1    › Number, Measure & Accuracy           Limits of Accuracy
  ∙07  ›  Measurements & Accuracy
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You should have completed WL01 (Limits of Accuracy) before attempting this exercise...

...in the animation, you saw the distressing situation of the sofa that wouldn't fit...

...I'm sorry if that left you traumatised!

You have a 2.4 m gap that needs a sofa

You find a sofa that is 2.4 m - you think it should be perfect; so you buy it

When it arrives, it doesn't quite fit! Why?

Well the gap was actually 2.35238442734...cm (that is 2.4 m, rounded to the nearest 10 cm)

and the sofa was actually 2.441248466...cm (which is also 2.4 m rounded to the nearest 10 cm)

But a 2.441248466...cm SOFT WON'T fit into a 2.35238442734...cm GAP

    It is not possible to take any measurement with 100% accuracy, so any measurement must be taken to some degree of accuracy.
    For instance, when measuring the length of a shelf; then measurement may be accurate to the nearest 10 cm, to 1 d.p, to 2 d.p, to 6 s.f. etc.
  • So, all measurements should be accompanied by a statement of the accuracy to which the measure is taken. I.e.
    1. 120,000 (nearest 10,000)
    2. 235 (nearest 5)
    3. 42 (nearest whole number)
    4. 12.6 (1 d.p.)
    5. 12.62 (2 d.p.)
    6. 0.35 (2 s.f.)
    7. 3500 (2 s.f.)
  • The information inside the brackets tells us the ‘error’ in the number:
    1. For: 120,000 (nearest 10,000), the ‘error’ is 10,000
    2. For: 235 (nearest 5), the ‘error’ is 5
    3. 42 (nearest whole number), the ‘error’ is 1
    4. 12.6 (1 d.p.), the ‘error’ is 1/10. Why? 
      	
       tens  units  tenths
         │      │     │
         └───┐  │     │
             ▼  ▼     ▼
             1  2  .  6             Because the 1st decimal place is in the ⅒ths column
                     ╘╤╛
               1st decimal place
	        
    5. 12.62 (2 d.p.), the ‘error’ is 1/100. Why? 
      	
      tens  units  tenths  hundredths
       │      └─┐     │      │
       └─────┐  │     │  ┌───┘
             ▼  ▼     ▼  ▼
             1  2  .  6  2          Because the 2nd decimal place is in the Ή⁄₁₀₀ths column
                        ╘╤╛
                  2nd decimal place
        
    6. 0.35 (2 s.f.), the ‘error’ is 1/100. Why? 
      	
          units  tenths  hundredths
            │      │        │
            └┐     │  ┌─────┘
             ▼     ▼  ▼
             0  .  3  5             Because the 2nd significant figure is in the Ή⁄₁₀₀ths column
                     ╘╤╛
             2nd significant figure
	        
    7. 3500 (2 s.f.), the ‘error’ is 100. Why? 
      	
 thousands hundreds tens units
      │       └────┐  │    │
      └─────────┐  │  │  ┌─┘
                ▼  ▼  ▼  ▼
                3  5  0  0          Because the 2nd significant figure is in the 100s column
                  ╘╤╛
          2nd significant figure
	        
  • Once we have found the ‘error’, we can find the ‘maximum absolute error’ (M.A.E.), the ‘upper bound’ (U.B.) and the ‘lower bound’ (L.B.) using:
    • M.A.E. = error χ 2
    • U.B. = number + M.A.E.
    • L.B. = number – M.A.E.
  • We can then say: L.B. < number < U.B.
  • This statement explains the range of possible numbers that could give rise to the original measurement
e.g. Find the U.B. and L.B. of: 4.2 (1 d.p.)

