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Select the outlier in the data set.\newline8,74,80,85,87,94,96,998, 74, 80, 85, 87, 94, 96, 99\newlineIf the outlier were removed from the data set, would the mean increase or decrease?\newlineChoices:\newline(A)increase\newline(B)decrease

Full solution

Q. Select the outlier in the data set.\newline8,74,80,85,87,94,96,998, 74, 80, 85, 87, 94, 96, 99\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. In this case, the number 88 stands out as it is much lower than all other numbers.
  2. Determine Outlier: Determine if 88 is an outlier using the interquartile range (IQR) method or by observing the gap between it and the closest number in the set.\newlineThe gap between 88 and the next closest number, 7474, is 6666, which is significantly larger than the gaps between any other two consecutive numbers in the set. This suggests that 88 is indeed an outlier.
  3. Calculate Mean: Calculate the mean of the original data set.\newlineMean = (8+74+80+85+87+94+96+99)/8(8 + 74 + 80 + 85 + 87 + 94 + 96 + 99) / 8\newlineMean = (623)/8(623) / 8\newlineMean = 77.87577.875
  4. Remove Outlier: Remove the outlier (8)(8) and calculate the new mean of the remaining data set.\newlineNew Mean =(74+80+85+87+94+96+99)/7= (74 + 80 + 85 + 87 + 94 + 96 + 99) / 7\newlineNew Mean =(615)/7= (615) / 7\newlineNew Mean =87.857= 87.857
  5. Compare Means: Compare the original mean to the new mean to determine if it would increase or decrease.\newlineThe original mean was 77.87577.875, and the new mean without the outlier is 87.85787.857. Since the new mean is higher, removing the outlier would increase the mean.

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