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Select the outlier in the data set.\newline12,22,28,46,51,56,60,90,83712, 22, 28, 46, 51, 56, 60, 90, 837\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.\newline12,22,28,46,51,56,60,90,83712, 22, 28, 46, 51, 56, 60, 90, 837\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 given data set.\newlineThe data set is: 12,22,28,46,51,56,60,90,83712, 22, 28, 46, 51, 56, 60, 90, 837.\newlineAn outlier is a data point that is significantly different from the rest of the data. We can visually inspect the data set and see that 837837 is much larger than all other values.
  2. Calculate Mean with Outlier: Calculate the mean of the data set with the outlier.\newlineTo find the mean, add all the numbers together and divide by the number of data points.\newlineMean = (12+22+28+46+51+56+60+90+837)/9(12 + 22 + 28 + 46 + 51 + 56 + 60 + 90 + 837) / 9\newlineMean = (1202)/9(1202) / 9\newlineMean ≈133.56\approx 133.56
  3. Remove Outlier: Remove the outlier and calculate the new mean.\newlineWithout the outlier 837837, the data set is: 1212, 2222, 2828, 4646, 5151, 5656, 6060, 9090.\newlineNow calculate the mean of this new data set.\newlineNew Mean = (12+22+28+46+51+56+60+90)/8(12 + 22 + 28 + 46 + 51 + 56 + 60 + 90) / 8\newlineNew Mean = 121200\newlineNew Mean 121211
  4. Compare Means: Compare the two means to determine if the mean would increase or decrease after removing the outlier. The original mean with the outlier was approximately 133.56133.56. The new mean without the outlier is approximately 45.62545.625. Since the new mean is less than the original mean, removing the outlier would decrease the mean.

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