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Select the outlier in the data set.\newline25,28,42,57,74,79,84,88,71125, 28, 42, 57, 74, 79, 84, 88, 711\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.\newline25,28,42,57,74,79,84,88,71125, 28, 42, 57, 74, 79, 84, 88, 711\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: 25,28,42,57,74,79,84,88,71125, 28, 42, 57, 74, 79, 84, 88, 711. To find an outlier, we can use the interquartile range (IQR) method or look for a value that is significantly different from the rest of the data points.
  2. Observation of Outlier: By observation, 711711 is significantly higher than all other values in the data set, which makes it a clear outlier without needing to calculate the IQR\text{IQR}.
  3. Calculate Mean with Outlier: Calculate the mean of the data set including the outlier: (25+28+42+57+74+79+84+88+711)/9(25 + 28 + 42 + 57 + 74 + 79 + 84 + 88 + 711) / 9.\newlineMean with outlier = (25+28+42+57+74+79+84+88+711)/9=1188/9=132(25 + 28 + 42 + 57 + 74 + 79 + 84 + 88 + 711) / 9 = 1188 / 9 = 132.
  4. Calculate Mean without Outlier: Calculate the mean of the data set without the outlier: (25+28+42+57+74+79+84+88)/8(25 + 28 + 42 + 57 + 74 + 79 + 84 + 88) / 8.\newlineMean without outlier = (25+28+42+57+74+79+84+88)/8=477/8=59.625(25 + 28 + 42 + 57 + 74 + 79 + 84 + 88) / 8 = 477 / 8 = 59.625.
  5. Compare Means: Compare the two means to determine if the mean would increase or decrease upon the removal of the outlier. The mean without the outlier (59.62559.625) is less than the mean with the outlier (132132), so the mean would decrease if the outlier were removed.

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