the fourth significant figure is $1$1, so we need to round to the thousandths place.

Do: $6.53126\times10^7$6.53126×107 to $4$4 significant figures = $6.531\times10^7$6.531×107 

Reflect: Notice we still keep the $10^7$107 but we only need $4$4 of the digits.

d) Round $6.00002\times10^8$6.00002×108 to $3$3 significant figures.

Think: Starting from the left we find that the second $0$0 is the third significant figure, so we need to round to the hundredths place.

Do: $6.00002\times10^8$6.00002×108 to $3$3 significant figures = $6.00\times10^8$6.00×108         

Reflect: Notice we have to write the zeros here - all the digits count as significant in scientific notation, even the zeros.

Practice questions

QUESTION 3

QUESTION 4

QUESTION 5

Accuracy in measurement (absolute error)

The aim of a measurement is to obtain the true value of a quantity, be it the height of a tree, the temperature of a room, the mass of a rock and so on. We can carefully design a measurement procedure so that the measured value is as close as possible to the true value, but there will always be some difference between the two.

The difference between the measured value and the true value is called the error:

$\text{Error }=\text{Measured value }-\text{True value }$Error =Measured value −True value

Notice that the error may be positive or negative depending on whether the measured value overshoots or undershoots the true value.

In general the true value and the error are both unknown to us. Instead we can use the measured value, together with the uncertainty of the measurement, to produce an interval within which the true value will likely lie. The lower value of the interval is called lower bound, and the higher value is the upper bound.

 

One useful value is the absolute error and is defined as half the distance between the lower bound and upper bound. Usually the measured value is centered between the lower bound and upper bound, so we can also find the absolute error by subtracting the measured value from the upper bound.

Measurements taken from any measuring device have an accuracy of $\pm\frac{1}{2}$±12​ of the smallest unit measured. Rounded measurements also have an accuracy of $\pm\frac{1}{2}$±12​ of the unit the measurement is being rounded to. So someone who has a height of $63$63 inches using a ruler, will have a lower bound of $62.5$62.5 cm and an upper bound of $63.5$63.5 cm. Conventionally when we use a measuring device, or are rounding, we naturally round up in the case that we have a tie. In other words, the lower bound is always included in the uncertainty range while the upper bound is always excluded from the uncertainty range.

Worked example

question 6

A person's height is measured to be $1.68$1.68 meters rounded to the nearest centimeter.

a) What is the lower bound of their height?

Think: What heights would be rounded up to $1.68$1.68 m? What is the lower bound of these values?

Do: The lower bound is $1.675$1.675 m, since any height $h$h in the range $1.675\le h<1.68$1.675≤h<1.68 will round up to $1.68$1.68.

Reflect: The difference between the measured height and the lower bound is $0.005$0.005 m, or $0.5$0.5 cm, which is half of the smallest unit of the measurement. This is the absolute error.

b) What is the upper bound of their height?

Think: What heights would be rounded down to $1.68$1.68 m? What is the upper bound of these values?

Do: The upper bound is $1.685$1.685 m, since any height $h$h in the range $1.681.68<h<1.685 will round down to $1.68$1.68.

Reflect: Even though $1.685$1.685 is the upper bound of the person's height, it does not belong to the uncertainty range since $1.685$1.685 rounds up to $1.69$1.69.

Question 7

The population of a species of animals is $95000$95000 to the nearest $1000$1000 animals.

a) What is the minimum possible population?

Think: What possible values round up to $95000$95000 to the nearest $1000$1000 animals?

Do: The lower bound is $94500$94500, since any population $p$p in the range $94500\le p<95000$94500≤p<95000 will round up to $95000$95000. Since the lower bound is always in the uncertainty region, the minimum possible population is $95000$95000 as well.

b) What is the maximum possible population?

Think: What possible values round down to $95000$95000 to the nearest $1000$1000 animals?

Do: The upper bound is $95500$95500, since any population $p$p in the range $9500095000<p<95500 will round down to $95000$95000. A calculation of $95500$95500 will round up to $96000$96000. However, since population is discrete, $95499$95499 is included in the uncertainty range and is the largest possible population. So the maximum possible population is $95499$95499.

Reflect: In this case, the upper bound is not necessarily the maximum possible value being measured.

Practice questions

Question 8

Question 9

question 10