Free-Standing Mathematics Activity
Errors
© Nuffield Foundation 2011
© Nuffield Foundation 2011
Errors
How certain are you about the accuracy of your measurements?
© Nuffield Foundation 2011
Errors
Measurements are subject to errors.
Length of swimming pool = 25 metres to nearest metre
Think about What is the shortest possible length?
What is the longest possible length?
24.5 m
Lower bound
24 m
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25.5 m
Upper bound
25 m
Maximum error
0.5 m
26 m
Weight of package
4.3 kg
to 1 decimal place
Think about: What is the smallest possible weight?
What is the largest possible weight?
4.25 kg
Lower bound
4.2 kg
4.35 kg
Upper bound
4.3 kg
Maximum error
0.05
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4.4 kg
Errors in measurements
Lower bound = lowest possible value
Upper bound = highest possible value
When a measure is expressed to a given unit,
the maximum error is half a unit.
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Errors in measurements
Accuracy
Maximum error
Nearest 100
50
Nearest 10
5
Nearest whole number
0.5
To 1 decimal place (nearest 0.1)
0.05
To 2 decimal places (nearest 0.01)
0.005
Think about
What is the maximum error?
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Length of journey = 250 miles to nearest 10 miles
Think about
What is the maximum error?
Upper bound
255 miles
Lower bound
245 miles
250 miles
–5
miles
+5
miles
Length of journey = 250 ± 5 miles
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Winning time in a race
36.32 seconds to nearest 0.01 second
Think about
What is the maximum error?
Maximum error = 0.005 seconds
Upper bound
= 36.32 + 0.005 = 36.325 seconds
Lower bound
= 36.32 – 0.005 = 36.315 seconds
Winning time
= 36.32 ± 0.005 seconds
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Temperature of furnace = 1400°C to 2 significant figures
Think about
What is the highest possible temperature?
What is the lowest possible temperature?
1350°C
Lower bound
1300°C
1450°C
Upper bound
1400°C
–50°C +50°C
Temperature = 1400 ± 50 °C
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1500°C
If temperature of furnace = 1400°C to 3 significant figures
Think about
What is the highest possible temperature now?
What is the lowest possible temperature?
1395°C
Lower bound
1390°C
1405°C
Upper bound
1400°C
– 5°C + 5°C
Temperature = 1400 ± 5 °C
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1410°C
45 m
Car park
Best estimate of
perimeter
24 + 56
= 80 m
= 45 + 24 + 83 + 56
+ 128 + 80
24 m
83 m
A
B
56 m
45 + 83 = 128 m
= 416 m
Best estimate of area
Area of A = 80  45 = 3600 m2
Area of B = 83  56 = 4648 m2
Total area = 3600 + 4648 = 8248 m2
© Nuffield Foundation 2011
Think about:
How accurate
are these estimates?
Car park upper bounds
Upper bound of perimeter
45.5 m
= 45.5 + 24.5 + 83.5
+ 56.5 + 129 + 81
24.5 m
= 420 m
Upper bound of area
24.5 + 56.5
= 81 m
83.5 m
A
B
56.5 m
45.5 + 83.5 = 129 m
Upper bound of area of A = 81  45.5 = 3685.5 m2
Upper bound of area of B = 83.5  56.5 = 4717.75 m2
Upper bound of total area = 3685.5 + 4717.75 = 8403.25 m2
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Car park lower bounds
Lower bound of perimeter
= 44.5 + 23.5 + 82.5
+ 55.5 + 127 + 79
= 412 m
Lower bound of area
44.5 m
23.5 m
23.5 + 55.5
= 79 m
82.5 m
A
B
44.5 + 82.5 = 127 m
Lower bound of area of A = 79  44.5 = 3515.5 m2
Lower bound of area of B = 82.5  55.5 = 4578.75 m2
Lower bound of total area = 3515.5 + 4578.75 = 8094.25 m2
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55.5 m
Car park
45 m
Perimeter
24 m
83 m
Best estimate = 416 m
Lower bound = 412 m
56 m
Upper bound = 420 m
Total area
Think about
What final answers should be given?
Best estimate = 8248 m2
Lower bound = 8094.25 m2
Upper bound = 8403.25 m2
Perimeter = 420 m (to 2 sf) Total area = 8200 m2 (to 2 sf)
© Nuffield Foundation 2011
Example: Find the volume and surface area of a cone with
radius r = 3.5 cm, height h = 5.2 cm (to 1 dp)
Volume V = 13r 2h
Best estimate V  13 π  3.52  5.2
= 66.7 cm3
Upper bound V  13 π  3.552  5.25 = 69.3 cm3
Lower bound V  13 π  3.452  5.15 = 64.2 cm3
Think about
WhatEstimate
final answer
should
given?
3 (to 2 sf)
Best
of Volume
= 67becm
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h
r
Surface area of cone r = 3.5 cm, h = 5.2 cm (to 2 sf)
Surface area S = r(r + l)
l  r 2  h2
Best estimate l  3.52  5.22 = 6.2682 cm
S  π  3.5   3.5  6.2682
= 107.41 cm2
l
h
Upper bound l  3.552  5.252 = 6.3376 cm
S  π  3.55  3.55  6.3376 
Lower bound
= 110.27
l  3.452  5.152
S  π  3.45  3.45  6.1988 
r
cm2
= 6.1988 cm
= 104.58 cm2
Think about
What final answer
should be given?
Best estimate of surface area = 110 cm2 (to 2 sf)
© Nuffield Foundation 2011
At the end
of the activity
12.0  0.1 cm
• What is the maximum value for the diameter of this CD?
• What is the minimum value for the diameter?
• What are the maximum and minimum values for the radius?
• Write the radius in the form a  b
• Work out a best estimate for the area of the top of the CD.
How accurately do you think you should give the answer?
• Work out the upper and lower bounds for the area.
Was the answer you gave reasonable?
• In general, what accuracy should you give in answers to
calculations involving measurements?
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