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© Nuffield Foundation 2011
© Nuffield Foundation 2011
ErrorsFree-Standing Mathematics Activity
© Nuffield Foundation 2011
Errors
How certain are you about the accuracy of your measurements?
© Nuffield Foundation 2011
ErrorsMeasurements are subject to errors.Length of swimming pool = 25 metres to nearest metre
25.5 mUpper bound
25 m 26 m24 m
24.5 mLower bound
Think about What is the shortest possible length? What is the longest possible length?
Maximum error0.5 m
© Nuffield Foundation 2011
4.35 kgUpper bound
4.3 kg 4.4 kg4.2 kg
4.25 kgLower bound
Maximum error0.05
Weight of package 4.3 kg to 1 decimal place
Think about: What is the smallest possible weight? What is the largest possible weight?
© Nuffield Foundation 2011
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.
© Nuffield Foundation 2011
Errors in measurements
Nearest 100 50
Accuracy Maximum error
Nearest 10 5
Nearest whole number 0.5
To 1 decimal place 0.05(nearest 0.1)
To 2 decimal places 0.005(nearest 0.01)
Think about What is the maximum error?
© Nuffield Foundation 2011
Length of journey = 250 miles to nearest 10 miles
Upper bound255 miles
250 miles
Lower bound245 miles
+ 5 miles
– 5 miles
Length of journey = 250 ± 5 miles
Think about What is the maximum error?
© Nuffield Foundation 2011
Winning time
Maximum error = 0.005 seconds
Lower bound = 36.32 – 0.005 = 36.315 seconds
Upper bound = 36.32 + 0.005 = 36.325 seconds
Winning time in a race
36.32 seconds to nearest 0.01 second
= 36.32 ± 0.005 seconds
Think aboutWhat is the maximum error?
© Nuffield Foundation 2011
Temperature of furnace = 1400C to 2 significant figures
= 1400 ± 50 °C Temperature
1450°CUpper bound
1400°C 1500°C1300°C
1350°CLower bound
–50°C +50°C
Think about What is the highest possible temperature?What is the lowest possible temperature?
© Nuffield Foundation 2011
If temperature of furnace = 1400°C to 3 significant figures
Temperature= 1400 ± 5 °C
1405°CUpper bound
1400°C 1410°C1390°C
1395°CLower bound
– 5°C + 5°C
Think about What is the highest possible temperature now?What is the lowest possible temperature?
© Nuffield Foundation 2011
56 m
83 m24 m
45 mCar park
Best estimate of perimeter
= 45 + 24 + 83 + 56
+ 128 + 80 45 + 83 = 128 m
24 + 56= 80 m
= 416 m
Best estimate of area
Area of A = 3600 m2
Area of B = 4648 m2
Total area = 3600 + 4648 = 8248 m2
Think about: How accurate are these estimates?
A B
= 80 45
= 83 56
© Nuffield Foundation 2011
Car park upper bounds
Upper bound of perimeter
= 45.5 + 24.5 + 83.5 + 56.5 + 129 + 81
45.5 + 83.5 = 129 m
24.5 + 56.5= 81 m= 420 m
Upper bound of area
Upper bound of area of A
A B
= 3685.5 m2
Upper bound of area of B = 4717.75 m2
Upper bound of total area = 3685.5 + 4717.75= 8403.25 m2
56.5 m
83.5 m24.5 m
45.5 m
= 81 45.5
= 83.5 56.5
© Nuffield Foundation 2011
55.5 m
82.5 m23.5 m
44.5 m
Car park lower bounds
Lower bound of perimeter
= 44.5 + 23.5 + 82.5 + 55.5 + 127 + 79= 412 m
Lower bound of area
Lower bound of area of A = 3515.5 m2
Lower bound of area of B = 4578.75 m2
Lower bound of total area = 3515.5 + 4578.75 = 8094.25 m2
44.5 + 82.5 = 127 m
23.5 + 55.5= 79 m A BA B
= 79 44.5
= 82.5 55.5
© Nuffield Foundation 2011
56 m
83 m24 m
45 mCar park
Perimeter
Total area
Best estimate = 8248 m2
Best estimate = 416 mLower bound = 412 m
Upper bound = 420 m
Lower bound = 8094.25 m2 Upper bound = 8403.25 m2
Perimeter = 420 m (to 2 sf) Total area = 8200 m2 (to 2 sf)
Think about What final answers should be given?
© 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)
hr231Volume V =
5.23.531 2 πVBest estimate
= 66.7 cm3
5.253.5531 2 πVUpper bound = 69.3 cm3
5.153.4531 2 πVLower bound = 64.2 cm3
Best Estimate of Volume= 67 cm3 (to 2 sf)
h
r
Think about What final answer should be given?
© Nuffield Foundation 2011
Surface area S = r(r + l)
6.26823.53.5πS = 107.41 cm2
Upper bound
22 hrl
Best estimate 22 5.23.5 l = 6.2682 cm
6.33763.553.55 πS = 110.27 cm2
22 5.253.55 l = 6.3376 cm
Surface area of cone r = 3.5 cm, h = 5.2 cm (to 2 sf)
Lower bound
6.19883.453.45 πS = 104.58 cm2
22 5.153.45 l = 6.1988 cm
Best estimate of surface area = 110 cm2 (to 2 sf)
h
r
l
Think about What final answer should be given?
© Nuffield Foundation 2011
• Write the radius in the form a b
• 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?
• 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?
12.0 0.1 cm
At the end of the activity