help - limits of accuracy! (1 Viewer)

[j92]

There was this question and i dunno how to figure it out since i don't know the limits of accuracy???????????????


5. A children's playground is measured to be 10.2m by 8.5m

a) find the limits of accuracy of each measurement

b) calculate the smallest and largest possible areas

c) calculate the maximum percentage error in the area of the playground.





i think i can do b and c but im not sure about the limits


would you express it like 0.5
or like 9.7-10.7 and 8-9

so confused?

 

[PC]

Each measurement is in metres and is given correct to 1 decimal place. This means the precision of each measurement is 0.1 m.

The absolute error is always half the precision.
Absolute error = 1/2 x 0.1 = 0.05 m

(a) Maximium possible length = 10.2 + 0.05 = 10.25 m
Minimum possible length = 10.2 – 0.05 = 10.15 m
Maximum possible length = 8.5 + 0.05 = 8.55 m
Minimum possible length = 8.5 – 0.05 = 8.45 m

(b) Maximum possible area = 10.25 x 8.55 = 87.6375 m²
Minimum possible area = 10.15 x 8.45 = 85.7675 m²

(c) Not sure about this one. Percentage error usually only applies to lengths, not areas or volumes, as the differences between the maximum area and calculated value, and minimum area and calculated value will be different. So I'm guessing, that it might be:

Calculated Area = 10.2 x 8.5 = 86.7 m²
Difference = 87.6375 – 86.7 = 0.9375
Percentage Error = Difference/Calculated Value x 100/1
= 0.9375/86.7 x 100/1
= 1.081314879%
= 1.08%

 

[j92]

so um...metres is 0.05 and cm is 0.5?

 

[PC]

Why worry about centimetres?

 

[j92]

so it doesnt matter on the...like.. cm or m or mm..

it matters on the decimal places

like 4.5 the error would be 0.05
but like 16 would be 0.5?

 

[PC]

Well, all your measurements were given in metres, so leave them in metres. Don't make life hard for yourself when you don't have to!

The measurements were all done in metres to 1 decimal place, so that means that the precision of the instrument was correct to the nearest 0.1 m, so the absolute error is half that, 0.05 m.

Keep everything in metres.

A different question might be the length of a pencil, given as 12 cm, correct to the nearest cm. This means that the precision is 1 cm, so the absolute error is half that, 0.5 cm. There's no need to change this to mm or even metres.

 

[rowdyroddy]

it is percentage error
whatever half the measure is, if its in cm, then half cm over and below