8.2: Estimation by Clustering

Last updated
Save as PDF

Page ID: 48879

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

Learning Objectives

understand the concept of clustering
be able to estimate the result of adding more than two numbers when clustering occurs using the clustering technique

Cluster
When more than two numbers are to be added, the sum may be estimated using the clustering technique. The rounding technique could also be used, but if several of the numbers are seen to cluster (are seen to be close to) one particular number, the clustering technique provides a quicker estimate. Consider a sum such as

$32 + 68 + 29 + 73$

Notice two things:

There are more than two numbers to be added.

Clustering occurs.

Both 68 and 73 cluster around 70, so $68 + 73$ is close to $80 + 70 = 2(70) = 140$.
$The expression 32 + 68 + 29 + 71. 32 and 29 are grouped together with a line, and 68 and 71 are grouped in the same way.$
Both 32 and 29 cluster around 30, so $32 + 29$ is close to $30 + 30 = 2(30) = 60$.

The sum may be estimated by

$\begin{array} {rcl} {(2 \cdot 30) + (2 \cdot 70)} & = & {60 + 140} \\ {} & = & {200} \end{array}$

In fact, $32 + 68 + 29 + 73 = 202$.

Sample Set A

Estimate each sum. Results may vary.

$27 + 48 + 31 + 52$.

Solution

27 and 31 cluster near 30. Their sum is about $2 \cdot 30 = 60$.

48 and 52 cluster near 50. Their sum is about $2 \cdot 50 = 100$.

Thus, $27 + 48 + 31 + 52$ is about $\begin{array} {rcl} {(2 \cdot 30) + (2 \cdot 50)} & = & {60 + 100} \\ {} & = & {160} \end{array}$

In fact, $27 + 48 + 31 + 52 = 158.$.

Sample Set A

$88 + 21 + 19 + 91$.

Solution

88 and 91 cluster near 90. Their sum is about $2 \cdot 90 = 180$.

21 and 19 cluster near 20. Their sum is about $2 \cdot 20 = 40$.

Thus, $88 + 21 + 19 + 91$ is about $\begin{array} {rcl} {(2 \cdot 90) + (2 \cdot 20)} & = & {180 + 40} \\ {} & = & {220} \end{array}$

In fact, $88 + 21 + 19 + 91 = 219$.

Sample Set A

$17 + 21 + 48 + 18$.

Solution

17, 21, and 18 cluster near 20. Their sum is about $3 \cdot 20 = 60$.

48 is about 50.

Thus, $17 + 21 + 48 + 18$ is about $\begin{array} {rcl} {(3 \cdot 20) + 50} & = & {60 + 50} \\ {} & = & {110} \end{array}$

In fact, $17 + 21 + 48 + 18 = 104.$.

Sample Set A

$61 + 48 + 39 + 57 + 52$.

Solution

61 and 57 cluster near 60. Their sum is about $2 \cdot 60 = 120$.

48, 49, and 52 cluster near 50. Their sum is about $3 \cdot 50 = 150$.

Thus, $61 + 48 + 39 + 57 + 52$ is about $\begin{array} {rcl} {(2 \cdot 60) + (3 \cdot 50)} & = & {120 + 150} \\ {} & = & {270} \end{array}$

In fact, $61 + 48 + 39 + 57 + 52 = 267.$.

Sample Set A

$706 + 321 + 293 + 684$.

Solution

706 and 684 cluster near 700. Their sum is about $2 \cdot 700 = 1,400$.

321 and 293 cluster near 300. Their sum is about $2 \cdot 300 = 600$.

Thus, $706 + 321 + 293 + 684$ is about $\begin{array} {rcl} {(2 \cdot 700) + (2 \cdot 300)} & = & {1,400 + 600} \\ {} & = & {2,000} \end{array}$

In fact, $706 + 321 + 293 + 684 = 2,004.$.

Practice Set A

Use the clustering method to estimate each sum.

$28 + 51 + 31 + 47$

Answer: $(2 \cdot 30) + (2 \cdot 50) = 60 + 100 = 160$

Practice Set A

$42 + 39 + 68 + 41$

Answer: $(3 \cdot 40) + 70 = 120 + 70 = 190$

Practice Set A

$37 + 39 + 83 + 42 + 79$

Answer: $(3 \cdot 40) + (2 \cdot 80) = 120 + 160 = 280$

Practice Set A

$612 + 585 + 830 + 794$

Answer: $(2 \cdot 600) + (2 \cdot 800) = 1,200 + 1,600 = 2,800$

Exercises

Use the clustering method to estimate each sum. Results may vary.

Exercise $\PageIndex{1}$

$28 + 51 + 31 + 47$

Answer: $2(30) + 2(50) = 160 (157)$

Exercise $\PageIndex{2}$

$42 + 19 + 39 + 23$

Exercise $\PageIndex{3}$

$88 + 62 + 59 + 90$

Answer: $2(90) + 2(60) = 300 (299)$

Exercise $\PageIndex{4}$

$76 + 29 + 33 + 82$

Exercise $\PageIndex{5}$

$19 + 23 + 87 + 21$

Answer: $3(20) + 90 = 150 (150)$

Exercise $\PageIndex{6}$

$41 + 28 + 42 + 37$

Exercise $\PageIndex{7}$

$89 + 32 + 89 + 93$

Answer: $3(90) + 30 = 300 (303)$

Exercise $\PageIndex{8}$

$73 + 72 + 27 + 71$

Exercise $\PageIndex{9}$

$43 + 62 + 61 + 55$

Answer: $40 + 3(60) = 220 (221)$

Exercise $\PageIndex{10}$

$31 + 77 + 31 + 27$

Exercise $\PageIndex{11}$

$57 + 34 + 28 + 61 + 62$

Answer: $3(60) + 2(30) = 240 (242)$

Exercise $\PageIndex{12}$

$94 + 18 + 23 + 91 + 19$

Exercise $\PageIndex{13}$

$103 + 72 + 66 + 97 + 99$

Answer: $3(100) + 2(70) = 440 (437)$

Exercise $\PageIndex{14}$

$42 + 121 + 119 + 124 + 41$

Exercise $\PageIndex{15}$

$19 + 24 + 87 + 23 + 91 + 93$

Answer: $3(20) + 3(90) = 330 (337)$

Exercise $\PageIndex{16}$

$108 + 61 + 63 + 96 + 57 + 99$

Exercise $\PageIndex{17}$

$518 + 721 + 493 + 689$

Answer: $2(500) + 2(700) = 2,400 (2,421)$

Exercise $\PageIndex{18}$

$981 + 1208 + 1214 + 1006$

Exercise $\PageIndex{19}$

$23 + 81 + 77 + 79 + 19 + 81$

Answer: $2(20) + 4(80) = 360 (360)$

Exercise $\PageIndex{20}$

$94 + 68 + 66 + 101 + 106 + 71 + 110$

Exercises for Review

Exercise $\PageIndex{21}$

Specify all the digits greater than 6.

Answer: 7, 8, 9

Exercise $\PageIndex{22}$

Find the product: $\dfrac{2}{3} \cdot \dfrac{9}{14} \cdot \dfrac{7}{12}$.

Exercise $\PageIndex{23}$

Convert 0.06 to a fraction.

Answer: $\dfrac{3}{50}$

Exercise $\PageIndex{24}$

Write the proportion in fractional form: "5 is to 8 as 25 is to 40."

Exercise $\PageIndex{25}$

Estimate the sum using the method of rounding: $4,882 + 2,704$.

Answer: $4,900 + 2,700 = 7,600 (7,586)$

Search

Text Color

Text Size

Margin Size

Font Type