In the data set below, what are the lower quartile, the median, and the upper quartile? 82, 82, 82, 82, 82, 83, 84, 86, 86, 87, 87 lower quartile = _____ median = _____ upper quartile = _____

Identify the median: Step 1: Identify the median of the data set 82, 82, 82, 82, 82, 83, 84, 86, 86, 87, 87. Since there are 11 numbers, the median is the 6th number. Calculation: Median = 83 Determine lower quartile: Step 2: Determine the lower quartile. For the lower quartile, consider the first half of the data set, excluding the median. First half: 82, 82, 82, 82, 82. Calculation: Lower quartile = 82 (middle value of the first half) Determine upper quartile: Step 3: Determine the upper quartile. For the upper quartile, consider the second half of the data set, excluding the median. Second half: 84, 86, 86, 87, 87. Calculation: Upper quartile = 86 (middle value of the second half)

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In the data set below, what are the lower quartile, the median, and the upper quartile? $\newline$ $82, 82, 82, 82, 82, 83, 84, 86, 86, 87, 87$ $\newline$ lower quartile $=$ $\_\_$ $\newline$ median $=$ $\_\_$ $\newline$ upper quartile $=$ $\_\_$

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Math Problems

Algebra 1

Calculate quartiles and interquartile range

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Q. In the data set below, what are the lower quartile, the median, and the upper quartile?

\newline

82, 82, 82, 82, 82, 83, 84, 86, 86, 87, 87

\newline

lower quartile

=

\_\_

\newline

median

=

\_\_

\newline

upper quartile

=

\_\_

Identify the median: Step $1$ : Identify the median of the data set $82, 82, 82, 82, 82, 83, 84, 86, 86, 87, 87$ . Since there are $11$ numbers, the median is the $6$ th number. Calculation: Median $= 83$
Determine lower quartile: Step $2$ : Determine the lower quartile. $\newline$ For the lower quartile, consider the first half of the data set, excluding the median. $\newline$ First half: $82, 82, 82, 82, 82$ . $\newline$ Calculation: Lower quartile = $82$ (middle value of the first half)
Determine upper quartile: Step $3$ : Determine the upper quartile. $\newline$ For the upper quartile, consider the second half of the data set, excluding the median. $\newline$ Second half: $84, 86, 86, 87, 87$ . $\newline$ Calculation: Upper quartile $= 86$ (middle value of the second half)