Firstly, let's determine the ERROR:

	
              units tenths
                │     │
                │     │
                ▼     ▼
                4  .  2             
                     ╘╤╛
               1st decimal place
            is in the tenths column
               So the ERROR is ⅒ or 0.1
	        

Next let's find the Maximum Absolute Error:

	
                       M.A.E. = 0.1 χ 2 = 0.05

Finally, let's find the Lower Bound and Upper Bound:

	
                                         M.A.E.
                               ╔═══════ + 0.05 ═══════► U.B. = 4.25
                 Number        ║
         ════════ 4.2 ═════════╣
                               ║
                               ╚═══════ - 0.05 ═══════► L.B. = 4.15
                                         M.A.E.
	        

So, we can say: 4.15 ≤ 4.2 (1 d.p.) < 4.25

Which means that 'any number between 4.15 and 4.25 (not including 4.25) could give the answer 4.2 (1 d.p.)'

 

e.g. Find the U.B. and L.B. of: 0.25 (2 d.p.)

Firstly, let's determine the ERROR:

	
              units tenths hundredths
                └┐     │       │
                 │     │  ┌────┘
                 ▼     ▼  ▼
                 0  .  2  5            
                 ╘╤╛
                   2nd decimal place
               is in the hundreths column
                   So the ERROR is Ή⁄₁₀₀ or 0.01

Next let's find the Maximum Absolute Error:

	
                     M.A.E. = 0.01 χ 2 = 0.005

Finally, let's find the Lower Bound and Upper Bound:

	
                                         M.A.E.
                               ╔══════ + 0.005 ═══════► U.B. = 0.255
                 Number        ║
         ════════ 0.25 ════════╣
                               ║
                               ╚══════ - 0.005 ═══════► L.B. = 0.245
                                         M.A.E.

So, we can say: 0.245 ≤ 0.25 (2 d.p.) < 0.255

Which means that 'any number between 0.245 and 0.255 (not including 0.255) could give the answer 0.25 (2 d.p.)'

 

e.g. Find the U.B. and L.B. of: 1380 (3 s.f.)

Firstly, let's determine the ERROR:

	
        thousands  hundreds  tens  units
              │       └────┐  │      │
              └─────────┐  │  │  ┌───┘
                        ▼  ▼  ▼  ▼
                        1  3  8  0            
                             ╘╤╛
                  2nd significant figure
                      So the ERROR is 10

Next let's find the Maximum Absolute Error:

	
                          M.A.E. = 10 χ 2 = 5

Finally, let's find the Lower Bound and Upper Bound:

	
                                         M.A.E.
                               ╔═════════ + 5 ═══════► U.B. = 1385
                 Number        ║
         ════════ 1380 ════════╣
                               ║
                               ╚═════════ - 5 ═══════► L.B. = 1375
                                         M.A.E.

So, we can say: 1375 ≤ 1380 (3 s.f.) < 1385

Which means that 'any number between 1375 and 1385 (but not including 1385) could give the answer 1380 (3 s.f.)'

 

Question 1: So, we need to round all of the numbers given to 1 d.p. and see which give 5.6 (1 d.p.): 

	
                                                    ┌───────────────────────────────┐
 5.00     5.15     5.49     5.50     5.51     5.53  │  5.55     5.56     .... 5.64  │  5.65     5.66
└─┬──┘   └─┬──┘   └─┬──┘   └─┬──┘   └─┬──┘   └─┬──┘ │ └─┬──┘   └─┬──┘   └────┬────┘ │ └─┬──┘   └─┬──┘
 5.0      5.2      5.5      5.5      5.5      5.5   │  5.6      5.6         5.6     │  5.7      5.7
                                                    └───────────────┬───────────────┘
                                                    ┌───────────────┴───────────────┐
                                                    │These are the numbers that,when│
                                                    │rounded to 1 d.p. give: 5.6 kg │
                                                    └───────────────────────────────┘

So, if the reading on the first (less accurate) scale was 5.6 kg (1 d.p.), then the ACTUAL mass of the sand could have been anything between 5.55 kg and 5.64 kg

Note: In fact, it could be even a little higher than 5.64 kg, because 5.649 kg ALSO rounds down to 5.6 kg - in fact anything up to (but not including 5.65 kg)

 

 

 

Question 2: You shouldn't have any problem rounding each of the numbers to the nearest £100...

