Download cor. to 3 sig. fig

3
Approximation and Errors
Case Study
3.1 Significant Figures
3.2 Scientific Notation
3.3 Approximation and Errors
Chapter Summary
Case Study
I guess there are 120
paper clips. What’s
your guess, Linda?
I guess there are 110
clips, John.
Suppose that there are actually
116 paper clips on the plate.
Did John or Linda make a better
estimation?
We need to find the difference
between each estimate and the
actual number.
The smaller the difference is, the better is the estimate.
John’s estimate  Actual number  120  116
4
Actual number  Linda’s estimate  116  110
6
Since 4 is less than 6, John made a better estimation.
P. 2
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 95
950 000. (correct to the nearest ten thousand)
 6 953 000. (correct to the nearest thousand)
The important digits of a number are called significant figures.
In most cases, the far left non-zero digit, having the largest place value
in a number, is the first significant figure.
This is also the most important figure.
Subsequent important digits are called the second significant figure,
the third significant figure, and so on.
P. 3
3.1 Significant Figures
A. Basic Concepts
For example, the number 1049 has 4 significant figures.
1
0
4 9
1st significant figure
2nd significant figure
3rd significant figure
4th significant figure
We also study significant
figures in decimals.
For example, the decimal
5.307 has 4 significant
figures.
For all numbers, any zeros between 2 non-zero digits are
significant figures.
P. 4
3.1 Significant Figures
A. Basic Concepts
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.
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.
P. 5
3.1 Significant Figures
A. Basic Concepts
The following table shows some examples. All the significant figures
are marked with green colour.
Numbers
Number of Significant Figures
4008
4
6.092
4
0.7
1
0.002 04
3
9.000
4
0.130
3
P. 6
3900 (correct to the
nearest integer) has 4
significant figures. But
3900 (correct to the
nearest ten) has 3
significant figures.
3.1 Significant Figures
B. Round off to the Required Significant Figures
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.
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.)
P. 7
The phrase ‘cor. to 3 sig.
fig.’ is the short form of
‘correct to 3 significant
figures’.
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 1
to the 3rd significant figure.
P. 8
3.1 Significant Figures
B. Round off to the Required Significant Figures
Example 3.1T
Round off 472 780 correct to
(a) 2 significant figures,
(b) 3 significant figures,
(c) 4 significant figures.
Solution:
(a) 470 000 (cor. to 2 sig. fig.)
(b) 473 000 (cor. to 3 sig. fig.)
(c) 472 800 (cor. to 4 sig. fig.)
P. 9
3.1 Significant Figures
B. Round off to the Required Significant Figures
Example 3.2T
Round 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. 10
3.1 Significant Figures
B. Round off to the Required Significant Figures
Example 3.3T
Round 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. 11
3.1 Significant Figures
B. Round off to the Required Significant Figures
Example 3.4T
The total price of 120 oranges is $250. Round off the average price
of each orange correct to 2 significant figures.
Solution:
The average price of each orange
 $(250  120)
 $2.0833
 $2.1 (cor. to 2 sig. fig.)
P. 12
3.2 Scientific Notation
A. Introduction
In science, we often deal with numbers which are very large or small:
 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
 0.000 000 000 000 000 000 000 03 g  3  1023 g
This kind of representation of numbers is called scientific notation.
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. 13
3.2 Scientific Notation
A. Introduction
Example 3.5T
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
Solution:
(a) 22 000  2.2  10 000
 2.2  104
(b) 0.000 000 7  7  0.000 000 1
 7  107
(c) 95 000  104  (9.5  104)  104
 9.5  108
(d) 0.008  102  (8  103)  102
 8  105
P. 14
Move the decimal point 4 places to the left
and then times 104
Move the decimal point 7 places to the right
and then times 107
am  an  am  n
3.2 Scientific Notation
A. Introduction
Example 3.6T
Convert the following numbers into integers or decimals.
(a) 1.002  107
(b) 6  105
Solution:
(a) 1.002  107  1.002  10 000 000
 10 020 000
(b) 6  105  6  0.000 01
 0.000 06
P. 15
3.2 Scientific Notation
A. Introduction
Example 3.7T
Without using a calculator, evaluate each of the following expressions.
Express the answers in scientific notation.
6 106
(a)
(b) (4  103)(3.3  105)
0.00015
Solution:
6  106
6  106
(a)

0.00015 1.5  104
6

106  ( 4)
1.5
 4  10 2
P. 16
(b) (4 103)(3.3 105)  4  3.3 103  5
 13.2 108
 1.32 109
3.2 Scientific Notation
A. Introduction
In calculator, we should key in ‘1.32
EXP
9’ to represent 1.32  109.
The calculator would display ‘1.32E9’ before execution.
If we key in ‘4
‘4
10
11’,
EXP
11
EXE
’, then the calculator would display
which means 4  1011.
Students should note
that the key-in sequences
of different calculators
may be different.
P. 17
3.2 Scientific Notation
B. Applications of 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.
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
P. 18
3.2 Scientific Notation
B. Applications of Scientific Notation
Example 3.8T
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. 19
3.2 Scientific Notation
B. Applications of Scientific Notation
Example 3.9T
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:
1.35  1010
The required ratio 
3.1 1011
13.5  1011

