PERCENTAGE ERROR
Subject:
MATHEMATICS
Term:
FIRST TERM
Week:
WEEK 2
Class:
SS 2
Topic:
PERCENTAGE ERROR
Previous lesson:
The pupils have previous knowledge of
Revision of Logarithm of Numbers Greater than One and Logarithm of Numbers Less than one; Reciprocal and Accuracy of Results Using Straight Calculation.
that was taught as a topic in the previous lesson
Behavioural objectives:
At the end of the lesson, the learners will be able to
- define percentage error
- Calculate simple sums on the calculation of percentage error
- explain percentage error (range of values via approximations)
- Calculate percentage error in relation to approximation
Instructional Materials:
- Wall charts
- Pictures
- Related Online Video
- Flash Cards
Methods of Teaching:
- Class Discussion
- Group Discussion
- Asking Questions
- Explanation
- Role Modelling
- Role Delegation
Reference Materials:
- Scheme of Work
- Online Information
- Textbooks
- Workbooks
Content:
WEEK TWO
TOPIC: PERCENTAGE ERROR
CONTENT
- Definition of percentage error
- Calculation of percentage error
- Percentage error (range of values via approximations)
- Calculations on percentage error in relation to approximation
Definition of Percentage Error
No measurement, however, carefully made is exact (accurate) i.e if the length of a classroom is measured as 2.8m to 2 s.f the actual length may be between 2.75 and 2.85, the error of this measurement is 2.75 – 2.8 or 2.85 – 2.8 = + 0.05.
2.75 2.85 2.8 2.9
Percentage error = error X 100
Actual measurement 1
error =+ 0.05
actual measurement 2.8
% error =0.05 x 100
2.8 1
=1.785% = 1.79%
Example 2
Suppose the length of the same room is measured to the nearest cm ,280cm i.e. (280cm) calculate the percentage error.
Measurement = 280cm.
The range of measurement will be between 279.5cm or 280cm
Error = 280 – 279.5 = 0.5cm
% error = error x 100
Measurement 1
% error = 0.5 x 100 = 0.178%
280 1
= 0.18% (2sf)
Example 3
The length of a field is measured as 500m; find the percentage error of the length if the room is measured to
i. nearest metre ii. nearest 10m iii. one significant figure.
Solutions
i. To the nearest metre
Measurement = 500m
Actual measurement = between 499.5 – 500.5
Error = + 0.5m
% error = error x 100
measurement
0.5 x 100 = 0.10%
500 1
= 0.10%
ii. Nearest 10m
Measurement = 500m,
range= 495m – 505m
error =+ 5m
error = 5 x 100 = 1%
500 1
iii. To 1 s.f.
measurement = 500m
range = 450 – 550
error = + 50
% error = 50 x 100 = 10%
500 1
Evaluation
1. The length of each side of a square is 3.6 cm to 2s.f. (a) Write down the smallest and the largest of each side. (b) Calculate the smallest and the largest values for the perimeter.
(c) Find the possible values of the area.
Percentage Error (range of values via approximations)
1. Range of values measured to the nearest whole number i.e. nearest tens, hundreds etc. e.g.
Find the range of values of N6000 to:
i. nearest naira = N5999.50 – 6000.50
ii. nearest N10 = N5995 – 6005
iii. nearest N100 = N5950 – 6050
iv. nearest N1000 = N5500 – 6,500
2. Range of values measured to a given significant figure. E.g. find the range of value of 6000 to
1 sf = 5500 – 6500
2 sf = 5950 – 6050
3 sf = 5995 – 6005
5 sf = 5999.95- 6000.05
3. Range of values measured to a given decimal places e.g. 39.8 to a 1d.p = 39.75 – 39.85.
Note: if it is 1 d.p, the range of values will be in 2 d.p, if 2 d.p, the range will be in 3 d.p etc. (i.e the range = d.p + 1). The same rule is also applicable to range of values to given significant figure.
Evaluation
Orally: From New General Mathematics SS 2 by J. B. Channon and Co 3rd edition exercise 46 no. 1a – f.
Calculations on percentage error:
Example:
Calculate the percentage error if
1. The capacity of a bucket is 7.5 litres to 1 d.p.
2. The mass of a student is 62kg to 2 s.f.
Solutions
1. Measurement = 7.5litres ( 1d.p)
Range of values = 7.45 – 7.55
Error = 7.5 – 7.45 = 0.05
% error = error x 100
measurement 1
0.05 x 100
7.51
= 0.67%
2. Measurement = 62kg (2 s.f)
Range of values = 61.5kg to 62.5kg
error = 6.2 – 61.5 = 0.5kg
% error = error x 100
measurement 1
0.5 x 100 = 0.81%
62 1
EVALUATION
1. Calculate correct to 2 s.f. the percentage error in approximately 0.375 to 0.4.
GENERAL EVALUATION / REVISION QUESTION
1. A metal rod was measured as 9.20 m. If the real length is 9.43 m, calculate the percentage error to 3 s.f
2.A student measures the radius of a circle as 1.46 cm instead of 1.38 cm. Calculate the percentage error.
3.The weight of sugar was recorded as 8.0 g instead of 8.2 g. What is the percentage error?
4.A student mistakenly approximated 0.03671 to 2 d.p instead of 2 s.f. What is the percentage error correct to 2 s.f
5.A man’s weight was measured as 81.5 kg instead of 80 kg. Find the percentage error in the measurement.
WEEKEND ASSIGNMENT
What is the error in the following measurement
1. The distance between two towns is 60km to the nearest km. (a) 5km (b) 0.5km (c) 8.3km (d) 0.83km
2. The area of a classroom is 400m2 to 2 s.f. (a) 50m2 (b) 1.25m2(c) 2.5m2 (d) 5m2
3. A sales girl gave a girl a balance of N1.15 to a customer instead of N1.25, calculate the % error.
4. A student measured the length of a room and obtained the measurement of 3.99m, if the percentage error of his measurement was 5% and his own measurement was smaller than the length, what is the length of the room?(a) 3.78m (b) 3.80m (c) 4.18m (d) 4.20m
5. A man is 1.5m tall to the nearest cm, calculate his percentage error.
(a) 0.05cm (b) 0.33% (c) 0.033% (d) 0.05cm
THEORY
1. A classroom is 10m by 10m; a student measured a side as 9.5m and the other side as 10m and uses his measurement to calculate the area of the classroom. Find the percentage error in a. the length of one of the sidesb. the area of the room
2. Instead of recording the number 1.23cm for the radius of a tube, a student recorded 1.32cm, find the percentage error correct to 1 d.p.
Reading Assignment
Essential Mathematics for SSS2, pages 13-22, Exercise 2.4
Conclusion
The class teacher wraps up or concludes the lesson by giving out short notes to summarize the topic that he or she has just taught.
The class teacher also goes round to make sure that the notes are well copied or well written by the pupils.
He or she makes the necessary corrections when and where the needs arise.
