Approximation and Errors3

3.1 Significant Figures

3.2 Scientific Notation

3.3 Approximation and Errors

Case Study

Chapter Summary

P. 2

Suppose that there are actually 116 paper clips on the plate.

Case Study

John’s estimate Actual number 120 116

Did John or Linda make a better estimation?

I guess there are 120 paper clips. What’s your guess, Linda?

I guess there are 110 clips, John.

We need to find the difference between each estimate and the actual number.

The smaller the difference is, the better is the estimate.

4Actual number Linda’s estimate 116 110

6

Since 4 is less than 6, John made a better estimation.

P. 3

3.1 Significant Figures

A. Basic Concepts

The population of Hong Kong at the end of 2007 is 7 000 000. (correct to the nearest million) 6 950 000. (correct to the nearest ten thousand) 6 953 000. (correct to the nearest thousand)

In most cases, the far left non-zero digit, having the largest place value in a number, is the first significant figure.

Subsequent important digits are called the second significant figure, the third significant figure, and so on.

The important digits of a number are called significant figures.

7 6 95 6 953

This is also the most important figure.

P. 4

1 0 4 9

1st significant figure 2nd significant figure 3rd significant figure 4th significant figure

For all numbers, any zeros between 2 non-zero digits are significant figures.

For example, the number 1049 has 4 significant figures.

3.1 Significant Figures

A. Basic Concepts

We also study significant figures in decimals.For example, the

decimal 5.307 has 4 significant figures.

P. 5

3.1 Significant Figures

A. Basic Concepts

0. 0 3 8 0

1st significant figure 2nd significant figure 3rd significant figure

For all decimals, any zeros after the last non-zero digit are significant figures.

For decimals less than 1, all zeros in front of the first non-zero digit are not significant figures. These zeros are called place holders.

For example, the number 0.0380 has only 3 significant figures.

P. 6

44008

Number of Significant FiguresNumbers

The following table shows some examples. All the significant figures are marked with green colour.

3.1 Significant Figures

A. Basic Concepts

3900 (correct to the nearest integer) has 4 significant figures. But 3900 (correct to the nearest ten) has 3 significant figures.

46.092

10.7

30.002 04

49.000

30.130

P. 7

Using the technique of rounding off for estimation,the number 47 054 can be expressed as:(a) 47 100 (cor. to the nearest hundred)

(b) 47 000 (cor. to the nearest thousand)

(c) 50 000 (cor. to the nearest ten thousand)

Alternatively, we can round off a number to a certain number of significant figures.

3.1 Significant Figures

B. Round off to the Required Significant Figures

We can express 47 054 as:(a) 47 100 (cor. to 3 sig. fig.)

(b) 47 000 (cor. to 2 sig. fig.)

(c) 50 000 (cor. to 1 sig. fig.)

The phrase ‘cor. to 3 sig. fig.’ is the short form of ‘correct to 3 significant figures’.

P. 8

3.1 Significant Figures

B. Round off to the Required Significant Figures

The following steps show how to round off 51 672, correct to 3 significant figures.

Step 1: Determine which digit is to be rounded off.51 672

3rd significant figure

Step 2: Round off according to the next significant figure.51 672 51 700 (cor. to 3 sig. fig.)

Since 7 is larger than 5, we add 1to the 3rd significant figure.

P. 9

Round off 472 780 correct to(a) 2 significant figures,(b) 3 significant figures,(c) 4 significant figures.

(a) 470 000 (cor. to 2 sig. fig.)

(b) 473 000 (cor. to 3 sig. fig.)

(c) 472 800 (cor. to 4 sig. fig.)

3.1 Significant Figures

B. Round off to the Required Significant Figures

Example 3.1T

Solution:

P. 10

3.1 Significant Figures

B. Round off to the Required Significant Figures

Example 3.2TRound off 0.300 649 correct to(a) 1 significant figure,(b) 2 significant figures,(c) 5 significant figures.

Solution:(a) 0.3 (cor. to 1 sig. fig.)

(b) 0.30 (cor. to 2 sig. fig.)

(c) 0.300 65 (cor. to 5 sig. fig.)

