Difference between revisions of "2017 AMC 12B Problems/Problem 5"
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==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 | + | 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 17:42, 27 September 2022
Problem
The data set has median
, first quartile
, and third quartile
. An outlier in a data set is a value that is more than
times the interquartile range below the first quartle (
) or more than
times the interquartile range above the third quartile (
), where the interquartile range is defined as
. How many outliers does this data set have?
Solution
The interquartile range is defined as , which is
.
times this value is
, so all values more than
below
=
is an outlier. The only one that fits this is
. All values more than
above
are also outliers, of which there are none so there is only
outlier in total.
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
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 |
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