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Select the outlier in the data set.\newline3,52,54,55,57,58,60,74,833, 52, 54, 55, 57, 58, 60, 74, 83\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.\newline3,52,54,55,57,58,60,74,833, 52, 54, 55, 57, 58, 60, 74, 83\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: 3,52,54,55,57,58,60,74,833, 52, 54, 55, 57, 58, 60, 74, 83. To find the outlier, we can use the interquartile range (IQR) method or look for a number that is significantly different from the rest of the data.
  2. Find Outlier: Looking at the data set, 33 stands out as significantly lower than all other numbers, which are all above 5050. Therefore, 33 is the outlier.
  3. Calculate Mean with Outlier: Now, calculate the mean of the data set with the outlier included:\newlineMean = (3+52+54+55+57+58+60+74+83)/9(3 + 52 + 54 + 55 + 57 + 58 + 60 + 74 + 83) / 9\newlineMean = (446)/9(446) / 9\newlineMean ≈49.56\approx 49.56
  4. Calculate Mean without Outlier: Next, calculate the mean of the data set without the outlier:\newlineMean without outlier = (52+54+55+57+58+60+74+83)/8(52 + 54 + 55 + 57 + 58 + 60 + 74 + 83) / 8\newlineMean without outlier = (493)/8(493) / 8\newlineMean without outlier ≈61.63\approx 61.63
  5. Compare Means: Compare the two means to determine if the mean would increase or decrease upon the removal of the outlier:\newlineMean with outlier ≈49.56\approx 49.56\newlineMean without outlier ≈61.63\approx 61.63\newlineSince the mean without the outlier is greater than the mean with the outlier, the mean would increase if the outlier were removed.

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