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Select the outlier in the data set.\newline4,58,60,61,62,64,65,67,724, 58, 60, 61, 62, 64, 65, 67, 72\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.\newline4,58,60,61,62,64,65,67,724, 58, 60, 61, 62, 64, 65, 67, 72\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 given data set.\newlineThe data set is: 4,58,60,61,62,64,65,67,724, 58, 60, 61, 62, 64, 65, 67, 72.\newlineAn outlier is a value that is significantly higher or lower than most of the other values in a data set. To identify an outlier, we can use the interquartile range (IQR) method or simply look for a value that is distinctly separate from the cluster of other values.\newlineIn this case, without calculating the IQR, we can see that the number 44 is much lower than all the other numbers, which are all 5858 or higher. Therefore, 44 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 then divide by the number of values.\newlineMean = (4+58+60+61+62+64+65+67+72)/9(4 + 58 + 60 + 61 + 62 + 64 + 65 + 67 + 72) / 9\newlineMean = 503/9503 / 9\newlineMean 55.89\approx 55.89
  3. Calculate Mean without Outlier: Calculate the mean of the data set without the outlier.\newlineRemove the outlier 44 and then calculate the new mean.\newlineNew Mean = (58+60+61+62+64+65+67+72)/8(58 + 60 + 61 + 62 + 64 + 65 + 67 + 72) / 8\newlineNew Mean = 509/8509 / 8\newlineNew Mean 63.63\approx 63.63
  4. Compare Mean Changes: Compare the two means to determine if the mean would increase or decrease upon the removal of the outlier. The original mean with the outlier was approximately 55.8955.89, and the new mean without the outlier is approximately 63.6363.63. Since the new mean is higher than the original mean, the mean would increase if the outlier were removed.

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