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In the data set below, what are the lower quartile, the median, and the upper quartile?\newline1,2,4,6,6,6,6,6,7,8,91, 2, 4, 6, 6, 6, 6, 6, 7, 8, 9\newlinelower quartile==__\_\_\newlinemedian==__\_\_\newlineupper quartile==__\_\_

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Q. In the data set below, what are the lower quartile, the median, and the upper quartile?\newline1,2,4,6,6,6,6,6,7,8,91, 2, 4, 6, 6, 6, 6, 6, 7, 8, 9\newlinelower quartile==__\_\_\newlinemedian==__\_\_\newlineupper quartile==__\_\_
  1. Sort Data Set: Arrange the data set in ascending order if it is not already sorted.\newlineData set: 1,2,4,6,6,6,6,6,7,8,91, 2, 4, 6, 6, 6, 6, 6, 7, 8, 9\newlineThe data set is already in ascending order.
  2. Find Median: Find the median of the data set.\newlineThere are 1111 numbers in the data set, which is an odd number, so the median will be the middle number.\newlineTo find the middle number: (11+1)/2=6(11 + 1) / 2 = 6th number\newlineThe 66th number in the data set is 66.\newlineMedian = 66
  3. Lower Quartile Data: Identify the data set for the lower quartile.\newlineFor the lower quartile, consider the first half of the data set, excluding the median if it is a part of the data.\newlineFirst half is 1,2,4,6,61, 2, 4, 6, 6 (the first five numbers).\newlineLower quartile data: 1,2,4,6,61, 2, 4, 6, 6
  4. Lower Quartile Value: Find the value of the lower quartile.\newlineSince there are 55 numbers in the lower quartile data set, the lower quartile will be the middle number.\newline(5+1)/2=3(5 + 1) / 2 = 3rd number\newlineThe 33rd number in the lower quartile data set is 44.\newlineLower quartile =4= 4
  5. Upper Quartile Data: Identify the data set for the upper quartile.\newlineFor the upper quartile, consider the second half of the data set, excluding the median if it is a part of the data.\newlineSecond half is 6,6,7,8,96, 6, 7, 8, 9 (the last five numbers).\newlineUpper quartile data: 6,6,7,8,96, 6, 7, 8, 9
  6. Upper Quartile Value: Find the value of the upper quartile.\newlineSince there are 55 numbers in the upper quartile data set, the upper quartile will be the middle number.\newline(5+1)/2=3(5 + 1) / 2 = 3rd number\newlineThe 33rd number in the upper quartile data set is 77.\newlineUpper quartile = 77

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