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Select the outlier in the data set.\newline15,25,57,58,59,82,94,57015, 25, 57, 58, 59, 82, 94, 570\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.\newline15,25,57,58,59,82,94,57015, 25, 57, 58, 59, 82, 94, 570\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: 15,25,57,58,59,82,94,57015, 25, 57, 58, 59, 82, 94, 570.\newlineAn outlier is a data point that is significantly different from the rest of the data. In this case, 570570 stands out as it is much larger than the other numbers.
  2. Calculate Mean with Outlier: Calculate the mean of the data set with the outlier included.\newlineMean = (15+25+57+58+59+82+94+570)/8(15 + 25 + 57 + 58 + 59 + 82 + 94 + 570) / 8\newlineMean = (960)/8(960) / 8\newlineMean = 120120
  3. Calculate Mean without Outlier: Calculate the mean of the data set without the outlier.\newlineMean without outlier = (15+25+57+58+59+82+94)/7(15 + 25 + 57 + 58 + 59 + 82 + 94) / 7\newlineMean without outlier = (390)/7(390) / 7\newlineMean without outlier = 55.7155.71 (rounded to two decimal places)
  4. Compare Means: Compare the means to determine if the mean would increase or decrease with the removal of the outlier. The mean with the outlier is 120120, and the mean without the outlier is approximately 55.7155.71. Removing the outlier significantly decreases the mean.

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