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Select the outlier in the data set.\newline6,50,52,53,56,58,60,64,696, 50, 52, 53, 56, 58, 60, 64, 69\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.\newline6,50,52,53,56,58,60,64,696, 50, 52, 53, 56, 58, 60, 64, 69\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: Step 11: Identify the outlier in the data set.\newlineData set: 6,50,52,53,56,58,60,64,696, 50, 52, 53, 56, 58, 60, 64, 69.\newlineTo find the outlier, calculate the mean and standard deviation. Then, determine which values lie significantly outside the typical range of the data.\newlineMean = (6+50+52+53+56+58+60+64+69)/9=52(6 + 50 + 52 + 53 + 56 + 58 + 60 + 64 + 69) / 9 = 52.\newlineStandard deviation calculation:\newlineVariance = [(6−52)2+(50−52)2+(52−52)2+(53−52)2+(56−52)2+(58−52)2+(60−52)2+(64−52)2+(69−52)2]/9[(6-52)^2 + (50-52)^2 + (52-52)^2 + (53-52)^2 + (56-52)^2 + (58-52)^2 + (60-52)^2 + (64-52)^2 + (69-52)^2] / 9\newline= [2116+4+0+1+16+36+64+144+289]/9[2116 + 4 + 0 + 1 + 16 + 36 + 64 + 144 + 289] / 9\newline= 2668/9=296.442668 / 9 = 296.44\newlineStandard deviation = 296.44≈17.22\sqrt{296.44} \approx 17.22\newlineOutlier check: 66 is more than 33 standard deviations away from the mean (52−3∗17.22=0.34)(52 - 3*17.22 = 0.34), so 66 is an outlier.
  2. Calculate Mean and Standard Deviation: Step 22: Determine if the mean would increase or decrease if the outlier were removed.\newlineNew data set without the outlier: 50,52,53,56,58,60,64,6950, 52, 53, 56, 58, 60, 64, 69.\newlineNew mean = (50+52+53+56+58+60+64+69)/8=57.75(50 + 52 + 53 + 56 + 58 + 60 + 64 + 69) / 8 = 57.75.\newlineSince the new mean (57.7557.75) is higher than the original mean (5252), removing the outlier increases the mean.

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