P. 11

3.1 Significant Figures

B. Round off to the Required Significant Figures

Example 3.3TRound off 9995 correct to(a) 1 significant figure,(b) 2 significant figures,(c) 3 significant figures.

Solution:(a) 10 000 (cor. to 1 sig. fig.)

(b) 10 000 (cor. to 2 sig. fig.)

(c) 10 000 (cor. to 3 sig. fig.)

P. 12

The total price of 120 oranges is $250. Round off the average price of each orange correct to 2 significant figures.

3.1 Significant Figures

B. Round off to the Required Significant Figures

Example 3.4T

The average price of each orange $(250 120) $2.0833

Solution:

$2.1 (cor. to 2 sig. fig.)

P. 13

3.2 Scientific Notation

A. Introduction

Our body produces about 173 000 000 000 red blood cells each day. The mass of 1 water molecule is about 0.000 000 000 000 000 000 000 03 g.

It is convenient to rewrite these numbers as follows:

173 000 000 000 1.73 100 000 000 000 1.73 1011

This kind of representation of numbers is called scientific notation.

In science, we often deal with numbers which are very large or small:

0.000 000 000 000 000 000 000 03 g 3 1023 g

A positive number is expressed in scientific notation if it is in the form a 10n, where 1 a 10 and n is an integer.

P. 14

Express each of the following numbers in scientific notation.(a) 22 000 (b) 0.000 000 7(c) 95 000 104 (d) 0.008 102

3.2 Scientific Notation

A. Introduction

Example 3.5T

Solution:(a) 22 000 2.2 10 000

(b) 0.000 000 7 7 0.000 000 1

(c) 95 000 104 (9.5 104) 104

(d) 0.008 102 (8 103) 102

2.2 104

7 107

9.5 108

8 105

Move the decimal point 4 places to the leftand then times 104

Move the decimal point 7 places to the rightand then times 107

am an am n

P. 15

3.2 Scientific Notation

A. Introduction

Example 3.6TConvert the following numbers into integers or decimals.(a) 1.002 107

(b) 6 105

Solution:(a) 1.002 107 1.002 10 000 000

(b) 6 105 6 0.000 01

10 020 000

0.000 06

P. 16

4

66

105.1

106

00015.0

106 (a)

3.2 Scientific Notation

A. Introduction

Example 3.7TWithout using a calculator, evaluate each of the following expressions. Express the answers in scientific notation.

(a) (b) (4 103)(3.3 105)00015.0

106 6

Solution:

)4( 6105.1

6

2104

5 353 103.34)103.3)(104( (b) 8102.13 91032.1

P. 17

3.2 Scientific Notation

A. Introduction

In calculator, we should key in ‘1.32 9’ to represent 1.32 109.EXP

The calculator would display ‘1.32E9’ before execution.

If we key in ‘4 11 ’, then the calculator would display

‘4 10 11’, which means 4 1011.

EXP EXE

Students should note that the key-in sequences of different calculators may be different.

P. 18

B. Applications of Scientific Notation

For example:1 228 550 000 m3 of water was consumed in Hong Kong in 2005 2006.

1 228 550 000 m3 1 230 000 000 m3 (cor. to 3 sig. fig.)

1.23 109 m3

3.2 Scientific Notation

When presenting estimates, we can express numbers in scientific notation and round off the value correct to a certain number of significant figures.

P. 19

Example 3.8T

B. Applications of Scientific Notation

3.2 Scientific Notation

36 084 people were injured in road accidents last year. Round off the number correct to 3 significant figures and express the answer in scientific notation.

Solution:36 084 36 100 (cor. to 3 sig. fig.)

3.61 104

P. 20

11

10

101.3

1035.1

11

11

101.3

105.13

Example 3.9T

B. Applications of Scientific Notation

3.2 Scientific Notation

The atomic radii of a helium atom and a gold atom are 3.1 1011 m and 1.35 1010 m respectively. How many times is the atomic radius of a gold atom to that of a helium atom? Give the answer correct to 2 significant figures.

Solution:The required ratio

4.4 (cor. to 2 sig. fig.)