3.1 1011
 4.4 (cor. to 2 sig. fig.)
P. 20
3.3 Approximation and Errors
In Book 1A Chapter 2, we learnt that no measurements give exact
values.
They are only approximations.
Hence, every measurement or estimation has errors.
P. 21
3.3 Approximation and Errors
A. Absolute Error
The difference between a measured value (or an estimated value)
and the actual value is called the absolute error.
If the actual value is larger than a measured value, then
absolute error  actual value  measured value
The absolute error is
always positive.
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. 22
3.3 Approximation and Errors
A. Absolute Error
Example 3.10T
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. 23
3.3 Approximation and Errors
B. Maximum Absolute Error
Measuring tools with smaller scale intervals can give estimations
with higher accuracy.
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.
The measured length is 15 cm, correct to the
nearest cm.
∴ The minimum and the maximum
actual lengths are 14.5 cm and 15.5 cm
respectively.
The absolute error of this measurement does not exceed 0.5 cm.
This figure is called the maximum absolute error.
P. 24
3.3 Approximation and Errors
B. Maximum Absolute Error
In general,
1
maximum absolute error   scale interval of the measuring tool
2
In the figure, the scale interval of the ruler is 1 cm.
 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.
Lower limit  Measured value  Maximum absolute error
Upper limit  Measured value  Maximum absolute error
P. 25
3.3 Approximation and Errors
B. Maximum Absolute Error
Example 3.11T
The capacity of a bottle is 5.38 L, correct to 3 significant figures.
What is the lower limit of the capacity?
Solution:
The scale interval of measurement  0.01 L
1
The maximum absolute error   0.01 L
2
 0.005 L
The lower limit of the capacity  (5.38  0.005) L
 5.375 L
P. 26
The 3rd significant
figure of the number
5.38 is 8. Hence the
capacity is correct to
the nearest 0.01 L.
3.3 Approximation and Errors
B. Maximum Absolute Error
Example 3.12T
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:
1
The maximum absolute error of measurement   1 cm  0.5 cm
2
Base length:
Height:
Lower limit  (8  0.5) cm  7.5 cm
Lower limit  10.5 cm
Upper limit  (8  0.5) cm  8.5 cm
Upper limit  11.5 cm
7.5 cm 10.5 cm
 39.375 cm2
Lower limit of the area 
2
8.5 cm 11.5 cm
 48.875 cm2
Upper limit of the area 
2
∴ The actual area lies between 39.375 cm2 and 48.875 cm2.
P. 27
3.3 Approximation and Errors
C. Relative Error
Consider the following set of data:
Animal Actual weight (kg) Measured weight (kg) Absolute error (kg)
Pig
63.42
63
0.42
Dog
5.19
4.77
0.42
Both 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.
P. 28
3.3 Approximation and Errors
C. Relative Error
Animal Actual weight (kg)
Pig
Dog
Absolute error (kg)
63.42
5.19
0.42
0.42
Absolute error
Actual weight
0.006 62
0.0809
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.
Relative error 
P. 29
Absolute error
Actual value
3.3 Approximation and Errors
C. Relative Error
The smaller the relative error is, the higher is the accuracy of a
measurement.
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.
Relative error 
P. 30
Maximum absolute error
Measured value
3.3 Approximation and Errors
C. Relative Error
Example 3.13T
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.
Solution:
(a) The length of the string  75 cm
1
(b) Maximum absolute error   5 cm
2
 2.5 cm
2.5 cm
(c) Relative error 
75 cm
 0.0333 (cor. to 3 sig. fig.)
P. 31
3.3 Approximation and Errors
C. Relative Error
Example 3.14T
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
1
error of the result is , find the lower limit of the actual result.
72
Solution:
Let x g be the weight of a package of sugar.
2.5 1

x 72
x  2.5  72
 180
∴ The weight of a package of sugar is 180 g.
∴ Lower limit of the actual weight  (180 – 2.5) g
 177.5 g
P. 32
First find out the
measured weight of the
package of sugar.
3.3 Approximation and Errors
D. Percentage Error
When the relative error is expressed as a percentage, it is called the
percentage error.
Percentage error  Relative error  100%
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.
P. 33
3.3 Approximation and Errors
D. Percentage Error
Example 3.15T
A university bought 1258 new computers last year. If the number
of computers bought is now correct to the nearest hundred, find the
percentage error of the estimation. (Give the answer correct to 3
significant figures.)
Solution:
1258  1300 (cor. to the nearest hundred)
Absolute error  1300 – 1258
 42
42
100%
Percentage error 
1258
 3.34% (cor. to 3 sig. fig.)
P. 34
3.3 Approximation and Errors
D. Percentage Error
Example 3.16T
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.
Solution:
For Christine’s measurement:
(a) For Sally’s measurement:
Maximum absolute error
Maximum absolute error
 50 mL  2  25 mL
 10 mL  2  5 mL
5
25
Percentage error 
Percentage error 
 100%
100%
300
1250
 1.67% (cor. to 3 sig. fig.)
 2%
(b) Since 1.67% is less than 2%, Sally measured more accurately.
P. 35
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. 36
Chapter Summary
3.2 Scientific Notation
All positive numbers can be expressed in the form a  10n, where
1  a  10 and n is an integer.
P. 37
Chapter Summary
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.
2. Maximum absolute error
(a)
Maximum absolute error
1
  Scale interval of the measuring tool
2
(b) Lower limit  Measured value  Maximum absolute error
(c) Upper limit  Measured value  Maximum absolute error
P. 38
Chapter Summary
3.3 Approximation and Errors
3. Relative error
Absolute error
(a) Relative error 
Actual value
Maximum absolute error
(b) Relative error 
Measured value
4. Percentage error
Percentage error  Relative error  100%
P. 39