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Many numbers come up as a result of a measurement
The measurements, however, are never exact
Sometimes the numbers we see are already rounded
Sometimes the numbers may be only rough estimates and/or projections
Today we learn how to operate with these uncertainties.
These are the digits that represent actual measurement.
There is a substantial difference between
"My weight is 154 pounds" and
"My weight is 154.00 pounds"
In the case of 154 pounds it may well be in fact 154.27, while
that cannot be the case when the number is 154.00
These two zeros are significant because they tell something.
While a non-zero digit is always significant,
a zero may be significant, and may be not.
That depends on what this zero wants to tell you.
A zero is \(not\) significant if it stays there only to keep the position for a correct order of magnitude. To check up: this non-significant zero disappears in scientific notation.
Otherwise a zero is significant.
Always significant:
nonzero digits;
zeros right from decimal point \(and\) right from a nonzero digit
example: 4.20, 3.00, but not 60, and not 0.01
Always non-significant:
Zeros to the left of the first non-zero digit. Example: 0.00036
Not significant unless stated otherwise:
zeros to the right of the last non-zero digit but before the decimal point. Examples: 500,000 or 340
Step 1 Decide about the last place to be kept.
Step 2 If the \(next\) digit is below 5, round down, while if the next digit is 5 and above, round up.
Example: 398.3 rounded to the nearest one is 398
Example: 398.3 rounded to the nearest ten is 400
These errors may be random and systematic.
Random errors are always with us. They are not predictable, but we typically can bound them.
Systematic errors are embedded in our measurement system. They are typically related to a problem with the system.
Is a two pound error of a scale big or small?
That may depend on what is weighted and why that thing is weighted
There are two ways to measure an error:
\( \text{absolute error} = \text{measured value} - \text{true value} \)
\( \text{relative error} = \frac{\text{measured value} - \text{true value}}{\text{true value}} \times 100\% = \frac{\text{absolute error}}{\text{true value} }\times 100\% \)
Both absolute and relative errors may be important.
Example: 2 pounds of absolute error is a small and acceptable error if you weight a truck because the relative error is small.
Example: 2 pounds of absolute error is too big and not acceptable if you weight a puppy because the relative error is big.
Example: 2 pounds of absolute error is too big and not acceptable if you weight an artificial satellite although the relative error may be small. It is the absolute error which may affect the trajectory badly.
Precision is about the amount of digits, while the accuracy is about being close to the true value.
Example. My true weight is 154 pounds.
A digital scale shows 151.23 pounds while a mechanical scale shows 155 pounds.
The digital scale is more precise,
while the mechanical is more accurate.
Rounding \( after \) the operations are performed.
Rules to follow
Addition/Subtraction:
round to the precision of the least precise number used.
Indeed, if you know one weight up to a pound, and another one up to a milligram,
you still know the sum no better than up to a pound.
Multiplication/division:
round such that the amount of significant digits in your answer is the same as the fewest in your operands.
Exercise 64, p. 161
At a hardware store, you buy a 55-kilogram bag of sand. You also buy a 1.25 kilograms of nails. What is the total weight of your purchases?
Solution. While clearly
\(55 + 1.25 = 56.25\),
this is not yet a good answer. One has to round it.
While you know the weight of the nails up to a hundredth of kilogram,
you know the weight of sand only up to a kilogram.
The least precise operand is the weight of sand,
and we have to round our sum \(55 + 1.25 = 56.25\) up to a kilogram.
Answer: 56 kilogram.
Each of 60,000 households in a city pays an average of $445.79 in property taxes. What is the total tax revenue?
Solution. We have to multiply two numbers.
While the number 445.79 has 5 significant digits,
the number
60,000= \(6 \times 10^4\) has only one significant digits.
Thus we have to round the product
\(60000 \times 445.79 = 26474000 \)
such that the result has only one significant digit.
Answer: \( \$ 30000000 = 3 \times 10^7 =30\) million dollars.