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Select the outlier in the data set. \newline2,66,70,75,82,84,88,912, 66, 70, 75, 82, 84, 88, 91

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Q. Select the outlier in the data set. \newline2,66,70,75,82,84,88,912, 66, 70, 75, 82, 84, 88, 91
  1. Identify Outlier: Data set: 2,66,70,75,82,84,88,912, 66, 70, 75, 82, 84, 88, 91 To find the outlier, we need to look for a number that is significantly different from the rest of the numbers in the set.
  2. Calculate Median: First, let's calculate the median of the data set to understand the central tendency. The median is the middle value when the numbers are arranged in order. Since there are 88 numbers, the median will be the average of the 44th and 55th numbers after arranging the data set in ascending order. The data set is already in ascending order, so the median is (75+82)/2=157/2=78.5(75 + 82) / 2 = 157 / 2 = 78.5
  3. Calculate Quartiles: Next, we calculate the first quartile (Q1Q_1) and the third quartile (Q3Q_3). These are the medians of the first half and the second half of the data set, respectively.\newlineFor Q1Q_1, we take the median of the first four numbers: (66+70)/2=136/2=68(66 + 70) / 2 = 136 / 2 = 68\newlineFor Q3Q_3, we take the median of the last four numbers: (84+88)/2=172/2=86(84 + 88) / 2 = 172 / 2 = 86
  4. Calculate IQR: Now, we calculate the interquartile range (IQR), which is Q3Q1Q3 - Q1. \newlineIQR=8668=18IQR = 86 - 68 = 18
  5. Find Bounds: To identify outliers, we can use the IQR to find the bounds outside of which a number is considered an outlier. The lower bound is Q11.5×IQRQ1 - 1.5 \times IQR, and the upper bound is Q3+1.5×IQRQ3 + 1.5 \times IQR. Lower bound = 68(1.5×18)=6827=4168 - (1.5 \times 18) = 68 - 27 = 41 Upper bound = 86+(1.5×18)=86+27=11386 + (1.5 \times 18) = 86 + 27 = 113
  6. Identify Outlier: Any number in the data set that is lower than the lower bound or higher than the upper bound is considered an outlier.\newlineIn our data set, the number 22 is below the lower bound of 4141.\newlineTherefore, 22 is the outlier in the data set.

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