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Select the outlier in the data set.\newline38,48,50,56,60,63,73,85,99738, 48, 50, 56, 60, 63, 73, 85, 997\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.\newline38,48,50,56,60,63,73,85,99738, 48, 50, 56, 60, 63, 73, 85, 997\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: 38,48,50,56,60,63,73,85,99738, 48, 50, 56, 60, 63, 73, 85, 997.\newlineAn outlier is a data point that is significantly different from the rest of the data. In this case, 997997 is much larger than all other numbers in the set.
  2. Calculate Mean with Outlier: Calculate the mean of the data set including the outlier.\newlineMean = (38+48+50+56+60+63+73+85+997)/9(38 + 48 + 50 + 56 + 60 + 63 + 73 + 85 + 997) / 9\newlineMean = (1470)/9(1470) / 9\newlineMean = 163.33163.33 (rounded to two decimal places)
  3. Remove Outlier, Calculate New Mean: Remove the outlier and calculate the new mean of the data set.\newlineNew data set without the outlier: 38,48,50,56,60,63,73,8538, 48, 50, 56, 60, 63, 73, 85.\newlineNew mean = (38+48+50+56+60+63+73+85)/8(38 + 48 + 50 + 56 + 60 + 63 + 73 + 85) / 8\newlineNew mean = (473)/8(473) / 8\newlineNew mean = 59.1359.13 (rounded to two decimal places)
  4. Compare Means: Compare the means to determine if the mean would increase or decrease upon removal of the outlier. The original mean with the outlier was 163.33163.33, and the new mean without the outlier is 59.1359.13. Since the new mean is lower than the original mean, removing the outlier would decrease the mean.

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