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Select the outlier in the data set.\newline34,36,44,46,48,52,56,62,76134, 36, 44, 46, 48, 52, 56, 62, 761\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.\newline34,36,44,46,48,52,56,62,76134, 36, 44, 46, 48, 52, 56, 62, 761\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 data set.\newlineTo find the outlier, we can look for a number that is significantly different from the rest of the numbers in the set. The data set is: 34,36,44,46,48,52,56,62,76134, 36, 44, 46, 48, 52, 56, 62, 761. At a glance, 761761 is much larger than all other numbers, which are relatively close to each other. Therefore, 761761 is the outlier.
  2. Calculate Mean with Outlier: Calculate the mean of the data set with the outlier.\newlineTo calculate the mean, add all the numbers together and divide by the number of values in the set. The sum of the data set is 34+36+44+46+48+52+56+62+761=113934 + 36 + 44 + 46 + 48 + 52 + 56 + 62 + 761 = 1139. There are 99 numbers in the set, so the mean is 11399=126.555\frac{1139}{9} = 126.555\ldots (rounded to three decimal places).
  3. Calculate Mean without Outlier: Calculate the mean of the data set without the outlier. Remove the outlier (761761) and calculate the new mean. The new sum is 1139761=3781139 - 761 = 378. There are now 88 numbers in the set, so the new mean is 378/8=47.25378 / 8 = 47.25.
  4. Compare Means: Compare the two means to determine if the mean would increase or decrease with the removal of the outlier. The original mean with the outlier was 126.555126.555\ldots, and the new mean without the outlier is 47.2547.25. Since the original mean is higher, removing the outlier will decrease the mean.

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