And, you should find that any of the numbers between £2,450 and £2,549 will all round to give £2,500 (nearest £100)

Note: In fact, it could even be a little higher than £2,549, because £2,549.99 also rounds down to £2,500 - so in fact, the ACTUAL profit could be anything up to £2,550 (but not including £2,550) 

 

 

 

Question 3: Your list will be like this: 

	
2.70   2.71    2.72    2.73    2.74    2.75    2.76    2.77    2.78    2.79    2.80    2.81  etc…

And when you highlight the numbers that round to give 2.8 (1 d.p.): 

      ┌──────────────────────────────────────────────────────────────────────────────┐
2.74  │2.75    2.76    2.77    2.78    2.79    2.80    2.81    2.82    2.83    2.84  │2.85    2.86
      └─┼────────────────────────────────────────────────────────────────────────────┘ │
        └────────────────────────────────┐              ┌──────────────────────────────┘
                                         ▼              ▼
         ╒══════════════════════════╕   2.75  ≤  x  <  2.85 	╒═══════════════════════════════╕
         │ 2.75 is the first number │  └─┬──┘         └─┬──┘  │ Even though 2.85 doesn't      │
         │ from our list that does  ├────┘              └─────┤ actually round to 2.8 (1 d.p.)│
         │ round up to 2.8 (1 d.p.) │                         │ we still put 2.85 at the end  │
         ╘══════════════════════════╛                         │ of our inequality… WHY?       │
                                                              ╘═══════════════╤═══════════════╛
                                                        ╒═════════════════════╧══════════════════════╕
                                                        │ The reason for writing 2.85 at the end is: │
                                                        │ Although 2.85 doesn't round to 2.8 (1.d.p.)│
                                                        │          2.849   DOES round to 2.8 (1 d.p.)│
                                                        │ So does: 2.8499                            │
                                                        │ So does: 2.84999                           │
                                                        │ And also 2.8499999999999999999……           │
                                                        │ So does anything up to (not including 2.85)│
                                                        ╘════════════════════════════════════════════╛

The inequality tells us something very important!

When a person measures a length and tell us it is 2.8 m (1 d.p.), then the inequality: 2.75 m ≤ x < 2.85 m tells us the ACTUAL length of the room might be anywhere between 2.75 m and (up to, but not including) 2.85 m

There is a better way of getting to this inequality than this long winded working with a big list of numbers that are rounded to see which do and which don't round to the value we are given:

The method is called: Limits of Accuracy:

Here's how it works: (using the value give in the question: x = 2.8 m (1 d.p.) )

The accuracy of the measure (in this case it is to: 1 d.p.) allows us to determine the ERROR: 

	
      units tenths
        │     │
        │     │                 ┌───────────────────────────────────────────────────────┐
        ▼     ▼                 │ Note: The fact that this DIGIT is ‘8’ is irrelevant   │
        2  .  8  ◄──────────────┤ ↝↝↝↝	We are only interested in which column it is in │
             ╘╤╛                │   ⟿	It is in the tenths column, so the error is ⅒	│
       1st decimal place        └───────────────────────────────────────────────────────┘
    is in the 1/10ths column:
     So the ERROR is 1/10 or 0.1

Next let's find the Maximum Absolute Error:

	
        M.A.E. = error χ 2
 
        M.A.E. =  0.1  χ 2 = 0.05

Finally, let's find the Lower Bound and Upper Bound:

	
                                        ┌───────┐
                                        │ Error │
                                        │   =   │
                                        │  0.1  │
                                        └───┬───┘
                                            ▼
                                            │χ2
                                            ▼
                                        ┌───┴───┐
                                        │ M.A.E.│
                                        │   =   │
                                        │  0.05 │
                                        └───┬───┘
                                            ▼
             Approximate                 ┌──┴──┐
                Number         ╔═════► + │0.05 │ ═══════► U.B. = 2.85
               ┌──┴──┐         ║         └─────┘
      ════►════│ 2.8 │════►════╣            │
               └─────┘         ║         ┌──┴──┐   
                               ╚═════► ─ │0.05 │ ═══════► L.B. = 2.75
                                         └─────┘

So, we can say: 2.75 m ≤  x < 2.85 m

Which means that any number between 2.75 m and 2.85 m (but not including 2.85 m) will round to give 2.8 m (1 d.p.)