P. 21

3.3 Approximation and Errors

They are only approximations.

Hence, every measurement or estimation has errors.

In Book 1A Chapter 2, we learnt that no measurements give exact values.

P. 22

A. Absolute Error

3.3 Approximation and Errors

The difference between a measured value (or an estimated value) and the actual value is called the absolute error.

The absolute error is always positive.

If the actual value is larger than a measured value, then

absolute error actual value measured value

If a measured value is larger than the actual value, then

absolute error measured value actual value

Example:A story book has 117 pages. Sara guessed that the book has 100 pages. ∴ The absolute error of Sara’s estimation is 17 pages.

P. 23

Example 3.10T

A. Absolute Error

3.3 Approximation and Errors

The actual weight of a pack of potato chips is 183.4 g. Find the absolute error if Julie corrects the weight to 1 significant figure.

Solution:183.4 g 200 g (cor. to 1 sig. fig.)

The absolute error (200 183.4) g 16.6

g

P. 24

Therefore, errors of measurements are related to the scale intervals of the tools.

In the figure, the scale interval of the ruler is 1 cm.

B. Maximum Absolute Error

3.3 Approximation and Errors

Measuring tools with smaller scale intervals can give estimations with higher accuracy.

∴ The minimum and the maximum actual lengths are 14.5 cm and 15.5 cm respectively.

The measured length is 15 cm, correct to the nearest cm.

The absolute error of this measurement does not exceed 0.5 cm. This figure is called the maximum absolute error.

P. 25

Lower limit ( 14.5 cm)Minimum possible value of measurement

Upper limit ( 15.5 cm)Maximum possible value of measurement

They can be found by the following formulas.

B. Maximum Absolute Error

3.3 Approximation and Errors

maximum absolute error scale interval of the measuring tool 2

1

In general,

In the figure, the scale interval of the ruler is 1 cm.

Lower limit Measured value Maximum absolute errorUpper limit Measured value Maximum absolute error

P. 26

Example 3.11T

B. Maximum Absolute Error

3.3 Approximation and Errors

The capacity of a bottle is 5.38 L, correct to 3 significant figures. What is the lower limit of the capacity?

Solution:

2

1The maximum absolute error 0.01 L

The lower limit of the capacity (5.38 0.005) L 5.375 L

0.005 L

The scale interval of measurement 0.01 LThe 3rd significant figure of the number 5.38 is 8. Hence the capacity is correct to the nearest 0.01 L.

P. 27

Lower limit (8 0.5) cm 7.5 cmUpper limit (8 0.5) cm 8.5 cm

Lower limit 10.5 cmUpper limit 11.5 cmLower limit of the area

Upper limit of the area

2cm 375.392

cm 5.10cm 5.7

∴ The actual area lies between 39.375 cm2 and 48.875 cm2.

Example 3.12T

B. Maximum Absolute Error

3.3 Approximation and Errors

Gordon measured the base length and height of a triangle to be 8 cm and 11 cm respectively, both correct to the nearest 1 cm. What is the range of the area of the triangle?

Solution:

2

1The maximum absolute error of measurement 1 cm 0.5 cm

2cm 875.482

cm 5.11cm 5.8

Base length: Height:

P. 28

Animal Actual weight (kg)

Measured weight (kg)

Absolute error (kg)

Pig 63.42 63 0.42

Dog 5.19 4.77 0.42Both of the absolute errors of the 2 measurements are the same, but they do not reflect the degree of accuracy.

To compare the accuracy of 2 estimations, we also have to determine the relative error.

C. Relative Error

3.3 Approximation and Errors

Consider the following set of data:

P. 29

The absolute error of the pig’s weight is small (insignificant) when compared to its actual weight.

On the other hand, the absolute error of the dog’s weight is significant because it is relatively large when compared to its actual weight.

In general, we use the relative error to determine the degree of accuracy of a measurement.

C. Relative Error

3.3 Approximation and Errors

0.08090.425.19Dog

0.006 620.4263.42Pig

Absolute error (kg)Actual weight (kg)Animal weightActual

error Absolute

Relative error valueActual

error Absolute

P. 30

For example, the degree of accuracy of the measurement of the pig’s weight is higher than that of the dog’s weight.

