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Select the outlier in the data set.\newline26,36,40,41,42,46,60,29826, 36, 40, 41, 42, 46, 60, 298\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.\newline26,36,40,41,42,46,60,29826, 36, 40, 41, 42, 46, 60, 298\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 data set. The data set is: 26,36,40,41,42,46,60,29826, 36, 40, 41, 42, 46, 60, 298. At a glance, 298298 is much larger than all other numbers, which are relatively close to each other. Therefore, 298298 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 data set. The sum of the data set is 26+36+40+41+42+46+60+298=58926 + 36 + 40 + 41 + 42 + 46 + 60 + 298 = 589. The number of values is 88. So, the mean is 589/8=73.625589 / 8 = 73.625.
  3. Calculate Mean without Outlier: Calculate the mean of the data set without the outlier. Remove the outlier 298298 and calculate the new sum: 26+36+40+41+42+46+60=29126 + 36 + 40 + 41 + 42 + 46 + 60 = 291. The new number of values is 77. So, the new mean is 2917=41.5714\frac{291}{7} = 41.5714 (rounded to four decimal places).
  4. Compare Means: Compare the means to determine if the mean would increase or decrease. The original mean with the outlier was 73.62573.625, and the new mean without the outlier is 41.571441.5714. Since the new mean is lower than the original mean, removing the outlier would decrease the mean.

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