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Select the outlier in the data set.\newline72,80,82,83,85,86,88,92,61072, 80, 82, 83, 85, 86, 88, 92, 610\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.\newline72,80,82,83,85,86,88,92,61072, 80, 82, 83, 85, 86, 88, 92, 610\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 look for a number that is significantly different from the rest of the numbers in the data set. In this case, 610610 stands out as it is much larger than all other values which are all below 100100.
  2. Determine Outlier: Determine if 610610 is an outlier using the interquartile range (IQR) method or by observing the gap between it and the rest of the data.\newlineThe gap between 610610 and the next highest number, 9292, is very large. This suggests that 610610 is an outlier without needing to calculate the IQR.
  3. Calculate Mean with Outlier: Calculate the mean of the data set with the outlier included.\newlineMean = (72+80+82+83+85+86+88+92+610)/9(72 + 80 + 82 + 83 + 85 + 86 + 88 + 92 + 610) / 9\newlineMean = (1278+610)/9(1278 + 610) / 9\newlineMean = 1888/91888 / 9\newlineMean 209.78\approx 209.78
  4. Calculate Mean without Outlier: Calculate the mean of the data set without the outlier.\newlineMean without outlier = (72+80+82+83+85+86+88+92)/8(72 + 80 + 82 + 83 + 85 + 86 + 88 + 92) / 8\newlineMean without outlier = 668/8668 / 8\newlineMean without outlier = 83.583.5
  5. Compare Mean: Compare the two means to determine if the mean would increase or decrease when the outlier is removed.\newlineThe mean with the outlier is approximately 209.78209.78, and the mean without the outlier is 83.583.5. Since 209.78209.78 is greater than 83.583.5, removing the outlier will decrease the mean.

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