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Select the outlier in the data set.\newline10,17,21,30,42,43,49,71,81610, 17, 21, 30, 42, 43, 49, 71, 816\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.\newline10,17,21,30,42,43,49,71,81610, 17, 21, 30, 42, 43, 49, 71, 816\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: Step 11: Identify the outlier in the data set.\newlineData set: 10,17,21,30,42,43,49,71,81610, 17, 21, 30, 42, 43, 49, 71, 816\newlineOutliers are values that are significantly higher or lower than the rest of the data. Here, 816816 stands out as it is much higher than the other values.
  2. Calculate Mean with Outlier: Step 22: Calculate the mean of the data set including the outlier.\newlineMean = (10+17+21+30+42+43+49+71+816)/9(10 + 17 + 21 + 30 + 42 + 43 + 49 + 71 + 816) / 9\newlineMean = 1099/91099 / 9\newlineMean = 122.11122.11
  3. Calculate Mean without Outlier: Step 33: Calculate the mean of the data set without the outlier.\newlineMean without outlier = (10+17+21+30+42+43+49+71)/8(10 + 17 + 21 + 30 + 42 + 43 + 49 + 71) / 8\newlineMean without outlier = 283/8283 / 8\newlineMean without outlier = 35.37535.375
  4. Compare Mean: Step 44: Compare the two means to determine if the mean increases or decreases when the outlier is removed.\newlineSince 35.37535.375 (mean without outlier) is less than 122.11122.11 (mean with outlier), removing the outlier decreases the mean.

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