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Select the outlier in the data set.\newline17,22,28,36,44,62,70,78,45817, 22, 28, 36, 44, 62, 70, 78, 458\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.\newline17,22,28,36,44,62,70,78,45817, 22, 28, 36, 44, 62, 70, 78, 458\newlineIf the outlier were removed from the data set, would the mean increase or decrease?\newlineChoices:\newline(A)increase\newline(B)decrease
  1. Identify the Outlier: Identify the outlier in the given data set.\newlineThe data set is: 17,22,28,36,44,62,70,78,45817, 22, 28, 36, 44, 62, 70, 78, 458.\newlineAn outlier is a data point that is significantly different from the other data points in a set. We can often spot an outlier by looking for a number that is much larger or smaller than the rest.\newlineIn this case, 458458 is much larger than all other numbers in the set, so it is the outlier.
  2. Calculate Mean with Outlier: Calculate the mean of the data set with the outlier included.\newlineTo find the mean, add all the numbers together and divide by the number of data points.\newlineMean = (17+22+28+36+44+62+70+78+458)/9(17 + 22 + 28 + 36 + 44 + 62 + 70 + 78 + 458) / 9\newlineMean = (815)/9(815) / 9\newlineMean 90.56\approx 90.56
  3. Calculate Mean without Outlier: Calculate the mean of the data set without the outlier.\newlineRemove the outlier 458458 and calculate the new mean.\newlineMean without outlier = (17+22+28+36+44+62+70+78)/8(17 + 22 + 28 + 36 + 44 + 62 + 70 + 78) / 8\newlineMean without outlier = (357)/8(357) / 8\newlineMean without outlier 44.63\approx 44.63
  4. Compare Mean Results: Compare the means to determine if the mean would increase or decrease with the removal of the outlier. The original mean with the outlier was approximately 90.5690.56, and the mean without the outlier is approximately 44.6344.63. Since the mean without the outlier is lower, the mean would decrease if the outlier were removed.

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