Sometimes it is impossible to find the actual values in some real-life situations.

So we can use the maximum absolute error and the measured value instead to find the relative error.

C. Relative Error

3.3 Approximation and Errors

The smaller the relative error is, the higher is the accuracy of a measurement.

Relative error valueMeasured

error absolute Maximum

P. 31

The figure below shows the length of a piece of string. Find(a) the length of the string,(b) the maximum absolute error of the length,(c) the relative error of the length, correct to 3 significant

figures.

(b) Maximum absolute error cm 52

1

cm 75

cm 2.5(c) Relative error

(cor. to 3 sig. fig.)

Example 3.13T

C. Relative Error

3.3 Approximation and Errors

Solution:

cm 2.5

0.0333

(a) The length of the string 75 cm

P. 32

Let x g be the weight of a package of sugar.

72

15.2 xx 2.5 72

180∴ The weight of a package of sugar is 180 g.

Example 3.14T

C. Relative Error

3.3 Approximation and Errors

72

1

Louis used a balance to measure the weight of a package of sugar. The maximum absolute error of the balance is 2.5 g. If the relative

error of the result is , find the lower limit of the actual result.

Solution:

∴ Lower limit of the actual weight (180 – 2.5) g 177.5 g

First find out the measured weight of the package of sugar.

P. 33

For example, since the relative errors of the pig’s and the dog’s weight are 0.006 62 and 0.0809 respectively.

The percentage errors of the pig’s and the dog’s weight are 0.662% and 8.09% respectively.

The smaller the percentage error is, the higher is the accuracy of a measurement.

D. Percentage Error

3.3 Approximation and Errors

When the relative error is expressed as a percentage, it is called the percentage error.

Percentage error Relative error 100%

P. 34

A university bought 1258 new computers last year. If the number of computers bought is now correct to the nearest hundred, find thepercentage error of the estimation. (Give the answer correct to 3 significant figures.)

Absolute error 1300 – 1258 42

Percentage error %1001258

42

(cor. to 3 sig. fig.)

Example 3.15T

D. Percentage Error

3.3 Approximation and Errors

Solution:1258 1300 (cor. to the nearest hundred)

3.34%

P. 35

Sally and Christine measured the capacity of 2 boxes separately. Sally’s result was 300 mL correct to 2 significant figures. Christine’s result was 1250 mL correct to the nearest 50 mL.(a) Find the percentage errors of their measurements.

(Give the answer correct to 3 significant figures if necessary.)(b) Hence determine who measured more accurately.

(cor. to 3 sig. fig.)

(a) For Sally’s measurement:

Example 3.16T

D. Percentage Error

3.3 Approximation and Errors

Solution:

Maximum absolute error 10 mL 2 5 mL

Percentage error

For Christine’s measurement: Maximum absolute error

50 mL 2 25 mL

Percentage error

(b) Since 1.67% is less than 2%, Sally measured more accurately.

100%300

5 100%1250

25

%67.1 %2

P. 36

Chapter Summary

3.1 Significant Figures

1. For all numbers, any zeros between 2 non-zero digits are significant figures.

2. For all decimals, any zeros after the last non-zero digit are significant figures.

P. 37

3.2 Scientific Notation

All positive numbers can be expressed in the form a 10n, where 1 a 10 and n is an integer.

Chapter Summary

P. 38

3.3 Approximation and Errors

1. Absolute error (a) If the actual value is larger than a measured value,

then the absolute error actual value measured value.

(b) If a measured value is larger than the actual value, then the absolute error measured value actual

value.

Chapter Summary

(b) Lower limit Measured value Maximum absolute error

2. Maximum absolute error (a) Maximum absolute error

Scale interval of the measuring tool2

1

(c) Upper limit Measured value Maximum absolute error

P. 39

3.3 Approximation and Errors

Chapter Summary

4. Percentage errorPercentage error Relative error 100%

valueActual

error Absolute error Relative (a)

valueMeasured

error absolute Maximum error Relative (b)

3. Relative error