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Select the outlier in the data set.\newline10,12,18,38,60,65,69,86510, 12, 18, 38, 60, 65, 69, 865\newlineIf the outlier were removed from the data set, would the mean increase or decrease?\newlineChoices:\newline(A)increase\newline(B)decrease

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Q. Select the outlier in the data set.\newline10,12,18,38,60,65,69,86510, 12, 18, 38, 60, 65, 69, 865\newlineIf the outlier were removed from the data set, would the mean increase or decrease?\newlineChoices:\newline(A)increase\newline(B)decrease
  1. Identify Outlier: Identify the outlier in the data set.\newlineTo find the outlier, we can look for a number that is significantly different from the rest of the numbers in the set. In this case, the number 865865 stands out as it is much larger than all other numbers.
  2. Determine Outlier: Determine if 865865 is an outlier using the interquartile range (IQR) method or by observing the gap between it and the other numbers.\newlineThe gap between 865865 and the next highest number, 6969, is very large. This suggests that 865865 is an outlier without doing a detailed IQR calculation.
  3. Calculate Mean with Outlier: Calculate the mean of the data set with the outlier included.\newlineMean = (10+12+18+38+60+65+69+865)/8(10 + 12 + 18 + 38 + 60 + 65 + 69 + 865) / 8\newlineMean = 1137/81137 / 8\newlineMean = 142.125142.125
  4. Calculate Mean without Outlier: Calculate the mean of the data set without the outlier.\newlineMean without outlier = (10+12+18+38+60+65+69)/7(10 + 12 + 18 + 38 + 60 + 65 + 69) / 7\newlineMean without outlier = 272/7272 / 7\newlineMean without outlier 38.857\approx 38.857
  5. Compare Means: Compare the two means to determine if the mean would increase or decrease when the outlier is removed.\newlineThe mean with the outlier is 142.125142.125, and the mean without the outlier is approximately 38.85738.857. Since the mean without the outlier is less than the mean with the outlier, removing the outlier would decrease the mean.

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