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Select the outlier in the data set.\newline10,11,17,35,45,69,89,91,49610, 11, 17, 35, 45, 69, 89, 91, 496\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.\newline10,11,17,35,45,69,89,91,49610, 11, 17, 35, 45, 69, 89, 91, 496\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: 10,11,17,35,45,69,89,91,49610, 11, 17, 35, 45, 69, 89, 91, 496\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.
  2. Spotting the outlier: Looking at the data set, we can see that the number 496496 is much larger than all the other numbers, which are relatively close to each other. Therefore, 496496 is the outlier.
  3. Calculate mean with outlier: Calculate the mean of the data set with the outlier included.\newlineMean = (10+11+17+35+45+69+89+91+496)/9(10 + 11 + 17 + 35 + 45 + 69 + 89 + 91 + 496) / 9\newlineMean = (863)/9(863) / 9\newlineMean 95.89\approx 95.89
  4. Calculate mean without outlier: Calculate the mean of the data set without the outlier.\newlineMean without outlier = (10+11+17+35+45+69+89+91)/8(10 + 11 + 17 + 35 + 45 + 69 + 89 + 91) / 8\newlineMean without outlier = (367)/8(367) / 8\newlineMean without outlier 45.88\approx 45.88
  5. Compare mean with and without outlier: Compare the two means to determine if the mean would increase or decrease upon the removal of the outlier. The mean with the outlier is approximately 95.8995.89, and the mean without the outlier is approximately 45.8845.88. Removing the outlier significantly decreases the mean.

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