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Select the outlier in the data set.\newline7,57,61,62,63,65,66,68,757, 57, 61, 62, 63, 65, 66, 68, 75\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.\newline7,57,61,62,63,65,66,68,757, 57, 61, 62, 63, 65, 66, 68, 75\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: 7,57,61,62,63,65,66,68,757, 57, 61, 62, 63, 65, 66, 68, 75. To find the outlier, we can use the interquartile range (IQR) method or simply look for a number that is significantly different from the rest of the data.
  2. Find Outlier: Looking at the data set, we can see that the number 77 is significantly lower than all other numbers, which are all above 5050. Therefore, 77 is the outlier.
  3. Calculate Mean Original: Calculate the mean of the original data set including the outlier.\newlineMean = (7+57+61+62+63+65+66+68+75)/9(7 + 57 + 61 + 62 + 63 + 65 + 66 + 68 + 75) / 9\newlineMean = 524/9524 / 9\newlineMean ≈58.22\approx 58.22
  4. Calculate Mean Without Outlier: Calculate the mean of the data set without the outlier.\newlineMean without outlier = (57+61+62+63+65+66+68+75)/8(57 + 61 + 62 + 63 + 65 + 66 + 68 + 75) / 8\newlineMean without outlier = 517/8517 / 8\newlineMean without outlier ≈64.63\approx 64.63
  5. Compare Means: Compare the two means to determine if the mean would increase or decrease upon the removal of the outlier. Since 64.6364.63 (mean without outlier) is greater than 58.2258.22 (mean with outlier), the mean would increase if the outlier were removed.

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