We always introduce errors in our measurements whenever we round or approximate numbers. In this lesson, we will explore how to quantify and describe measurement errors.

Suppose that a particular rope measures precisely 48.9\,\textrm{cm}. Now, imagine we take a meterstick and record a measurement of 50\,\textrm{cm} to the nearest 10 centimeters. Let's mark this approximation on our meterstick.

The error of a measurement equals the difference between the rounded estimate and the "true" value. So, in this example, the error of our approximation equals

50\,\textrm{cm} - 48.9\,\textrm{cm} = 1.1\textrm{cm}.

We often don't know the "true" measurement in practical problems. So, the next best thing is to determine a lower bound and an upper bound for our measurement:

  • The lower bound of a measurement is its smallest possible value.

  • The upper bound of a measurement is its largest possible value.

Let's figure out the lower and upper bounds of our 50\,\textrm{cm} approximation:

  • The lower bound is 45 \, \textrm{cm} because this is the smallest value that gives 50 \, \textrm{cm} when rounded to the nearest 10\,\textrm{cm}.
  • The upper bound is 55 \, \textrm{cm} because this is (almost) the largest value that gives 50 \, \textrm{cm} when rounded to the nearest 10\,\textrm{cm}.

Let's use the letter l to denote the "true" length of the rope. Then, we can combine our lower and upper bound measurements into a single statement, as follows:

45\,\textrm{cm} \leq l \lt 55\,\textrm{cm}

Notice that we used a strict "<" inequality for the upper bound. We cannot include 55\textrm{cm} in our set of values because this number rounds to 60\,\textrm{cm}, not 50\,\textrm{cm} .

We can visualize the lower and upper bounds on our diagram as follows:

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So, we've seen how to calculate lower and upper bounds. Before we consider another example, note the following:

  • The lower and upper bound depend on our approximation's degree of accuracy.

  • In the last example, the degree of accuracy was 10\,\textrm{cm} because the rope was measured to the nearest 10 centimeters.

  • If we change the degree of accuracy, the lower and upper bounds will also change. So, for example, if we measured the rope to the nearest 1\,\textrm{cm}, the lower and upper bounds would be different.

  • The greatest possible error is the largest amount the "true" value can be from the approximation.

  • This greatest possible error is always one-half of the degree of accuracy. So, in the last example, the greatest possible error in our measurement was

\dfrac{10\,\textrm{cm}}{2} = {\color{blue}{5\,\textrm{cm}}}.

We can use the greatest possible error to compute the lower and upper bounds quickly:

  • To compute the lower bound of an approximation, we subtract the greatest possible error from the approximation. So, for the previous example, we have \textrm{lower bound} = 50\,\textrm{cm} - {\color{blue}{5\,\textrm{cm}}} = 45\,\textrm{cm.}

  • To compute the upper bound of an approximation, we add the greatest possible error to the approximation. So, for the previous example, we have \textrm{upper bound} = 50\,\textrm{cm} + {\color{blue}{5\,\textrm{cm}}} = 55\,\textrm{cm.}

Let's see another example.

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The height h of a lamppost is 30\,\textrm{ft} to the nearest 1\,\textrm{ft}. Determine the upper and lower bounds of this measurement.

EXPLANATION

Our measurement is given to the nearest 1\,\textrm{ft}. So, the greatest possible error of this measurement is

\dfrac{1\,\textrm{ft}}{2} = 0.5\,\textrm{ft}.

Therefore, we can calculate the lower and upper bounds of this measurement as follows:

\begin{align*} \textrm{lower bound} &= 30\,\textrm{ft} - 0.5\,\textrm{ft}\\[5pt] & = 29.5\,\textrm{ft}\\[10pt] \textrm{upper bound} &= 30\,\textrm{ft} + 0.5\,\textrm{ft} \\[5pt] &= 30.5\,\textrm{ft} \end{align*}

Since the exact value of h lies between the lower and upper bound, we conclude that

29.5\,\textrm{ft} \leq h < 30.5\,\textrm{ft}.

