Accuracy and Error
Upper and lower bounds
A table is measured to two decimal places.
It is 1.86m tall.
So the smallest this table could have been
is 1.855m since this
rounds up to 1.86 to two decimal places. Any smaller, and the number rounds
down.
The largest the table could have been
<1.865
since 1.865 itself rounds UP.
Hence the true length lies between these
upper and lower bounds. Mathematicians express this in two ways.
1.855 ≤ L < 1.865
or
1.86^{ +} 0.005
Note that the upper bound can never be
reached and so we write < instead of ≤, however when a question asks you for the
"upper bound" the answer is 1.865 and not <1.865
Relative and percentage error
The second way of representing an error in
a measurement tells us that the "absolute error" of our measurement is 0.005.
These are far better ways of representing
errors since they take into account the importance of the error compared to the
size of the object you are measuring. For example,
a three meter error could be dangerous when building a bridge (especially if it
is three meters too short!) but is negligible if measuring the distance from the
earth to the sun.
Diagrams
You can represent upper and lower bounds
using a diagram like the one below. You should know about this but it is rarely
asked at GCSE
Significant figures and decimal
places
When you record any long number you can
write it down to a chosen number of decimal places or significant
figures.
When you are dealing with decimal places
it is important to remember that 1.372000
is different (more accurate) to 1.372.
When dealing with significant figures, the
examiner will not always be able to tell how many s.f. you have written
something to. For example 1200 to one s.f. is "1000" but to two s.f. it is
"1200", which is the same as to three and four s.f.
Trick question- complication with
significant figures
Note that the upper and lower bounds a
measurement of 10cm to 2 s.f. are not 9.5 and 10.5 but 9.95 and 10.5.
This is because 9.5 to 2 s.f. is still 9.5! However 9.95 to 2 s.f. rounds to 10.
Calculations involving upper and
lower bounds
At every stage in a calculation use the
upper and lower bounds as appropriate
Question: I cut 20.7 cm off a piece of
wood measuring 32.2 cm. Find the upper and lower bounds for the length of the
remaining wood.
Answer
The upper bound of this new piece of wood
is the longest it could ever be. Therefore I need to assume the original piece
was (32.2 + 0.05 upper bound) and then cut as little as possible
off. Cut (20.7 - 0.05 lower bound) off. So the upper bound of the
remaining wood is 32.25 - 20.65 = 11.6
The lower bound of this new piece of wood
is the shortest it could ever be. Therefore I need to assume the original piece
was (32.2 - 0.05 lower bound) and then cut as much as possible
off. Cut (20.7 + 0.05 upper bound) off. So the lower bound of the
remaining wood is 32.15 - 20.75 = 11.4
Hence the actual value can only be given
as "10 to 1s.f." since this is the most accurate value that both 11.4 and 11.6
round to. If we tried to say that the true value can be written as "11 to 2s.f."
this would be wrong because 11.6 rounds to 12 to 2 significant figures.