 

 

 

Question 4: So, we can use the same method (Limits of Accuracy) as we did in question 3:

t = 12.5 m (1 d.p.) 

The accuracy of the measure (in this case it is to: 1 d.p.) allows us to determine the ERROR: 

   tens units tenths
     │    │     │
     │    │     │               ┌───────────────────────────────────────────────────────┐
     ▼    ▼     ▼               │ Note: The fact that this DIGIT is ‘5’ is irrelevant   │
     1    2  .  5  ◄────────────┤ ↝↝↝↝	We are only interested in which column it is in	│
               ╘╤╛              │   ⟿ 	It is in the tenths column, so the error is ⅒	│
         1st decimal place      └───────────────────────────────────────────────────────┘
      is in the 1/10ths column:
       So the ERROR is ⅒ or 0.1

Next let's find the Maximum Absolute Error:

        M.A.E. = error χ 2
 
        M.A.E. =  0.1  χ 2 = 0.05

Finally, let's find the Lower Bound and Upper Bound:

                                        ┌───────┐
                                        │ Error │
                                        │   =   │
                                        │  0.1  │
                                        └───┬───┘
                                            ▼
                                            │χ2
                                            ▼
                                        ┌───┴───┐
                                        │ M.A.E.│
                                        │   =   │
                                        │  0.05 │
                                        └───┬───┘
                                            ▼
                                         ┌──┴──┐
                Number          ╔════► + │0.05 │ ═══════► U.B. = 12.55
               ┌──┴───┐         ║        └─────┘
      ════►════│ 12.5 │════►════╣           │
               └──────┘         ║        ┌──┴──┐   
                                ╚════► ─ │0.05 │ ═══════► L.B. = ....
                                         └─────┘

So, we can say: .... m ≤  t < 12.55 m

Which means that any number between ... m and 12.55 m (but not including 12.55 m) will round to give 12.5 m (1 d.p.)

 

 

 

Question 5, part (a): Firstly - the NUMBER OF SPECTATORS is really something you ‘COUNT’, not something you ‘MEASURE’ - which means, in theory (but in this case, not in practice), it is possible to know the answer with 100% precision (measurements can never be made with 100% precision)...

...But, as I said, in this case, how could you realistically know how exactly people watched the football?

x = 75,000 (nearest 1000)

Firstly, we need to determine the ERROR:

Note: When ever the accuracy is quoted as: nearest ......, then the number in the brackets (after the word nearest) is the ERROR

So, in this case: x = 75,000 (nearest 1000) - the ERROR is 1000

Now we use the same method as we did in Question 4..

 

 

Question 5, part (b): There's two things we need to consider here:

1) How did the 2nd newspaper know the number of spectators so precisely?
Remember, ‘spectators’ means anyone wot watched the match, either at the stadium, or on TV/streaming services.  How can they know if I watched it? Can they tell if my T.V. is displaying the football or not? And, even if they can, how can they tell if my friend came round to watch it with me? Does the T.V. have a spy camera in it? So the answer is - unless we live in communist China, this is ‘SPURIOUS’ accuracy (i.e. pretending a level of precision that is actually unattainable)

2) In any case, does this value fit within the error range of the previous value: If so, at least they are consistent with each other...

 

 

 

Question 6: d = 1.25 mm (2 d.p.)

