Precision 2 | | DJS

Precision 2


The use of implied precision can be a minefield. Spoken numbers need rounding to something like 3 significant figures (on this page written as s.f.) and any extremely large or small number should be given in Standard Form.

1  A length is measured as 15 metres. Using the implied 2 s.f. precision, give both the shortest and longest possible lengths to the nearest millimetre, i.e. to 5 s.f.

2  An area is measured as 1.5 metres by 1.2 metres. Using the implied precision, give both the shortest and longest possible areas to the nearest square centimetre, i.e. to 4 s.f.

3  A square is measured as being 60m² . If this is assumed correct to 1s.f., give the lowest and highest values for the area and therefore the values of the length of one side to 2 s.f.

4  There are particularly difficult problems with rounding to a given precision when the significant figure part is unity.  1 tonne is 1000kg. If this mass has been given to three s.f. then it lies between 999.5 and 1005 kg. Notice that these two numbers each have 4 s.f., but that the distance from 1000 is different. What then are the limits if the tonne is given to    
 a)  2 s.f.    b) 4s.f.     c) 1 s.f.?

5  Dad thinks he put 32 litres in the fuel tank of the car shortly after the fuel gauge flashed at him, and after 305 miles it again demands attention. What do you think the fuel consumption is? What range of figures are valid answers to this question? See Going to Cambridge, too

6  Repeat the last question for 40 litres and 400 miles, paying particular attention to the range of possible answers.

Hopefully you now see the point of using a suitable level of precision.

Here is a question to ponder, which I don’t think has an easy answer:

What dates might conceivably be described as a year away from 2003? If you say you did something last year, when did you mean, if this is March? Might you have meant something different if this is October?

Answers: note the use of ≤ and <.
1.   14500, 15499
2.  1.450x1.150=1.6675, 1.55⁻x1.25⁻=1.9375⁻⁻ So to 4s.f. 1.668<area<1.937
3.  55m² ≤ area < 65m² => 7.417 < side < 8.062
4.  950kg ≤ tonne to 1s.f. < 1500kg,  995kg ≤ tonne to 2s.f. < 1050kg, 999.5kg≤tonne to 3s.f.<1005kg,  999.95kg ≤ tonne to 4s.f. < 1000.5kg,  
5.  31.5≤”32”<32.5;  304.5≤”305”<305.5;     304.5/32.5=9.37  < miles per litre @ 3s.f.<  305.5/31.5=9.70
6.  35≤"40”<45; 350≤“400”<450 [or possibly 395≤”400”<405]. gives   
    350/45 =  7.78 ≤miles per litre< 450/35 = 12.86.  [395/45 =  8.78 ≤miles per litre< 405/35 = 11.57] 

Q5 and Q6 could be in inverted units, litres per 100km, 100/answers shown => [10.3,10.67];   [7.78, 12.85] is amazingly close to the paired figures of [7.78,12.86]; [8.64, 11.38].

Dates: you might say 6 months either way, 20020601<“2003”<20040531. Many would call this wrong.Many would say that a year from the start of 2003 would be 20020101, so the interval woudl then be [20020101, 20041231]. But many poeople woudl say that an interval of three years cannot be right.what might “within a year of 2003" mean? If a substitute phrase was “around 2003” then having an error of half a year or all of a year seems reasonable. The issue here is taht we don’t have any idea of implied precision. Just becasue the year is an integer does not mean that integer precision is intended.

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