Question

thirteen students in a statistics class were asked how many siblings they have. their responses are listed below. 0, 1, 2, 3, 3, 3, 3, 3, 3, 6, 5, 6 which of the following boxplots correctly displays the data? number of siblings

Explanation:

Step1: Find the median

The data set is \(0,1,2,3,3,3,3,3,3,6,5,6\). Sorting it (already sorted except for the last two elements, which we can sort as \(0,1,2,3,3,3,3,3,3,5,6,6\)). Since \(n = 12\) (number of data points), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th values. \(\frac{n}{2}=6\), \(\frac{n}{2}+1 = 7\). The 6th value is \(3\) and the 7th value is \(3\), so the median \(Q_2=3\).

Step2: Find the lower quartile \(Q_1\)

The lower half of the data is \(0,1,2,3,3,3\). The number of data points in the lower half \(n_1 = 6\). The lower quartile \(Q_1\) is the average of the \(\frac{n_1}{2}\)th and \((\frac{n_1}{2}+1)\)th values. \(\frac{n_1}{2}=3\), \(\frac{n_1}{2}+1=4\). The 3rd value is \(2\) and the 4th value is \(3\), so \(Q_1=\frac{2 + 3}{2}=2.5\).

Step3: Find the upper quartile \(Q_3\)

The upper half of the data is \(3,3,5,6,6\). The number of data points in the upper half \(n_2=5\). The upper quartile \(Q_3\) is the \(\frac{3(n + 1)}{4}\)th value. Using the formula for \(n = 12\), \(\frac{3(n + 1)}{4}=\frac{3\times(12 + 1)}{4}=9.75\). Interpolating (or since \(n_2 = 5\), for \(n_2\) odd, \(Q_3\) is the \(\frac{n_2+1}{2}\)th value of the upper - half. \(\frac{n_2+1}{2}=3\) (for the upper - half \(3,3,5,6,6\)), so \(Q_3 = 5\).

Step4: Check for outliers

The inter - quartile range \(IQR=Q_3-Q_1=5 - 2.5 = 2.5\). The lower fence \(=Q_1-1.5\times IQR=2.5-1.5\times2.5=2.5-3.75=-1.25\). The upper fence \(=Q_3 + 1.5\times IQR=5+1.5\times2.5=5 + 3.75 = 8.75\). The value \(0\) is within \([-1.25,8.75]\) and \(6\) is within \([-1.25,8.75]\).

The box - plot should have the box from \(Q_1 = 2.5\) to \(Q_3=5\) with a line at \(Q_2 = 3\). The whiskers should extend to the non - outlier minimum and maximum. The non - outlier minimum is \(0\) and the non - outlier maximum is \(6\).

Answer:

The box - plot where the box is centered around \(3\) (median), from approximately \(2.5\) (lower quartile) to \(5\) (upper quartile) with whiskers extending to \(0\) and \(6\) (assuming the second box - plot in the image has the whisker on the right - hand side extending to \(6\) and the first one has an incorrect left - hand side whisker placement).