Firstly, let's determine the ERROR:

      units tenths hundredths
        └┐     │       │
         │     │    ┌──┘                ┌───────────────────────────────────────────────────────┐
         ▼     ▼    ▼                   │ Note: The fact that this DIGIT is ‘5’ is irrelevant   │
         1  .  2    5  ◄────────────────┤ ↝↝↝↝	We're only interested in which column it is in  │
                   ╘╤╛                  │   ⟿	It's in the hundreths column: the error is Ή⁄₁₀	│
            2nd decimal place           └───────────────────────────────────────────────────────┘
         is in the 1/100th column
           So the ERROR is 1/100 or 0.01

So the M.A.E. = … 

Which makes the L.B. = 1.245 and the U.B. = …

The range of possible values is: 1.245 mm < d < ...... mm

 

 

 

Question 7: x = 40 (nearest 10)

So, using the rule we discovered in question 5, the ERROR must be 10

 

 

Question 9: We need to consider two things:

1) How many cars we can load onto the transporter without it breaking because we overloaded it!

2) How many cars we can actually fit along the length of the transporter

Hypothetically, if it turns out we can fit 14 cars onto it, but that it will break if we try to load it will anything more than 12 cars, then the answer would be 12 (not 14) - I'm hoping that's obvious...

Now - lets start by considering how many cars we can dump on it without overloading it:

Now, if we didn't know about LIMITS of ACCURACY, and we'd use this calculation:

	
                            1 car ————————————————► 1700 kg
                 Χ14000	                             
                   1700	 ___ cars ◄————————————— 14,000 kg
  	 

So the calculation would be: \(\color{#62be53}{1} \times \,\,\frac{\color{#2b83c3}{14000}}{\color{#bd398c}{1700}}\)...

...BUT, we now need to consider what

	
                                                        ╒════════════════════╕
      ╒═════════════╕                                    This is approximate:
       This is EXACT                                     LB = 1650, UB = 1750
      ╘══════╦══════╛                                   ╘═════════╦══════════╛
             ╚═══════════════╗                        ╔═══════════╝
                             ▼                        ▼    
                            1 car ————————————————► 1700 kg
                 Χ14000	                             
                   1700	 ___ cars ◄————————————— 14,000 kg
                                                      ▲    
                                                      ╚═══════════╗
                                                         ╒════════╩═════════╕
                                                          And this: It has a
                                                          L.B. & an U.B. too 
                                                         ╘══════════════════╛  	 

So, we need to decide what values (L.B. or U.B.) to use in place of each of the numbers in our calculation...

...we are after the SMALLEST possible answer (that way, we can be certain the transporter will be able to cope with more than we give it...)

 

Now, we also need to do the calculation for how many cars will fit along the length of the transporter...

 

 

 

 

Question 10: If we didn't know about LIMITS of ACCURACY, and we wanted to calculate the length of wool in the larger pack, we'd use the Scale Factor Method: 

	
                            10 g ———————————————► 20 m
                                                       	Χ50
                            50 g ———————————————► _____	 10

So the calculation would be: \(\color{#62be53}{20} \times \,\,\frac{\color{#2b83c3}{50}}{\color{#bd398c}{10}}\)...

...BUT, we now need to consider what happens when these numbers are approximate:

	
   ╒════════════════════╕                          ╒════════════════════════╕
    This is approximate:                            This is approximate too:
    LB = 9.5 & UB = 10.5                            L.B. = 19.5, U.B. = 20.5
   ╘═════════╦══════════╛                          ╘═══════════╦════════════╛
             ╚═══════════════╗                     ╔═══════════╝
                             ▼                     ▼                    
                            10 g ———————————————► 20 m
                                                       	Χ50
                            50 g ———————————————► _____	 10 
                             ▲    
             ╔═══════════════╝
    ╒════════╩═════════╕
     And this: It has a
     L.B. & an U.B. too 
    ╘══════════════════╛
	  

So, we need to decide what values (L.B. or U.B.) to use in place of each of the numbers in our calculation...

...we are after the SMALLEST possible answer (that way, we can be certain it will be longer than we expect...)

Once you've done that - there's one more calculation to find how many belts you can knit...