Question

6 the data set represents the number of minutes ben walked each day for twelve days. 25, 68, 10, 27, 80, 42, 36, 20, 26, 38, 42, 26 ben created a boxplot to display this data. choose which components of the boxplot, if any, that ben got wrong. a all are correct b median c upper extreme d upper quartile e lower extreme f lower quartile

Explanation:

Step1: Sort the data

Sort the data set \(25,68,10,27,80,42,36,20,26,38,42,26\) in ascending order: \(10,20,25,26,26,27,36,38,42,42,68,80\).

Step2: Find the lower extreme

The lower extreme (minimum value) is \(10\). Ben's box - plot has the lower end of the whisker at a value greater than \(10\), so the lower extreme is wrong.

Step3: Find the lower quartile

For \(n = 12\) data points, the lower quartile \(Q_1\) is the value at the \(\frac{n + 1}{4}=\frac{12+1}{4}=3.25^{th}\) position. Using linear interpolation: \(Q_1=25+(26 - 25)\times0.25=25.25\). Ben's box - plot has the left - hand side of the box at a value greater than \(25.25\), so the lower quartile is wrong.

Step4: Find the median

For \(n = 12\) data points, the median \(Q_2\) is the average of the \(6^{th}\) and \(7^{th}\) values. \(Q_2=\frac{27 + 36}{2}=31.5\). Ben's box - plot has the middle line of the box at a value greater than \(31.5\), so the median is wrong.

Step5: Find the upper quartile

The upper quartile \(Q_3\) is the value at the \(\frac{3(n + 1)}{4}=\frac{3\times(12 + 1)}{4}=9.75^{th}\) position. Using linear interpolation: \(Q_3=42+(42 - 42)\times0.75 = 42\). Ben's box - plot has the right - hand side of the box at a value less than \(42\), so the upper quartile is wrong.

Step6: Find the upper extreme

The upper extreme (maximum value) is \(80\). Ben's box - plot has the upper end of the whisker at \(80\), so the upper extreme is correct.

Answer:

B. Median, D. Upper Quartile, E. Lower Extreme, F. Lower Quartile