
Concept explainers
To calculate: Find the measures of variability and any outliers for the set of data.

Answer to Problem 21STP
Therefore, range=68,median=46,First Quartile=27,Third Quartile=46,Inter-Quartile range=19,Outlier=80 is the solution of data set.
Explanation of Solution
Given information: The given set of data is,
30,62,35,80,12,24,30,39,53,38
Formula Used:
mean=sum of the datanumber of datamedian={(n+1)2}thterm for odd number of datamedian=mean of{(n)2}thterm+ {(n+1)2}thterm for even number of datamode=maximum occurence of particular data
Calculation: The given data can be evaluated as,
30,62,35,80,12,24,30,39,53,38Rearranging the data in ascending order,12,24,30,30,35,38,39,53,62,80range=80−12=68median=39+532median=4612,24,30,30,35,38,39,53,62,80First Quartile=24+302First Quartile=27Third Quartile=39+532Third Quartile=46Inter-Quartile range=46-27Inter-Quartile range=19Calculation for Outliers,Inter-Quartile range×1.5=19×1.5=28.5Determine Outliers,27-28.5=−1.546+28.5=74.5Therefore, 80 is outlier.
Therefore, range=68,median=46,First Quartile=27,Third Quartile=46,Inter-Quartile range=19,Outlier=80 is the solution of data set.
Chapter 10 Solutions
Glencoe Math Accelerated, Student Edition
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