Significant Figures, Estimating and Error Bounds



Significant Figures, Estimating and Error Bounds

Significant Figures (sig. figs.) are an incredibly useful tool, as if they are used correctly; you can avoid ‘rounding errors’ in multi-stage calculations. For example, if all your working is accurate to 4 sig. figs. then your answer will be correct to 3 sig. figs.

To round to any given number of significant figures; find the first ‘non-zero’ digit in your number – this is your first significant figure (and every digit after this is significant, even zeros). Count along the correct number of digits (same as the number of significant figures you are rounding too) and round normally.

Check your answer makes sense – if you have rounded 32745 to 2 sig. figs. and got an answer of 38 that isn’t sensible. Your answers should be of a similar size (magnitude) to your original number.

Examples

Important – always state the accuracy of your answer

32.14 = 32.1 (to 3 sig. figs.), don’t just write 32.14 = 32.1 as this is not correct.

Round 3471 to 1 significant figure

First sig. fig. is 3 which is in the 1000s column so we round to the nearest 1000

3471 = 3000 (to 1 sig. fig.)

Rouind 305 to 2 sig. figs.

First sig. fig. is 3. The second sig. fig. is ‘0’ (remember all figures are significant after the first one) which is in the tens column so we round to the nearest 10

305 = 310 (to 2 sig. figs.)

Round 0.02304 to 2 sig. figs.

First sig. fig. is the ‘2’ (remember zeros don’t count until you have found your first ‘non-zero’ digit) and the second sig. fig. is the ‘3’ in the 1000th s column so we are rounding to the nearest 1000th (or in effect to 3 d. p.)

0.02304 = 0.023 (to 2 sig. figs.)

Now a few without the explanation:

2.0048 = 2.00 (to 3 sig. figs.) 0.00000482 = 0.000005 (to 1 sig. fig.)

8.99999999 = 9.00 (to 3 sig. figs.) 0.010456 = 0.010 (to 2 sig. figs.)

10006 = 10010 (to 4 sig. figs.) 7041.145 = 70 (to 2 sig. figs.)

Top Tip – If you are having difficulty rounding the small decimal numbers, write your answer in Standard Form, round. and then write out ‘in full’.

E.g. Round 0.002547 to 2 significant figures

0.002547 = 2.547 x 10-3

= 2.5 x 10-3 (to 2 sig. figs.)

= 0.0025 (to 2 sig. figs.)

Past Paper Questions

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a) [pic]

b) 23.00559932 = 23.0 (to 3 sig. figs.)

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a) [pic]

b) 2.356213282 = 2.36 (to 3 sig. figs.)

Estimation

This is a skill we use daily without ever realising it (how many times do you go to the shops and not have enough money to pay? – not often I would guess).

The whole point of estimation is to give us an easy way to find an approximate answer which can be used to check a calculation.

We do this by rounding all the components to 1 significant figure and then performing the calculation.

Very Important – you must show your working or you will get no marks awarded for your answer.

Past Paper Questions

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18 x £2.15 ≈ 20 x £2

≈ £40

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71 x 19p ≈ 70 x 20p

≈ 140p

Error Bounds

When we measure things, we often work to a given accuracy (e.g. nearest centimetre). This called the tolerance.

We need to know what the smallest and largest possible values can be.

Example

A man is 180cm tall to the nearest 10cm. What are the least and greatest possible heights of the man?

The smallest he can be is 175cm, anything less than this will round down to 170cm.

The tallest he can be is 184.999999999999…..cm.

However from the chapter on rational numbers you should be able to show that: 184.9999999…. = 185 cm (bizarre isn’t it?) so the tallest he can be is 185cm (I know this should round up to 190cm but 184.9999999….would round down).

The difference between the maximum and minimum heights is 10cm – the same as the tolerance (works with continuous data).

Try this one……

A worm is 6.3cm long measured to the nearest millimetre. Find the minimum and maximum possible lengths of the worm.

Minimum length = 6.25cm

Maximum length = 6.35cm (same as 6.344444444444444……cm)

Past Paper Questions

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For 1 packet:

Minimum weight = 345 grams

Maximum weight = 355 grams

For six packets:

Minimum weight = 6 x 345 grams = 2070 grams

Maximum weight = 6 x 255 grams = 2130 grams

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i) least possible number of students = 6250

ii) greatest possible number of students = 6349

(note – can not be 6350 as we are dealing with discrete not continuous data)

Higher GCSE standard questions

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Expression is smallest when numerator has its smallest possible value and denominator has its largest possible value.

Denominator has largest possible value when it is ‘largest possible – smallest possible’

Smallest value [pic]

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If the lift can safely carry the boxes then the maximum weight of the boxes must be less than the smallest possible value for the maximum safe load

Maximum weight of boxes = 3 x 140.5 + 7 x 149.5

= 421.5 + 1046.5

= 1468 kg

Minimum value of the ‘maximum safe load’ = 1450kg

The maximum weight of the boxes could exceed the minimum value for the maximum safe load so they could not be lifted safely.

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Maximum possible difference occurs when Mark is at maximum possible height and Eileen is at minimum possible height

Maximum possible difference = 203.5 – 184.5

= 19cm

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Maximum price for one cubic metre of soil occurs when he gets the least volume of soil at the highest possible cost

Max. price per cubic metre = £82.49 ÷ 3.5

= £23.57

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Minimum safe load weight for lorry is 5150kg

Maximum possible weight of box 115.5kg

Greatest number of boxes which can be carried safely = 5150 ÷ 115.5

= 44 boxes

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Speed = distance ÷ time

Maximum speed when distance is at maximum and time is at minimum.

Maximum distance in metres = 405

Maximum distance in miles = 405 x 100 ÷ (12 x 2.54) ÷ 5280

(change from metres to centimetres to feet and finally into miles)

Minimum possible time in seconds = 45.5

Minimum possible time in hours = 45.5 ÷ 60 ÷ 60

(change from seconds to minutes to hours)

Maximum possible speed = (405 x 100 ÷ (12 x 2.54) ÷ 5280) ÷ (45.5 ÷ 60 ÷ 60)

= 19.9 m.p.h.

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