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Select the outlier in the data set.\newline29,39,49,68,83,95,99,74429, 39, 49, 68, 83, 95, 99, 744\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.\newline29,39,49,68,83,95,99,74429, 39, 49, 68, 83, 95, 99, 744\newlineIf the outlier were removed from the data set, would the mean increase or decrease?\newlineChoices:\newline(A)increase\newline(B)decrease
  1. Calculate Mean: Calculate the mean of the data set.\newlineMean = (29+39+49+68+83+95+99+744)/8(29 + 39 + 49 + 68 + 83 + 95 + 99 + 744) / 8\newlineMean = (1106)/8(1106) / 8\newlineMean = 138.25138.25
  2. Calculate Standard Deviation: Calculate the standard deviation of the data set.\newlineFirst, find the squared differences from the mean for each data point.\newlineSquared differences: (29−138.25)2(29-138.25)^2, (39−138.25)2(39-138.25)^2, (49−138.25)2(49-138.25)^2, (68−138.25)2(68-138.25)^2, (83−138.25)2(83-138.25)^2, (95−138.25)2(95-138.25)^2, (99−138.25)2(99-138.25)^2, (744−138.25)2(744-138.25)^2\newlineCalculating each: 11934.562511934.5625, 9801.06259801.0625, (39−138.25)2(39-138.25)^200, (39−138.25)2(39-138.25)^211, (39−138.25)2(39-138.25)^222, (39−138.25)2(39-138.25)^233, (39−138.25)2(39-138.25)^244, (39−138.25)2(39-138.25)^255\newlineSum of squared differences: (39−138.25)2(39-138.25)^266\newlineSum of squared differences = (39−138.25)2(39-138.25)^277\newlineVariance = Sum of squared differences / (number of data points - 11)\newlineVariance = (39−138.25)2(39-138.25)^288\newlineVariance = (39−138.25)2(39-138.25)^299\newlineStandard deviation = (49−138.25)2(49-138.25)^200\newlineStandard deviation (49−138.25)2(49-138.25)^211\newlineStandard deviation (49−138.25)2(49-138.25)^222
  3. Identify Outliers: Use the standard deviation to determine if there are any outliers.\newlineTypically, an outlier is a data point that is more than 1.51.5 times the interquartile range (IQR) above the third quartile or below the first quartile. However, since we do not have the quartiles, we can consider a data point that lies more than 33 standard deviations from the mean as an outlier.\newlineOutlier threshold: Mean ±3×\pm 3 \times Standard deviation\newlineLower threshold: 138.25−3×239.71≈−580.88138.25 - 3 \times 239.71 \approx -580.88 (which is not possible since all data points are positive)\newlineUpper threshold: 138.25+3×239.71≈857.38138.25 + 3 \times 239.71 \approx 857.38\newlineThe data point 744744 is above the upper threshold and is considered an outlier.
  4. Effect of Removing Outlier: Determine the effect on the mean if the outlier is removed. Removing the outlier will decrease the sum of the data set without significantly reducing the number of data points, which will result in a lower mean.

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