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Select the outlier in the data set.\newline4,58,60,62,63,64,65,67,724, 58, 60, 62, 63, 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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Identify an outlier and describe the effect of removing itright chevron icon

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Q. Select the outlier in the data set.\newline4,58,60,62,63,64,65,67,724, 58, 60, 62, 63, 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: 4,58,60,62,63,64,65,67,724, 58, 60, 62, 63, 64, 65, 67, 72. An outlier is a data point that is significantly different from the rest of the data. We can often spot an outlier by looking for a number that is much smaller or larger than the rest.
  2. Spot Outlier: Looking at the data set, we can see that the number 44 is much smaller than all the other numbers, which are all above 5050. Therefore, 44 is the outlier in this data set.
  3. Calculate Mean with Outlier: Now, let's calculate the mean of the data set with and without the outlier to determine if the mean would increase or decrease upon its removal.\newlineFirst, calculate the mean with the outlier included:\newlineMean = (4+58+60+62+63+64+65+67+72)/9(4 + 58 + 60 + 62 + 63 + 64 + 65 + 67 + 72) / 9\newlineMean = 505/9505 / 9\newlineMean ≈56.11\approx 56.11
  4. Calculate Mean without Outlier: Next, calculate the mean without the outlier:\newlineMean = (58+60+62+63+64+65+67+72)/8(58 + 60 + 62 + 63 + 64 + 65 + 67 + 72) / 8\newlineMean = 501/8501 / 8\newlineMean ≈62.63\approx 62.63
  5. Compare Means: Comparing the two means, we can see that the mean without the outlier (≈62.63\approx 62.63) is greater than the mean with the outlier (≈56.11\approx 56.11). Therefore, if the outlier were removed from the data set, the mean would increase.

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