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Difference between revisions of "2017 AMC 12B Problems/Problem 5"

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==Solution==
 
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The interquartile range is defined as <math>Q3 - Q1</math>, which is <math>43 - 33 = 10</math>. <math>1.5</math> times this value is <math>15</math>, so all values more than <math>15</math> below <math>Q1</math> = <math>33 - 15 = 18</math> is an outlier. The only one that fits this is <math>6</math>. All values more than <math>15</math> above <math>Q3</math> = <math>43 + 15 = 58</math> are also outliers, of which there are none so there is only <math>1 \boxed{\textbf{(B) }}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=B|num-b=4|num-a=6}}
 
{{AMC12 box|year=2017|ab=B|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:19, 16 February 2017

Problem 5

The data set $[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]$ has median $Q_2 = 40$, first quartile $Q_1 = 33$, and third quartile $Q_3 = 43$. An outlier in a data set is a value that is more than $1.5$ times the interquartile range below the first quartle ($Q_1$) or more than $1.5$ times the interquartile range above the third quartile ($Q_3$), where the interquartile range is defined as $Q_3 - Q_1$. How many outliers does this data set have?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

The interquartile range is defined as $Q3 - Q1$, which is $43 - 33 = 10$. $1.5$ times this value is $15$, so all values more than $15$ below $Q1$ = $33 - 15 = 18$ is an outlier. The only one that fits this is $6$. All values more than $15$ above $Q3$ = $43 + 15 = 58$ are also outliers, of which there are none so there is only $1 \boxed{\textbf{(B) }}$

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AMC 12 Problems and Solutions

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