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The distance $d$ between two cities is $10\,000\,\textrm{km}$ to the nearest $100\,\textrm{km}.$ Determine the upper and lower bounds of this measurement.

a
$9\,950\,\textrm{km} \leq d < 10\,050\,\textrm{km}$
b
$9\,500\,\textrm{km} \leq d < 10\,500\,\textrm{km}$
c
$9\,900\,\textrm{km} \leq d < 10\,100\,\textrm{km}$
d
$9\,000\,\textrm{km} \leq d < 11\,000\,\textrm{km}$
e
$9\,995\,\textrm{km} \leq d < 10\,005\,\textrm{km}$

The weight $w$ of a box of apples is $20\,\textrm{oz}$ to the nearest $1\,\textrm{oz}.$ Determine the upper and lower bounds of this measurement.

a
b
c
d
e

The area a of a golf course measures 0.6\,\textrm{km}^2 to the nearest tenth of a square kilometer. Determine the upper and lower bounds of this measurement.

EXPLANATION

Our measurement is given to the nearest 0.1\,\textrm{km}^2. So, the greatest possible error of this measurement is

\dfrac{0.1\,\textrm{km}^2}{2} = 0.05\,\textrm{km}^2.

Therefore, we can calculate the lower and upper bounds of this measurement as follows:

\begin{align*} \textrm{lower bound} &= 0.6\,\textrm{km}^2 - 0.05\,\textrm{km}^2\\[5pt] & = 0.55\,\textrm{km}^2\\[10pt] \textrm{upper bound} &= 0.6\,\textrm{km}^2 + 0.05\,\textrm{km}^2 \\[5pt] &= 0.65\,\textrm{km}^2 \end{align*}

Since the exact value of a lies between the lower and upper bound, we conclude that

0.55\,\textrm{km}^2 \leq a < 0.65\,\textrm{km}^2.

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The shoulder height of a moose $h$ measures $2.1\,\textrm{m}$ to the nearest tenth of a meter. Determine the upper and lower bounds of this measurement.

a
$1.1\,\textrm{m} \leq h < 3.1\,\textrm{m}$
b
$2.09\,\textrm{m} \leq h < 2.11\,\textrm{m}$
c
$2.095\,\textrm{m} \leq h < 2.105\,\textrm{m}$
d
$2.05\,\textrm{m} \leq h < 2.15\,\textrm{m}$
e
$2.0\,\textrm{m} \leq h < 2.2\,\textrm{m}$

The volume $v$ of acid in a test tube measures $10.25\,\textrm{mL}$ to the nearest hundredth of a milliliter. Determine the upper and lower bounds of this measurement.

a
b
c
d
e

The dimensions of a parking lot are 90\, \textrm{m} \times 50\, \textrm{m} , where each dimension is measured to the nearest 10\, \textrm{m}. Rounded to the nearest ten square meters, what is the upper bound for the area of the parking lot?

EXPLANATION

Each dimension is given to the nearest 10\,\textrm{m}. So, the greatest possible error of these measurements is

\dfrac{10\,\textrm{m}}{2} = 5\,\textrm{m}.

We wish to find the upper bound for the area of the parking lot. Therefore, we require the upper bound of each dimension.

The upper bounds for each dimension are as follows:

\begin{align*} & 90 \,\textrm{m} + 5 \,\textrm{m} = {\color{blue}{95\,\textrm{m}}} \\[5pt] & 50 \,\textrm{m} +5 \,\textrm{m} = {\color{red}{55\,\textrm{m}}} \end{align*}

So, the upper bound for the area of the parking lot is

\begin{align*} A &= {\color{blue}{95\,\textrm{m}}} \times {\color{red}{55\,\textrm{m}}} \\[5pt] &= 5\,225 \\[5pt] &\approx 5\,230 \, \textrm{m}^2 , \end{align*}

rounded to the nearest ten square meters.

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The dimensions of a rectangular soccer field are $50\, \textrm{m} \times 20\, \textrm{m}$, where each dimension is measured to the nearest $10\, \textrm{m}.$ What is the upper bound for the area of the soccer field?

a
$1\,250\textrm{ m}^2$
b
$1\,675\textrm{ m}^2$
c
$975\textrm{ m}^2$
d
$1\,125\textrm{ m}^2$
e
$1\,375\textrm{ m}^2$

The dimensions of a rectangular floor are $12\, \textrm{m} \times 3\,\textrm{m}$, where each dimension is measured to the nearest $1\, \textrm{m}.$ Rounded to the nearest square meter, what is the smallest possible area of the floor?

a
b
c
d
e

The dimensions of a package in the shape of a rectangular solid are $0.9 \,\textrm{m} \times 0.9 \,\textrm{m}\times 1.2 \,\textrm{m},$ where each dimension was measured to the nearest $0.1\, \textrm{m}.$ Rounded to the nearest tenth of a cubic meter, what is the upper bound for the volume of the package?

a
b
c
d
e
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