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Select the outlier in the data set.\newline11,16,24,36,80,95,99,54711, 16, 24, 36, 80, 95, 99, 547\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.\newline11,16,24,36,80,95,99,54711, 16, 24, 36, 80, 95, 99, 547\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: 11,16,24,36,80,95,99,54711, 16, 24, 36, 80, 95, 99, 547.\newlineTo find the outlier, look for a number that is significantly different from the rest. Here, 547547 stands out as it is much larger than the other numbers.
  2. Calculate Mean with Outlier: Step 22: Calculate the mean of the data set including the outlier.\newlineMean = (11+16+24+36+80+95+99+547)/8(11 + 16 + 24 + 36 + 80 + 95 + 99 + 547) / 8\newlineMean = 908/8908 / 8\newlineMean = 113.5113.5
  3. Calculate Mean without Outlier: Step 33: Calculate the mean of the data set without the outlier.\newlineMean without 547=(11+16+24+36+80+95+99)/7547 = (11 + 16 + 24 + 36 + 80 + 95 + 99) / 7\newlineMean without 547=361/7547 = 361 / 7\newlineMean without 547=51.57547 = 51.57
  4. Compare Mean Changes: Step 44: Compare the two means to determine if the mean increases or decreases when the outlier is removed.\newlineMean with outlier = 113.5113.5\newlineMean without outlier = 51.5751.57\newlineSince 113.5113.5 is greater than 51.5751.57, removing the outlier decreases the mean.

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