Timetable › Forums › Standard › Measurement › Measurement Error
Tagged: Accuracy, error, measurement error, percentage error
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All measurements are actually approximations, so there will always be some error in measurement. This measurement error is calculated using the scale with which the measurements are made, and then finding half of that.
Measurement Error = ±1/2 of the smallest division of a scale
Absolute error = ±½ × accuracy/precision
Lower bound = measurement – absolute error
Upper bound = measurement + absolute error
Percentage error = $$\frac{error}{measurement}\times 100$$
Some examples:
- measuring to the nearest cm gives a measurement error of ±1/2 cm or 5mm
- measuring to the nearest m gives a measurement error of ±1/2 m or 50cm
- measuring to the nearest km gives a measurement error of ±1/2 km or 500m
- measuring to the nearest 0.1cm gives a measurement error of ±1/2 0.1cm or 0.05cm
- measuring to the nearest 10 gives a measurement error of ±1/2 of 10 or 5
Rachel measures her height and finds that she is 171cm tall, to the nearest millimetre, the tallest she can be is:
a. 172.00cm b. 171.60cm c. 170.55cm d. 171.40cma will round to 172 b will round to 172 c will round to 171 d will round to 171 so the tallest that will round to 171 is D
A rectangle has a length of 18cm and a width of 12cm.
a. What is the largest possible error?
b. What is the smallest possible area?a. accuracy = 1cm
error =± 0.5cm
lower limits = 18cm – 0.5cm = 17.5cm
and
12cm – 0.5cm = 11.5cmupper limits = 18cm + 0.5cm = 15.5cm
and
12cm + 0.5cm = 12.5cmerror =± 0.5cm
b. consider the lower limits = 17.5cm × 11.5cm= 201.25cm2
A rectangular paddock is 36 metres long and 25 metres wide, measured to the nearest metre. It requires fencing along all sides.
a. Between what two measurements does the length lie?
b. Between what two measurements does the width lie?
c. What is the smallest possible length of fencing needed?
d. What is the largest possible length of fencing needed?
e. What is the maximum percentage error in the length of fencing needed?a. accuracy = 1m
error = 0.5m
length 36 – 0.5 and 36 + 0.5
35.5m – 36.5mb. width: 25 – 0.5 and 25 + 0.5
24.5m – 25.5mc. smallest length of fencing will be the perimeter using the lower limits
P = 35.5 + 24.5 + 35.5 + 24. 5
= 120md. largest length of fencing will be the perimeter using the upper limits
P = 36.5 + 26.5 + 36.5 + 26. 5= 124m
e. total length of fencing = 36 + 25 + 36 + 25
= 122mtotal error = 4 lots of 0.5m
= 2m (since there would be 0.5m per measurement)
% error = $$\frac{2}{122}\times100= 1.6%$$Tom measures a piece of fabric with a ruler that is only marked in centimetres. He measures the length to be 60cm and the width 90cm. The area of the fabric is: a. between 5399.5cm2 and 5400.5cm2
b. between 5384.75cm2 and 5414.75cm2
c. between 5325.25cm2 and 5475.25cm2
d. between 5251cm2 and 5551cm2accuracy = 1cm
error = 0.5cm
60 – 0.5 = 59.5
60 + 0.5 = 60.05
90 – 0.5 = 89.5
90 + 0.5 = 90.05
lower area = 59.5 × 89.5
= 5325.25cm2
upper area = 60.5 × 90.5
= 5475.25cm2
∴ C
A roast is to cook for 1 hour and 30 minutes correct to the nearest 10 minutes. The percentage error is?
accuracy = 10 minutes
error = 5 minutes (ie half the accuracy)
% error = error/measurement × 100
5/90 × 100 = 5.6%
(don’t forget you have to change the 1 hour 30 mins to minutes only)The top of the Sydney Harbour Bridge is measured to be 138.4m above sea level. What is the percentage error in this measurement?
accuracy is to 1 decimal place
accuracy = 0.1error = 0.05 (half of accuracy)
% error = $$\pm\frac{0.05}{138.4}\times100$$
% error = ±0.036%
For a measurement of 10cm, find:
a. the limits of measurement.
b. the percentage error.a. accuracy = 1cm For a measurement of 10cm Accuracy = 1cm, since it is to the nearest cm Error = ½ × 1
= 0.5cmLimits of measurement will be the measurement ±1/2 unit 10 – 0.5cm = 9.5cm
10 + 0.5cm = 10.5cmto find the lower limit, subtract the error from the measurement
to find the upper limit, add the error to the measurement9.5cm – 10.5cm b. error = error/measurement × 100
= 0.5/10 × 100
= 5%
accuracy= 1cm
error = ±1/2cm
put the error over the measurement and multiply by 100
A length is measured to be 13.5cm, correct to the nearest millimeter. Find the percentage error, to 2 decimal places
accuracy of measurement = 1mm
find the accuracy (or precision) error of measurement = 0.5mm
halve the accuracy to find the error % error = 0.05/13.5 × 100
= 0.37%
since we are in cm and mm, convert to the same unit
0.5mm = 0.05cm -
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