SOLUTION: This is in reference to the Chip's OHoy Challenge. I would appreciate if you could tell me if this part is correct. I am determining Quartiles: 1087, 1098, 1103, 1121, 1132, 113

Algebra ->  Probability-and-statistics -> SOLUTION: This is in reference to the Chip's OHoy Challenge. I would appreciate if you could tell me if this part is correct. I am determining Quartiles: 1087, 1098, 1103, 1121, 1132, 113      Log On


   

Question 346669: This is in reference to the Chip's OHoy Challenge. I would appreciate if you could tell me if this part is correct. I am determining Quartiles:
1087, 1098, 1103, 1121, 1132, 1135, 1137,1143,1154,1166,1185,1191,1199,1200,1213,1214,1215,1219,1219,1228,1239,1244,1247,1258,1269,1270,1279,1293,1294,1295,1307,1325,1345,1356,1363,1377,1402,1419,1440,1514,1545,1546
I calculated the following
min: 1087
Q1 = 1185
Q2 median = 1241.5
Q3 is 1325
Max 1546
IQR = 1325-1185 = 140
1185-1.5*140=975
1325+1.5*140=1535
Can you tell me if I am doing this correctly? thank you so much for your help.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You are correct. Good job.


I'm assuming that the last two calculations are referring to the boundaries which define where the outliers lie.