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

(Problem 5)
m (Solution)
 
(5 intermediate revisions by 4 users not shown)
Line 1: Line 1:
==Problem 5==
+
==Problem==
  
 
The data set <math>[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3 = 43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartle (<math>Q_1</math>) or more than <math>1.5</math> times the interquartile range above the third quartile (<math>Q_3</math>), where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?
 
The data set <math>[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3 = 43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartle (<math>Q_1</math>) or more than <math>1.5</math> times the interquartile range above the third quartile (<math>Q_3</math>), where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?
Line 6: Line 6:
  
 
==Solution==
 
==Solution==
 +
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 = 43 + 15 = 58</math> are also outliers, of which there are none so there is only <math>\boxed{\textbf{(B) 1}}</math> outlier in total.
  
 
==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}}

Latest revision as of 16:42, 27 September 2022

Problem

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 $\boxed{\textbf{(B) 1}}$ outlier in total.

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png