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e https://wwwmandomst dent/playerTestasportesed-t?244470Aoentenena Army jenriSI 7/6/18 5:47 PM Online Statistics Summer 2018 est Test #2 Chapter 3 Test Submit Test 3 of 10 (2 complete ? This Test: 10 pts possible This Question: 1 pt of central tendency is the tiwnsed mean Ii is computed by determining the mean of a data sed after dolsting the smalest and largest observed the trimmed mean for the data given in the accompanying table Is the trimmed mean resislant to changes in the extreme values in the given dala? alues Compute ?] Cick the iconto view data iable The tinmed mean is Round to the nearest thousandth as needed ) Is the trimmed mean resistant to changes in the extreme values for the gven data? O No because the timmed mean varies subetantialy when te es eme values change Dala Table resistant to changes in the extreme vatues for the data 0930829 0 0B4 084 091 0s4 086 05085 088 087 069 9 085 087 088 083 096 087 093 091 085 O Yes, because changing the estreme values does not change the tinmmed mean 0 091086 089 087 084 088 088 089 077 082 083 090 0.88 084 093 0.01?90 :088 032 085?84 084 086 Print Done Cick to select vour answers Sample Tests an search DOLL 6 O PExplanation / Answer
Solution :
Given that 0.93,0.91,0.89,0.83,0.91,0.88,0.90,0.88,0.88,0.94,0.91,0.96,0.91,0.88,0.88,0.92,0.82,0.86,0.86,0.87,0.86,0.89,0.84,0.85,0.90,0.86,0.87,0.93,0.89,0.77,0.93,0.84,0.90,0.86,0.93,0.91,0.87,0.82,0.81,0.84,0.84,0.88,0.88,0.85,0.81,0.83,0.90,0.86,0.84,0.87
=> ordered form is 0.77,0.81,0.81,0.82,0.82,0.83,0.83,0.84,0.84,0.84,0.84,0.84,0.85,0.85,0.86,0.86,0.86,0.86,0.86,0.86,0.87,0.87,0.87,0.87,0.88,0.88,0.88,0.88,0.88,0.88,0.88,0.89,0.89,0.89,0.90,0.90,0.90,0.90,0.91,0.91,0.91,0.91,0.91,0.92,0.93,0.93,0.93,0.93,0.94,0.96
for trimmed mean the deleting the smallest and largest observed value then the trimmed set is 0.81,0.81,0.82,0.82,0.83,0.83,0.84,0.84,0.84,0.84,0.84,0.85,0.85,0.86,0.86,0.86,0.86,0.86,0.86,0.87,0.87,0.87,0.87,0.88,0.88,0.88,0.88,0.88,0.88,0.88,0.89,0.89,0.89,0.90,0.90,0.90,0.90,0.91,0.91,0.91,0.91,0.91,0.92,0.93,0.93,0.93,0.93,0.94
=> Trimmed mean ? = ? X / n where ? X - Sum of your Trimmed Set
n - Total Numbers in Trimmed set
=> ? X = sum of trimmed set = 0.81 + 0.81 + 0.82 + 0.82 + 0.83 +.....+ 0.94 = 42.02
=> n = 48
=> Trimmed mean ? = 42.02/48
= 0.875
=> option b. Yes, because changing the extreme values does not change the trimmed mean.
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